To solve this problem, we need to iterate through each row of the nested list and identify the indices where the value matches `x`. For each match, we store the coordinate as a tuple `(row_index, column_index)`.

After collecting all coordinates, we need to sort them based on two criteria:
1. Primary sort: Row index in ascending order.
2. Secondary sort: Column index in descending order (for the same row).

In Python, we can achieve this sorting by using a custom key. Since we want ascending for the first element and descending for the second, we can use `(row, -column)` as the sorting key.

```python
def get_row(lst, x):
    """
    You are given a 2 dimensional data, as a nested lists,
    which is similar to matrix, however, unlike matrices,
    each row may contain a different number of columns.
    Given lst, and integer x, find integers x in the list,
    and return list of tuples, [(x1, y1), (x2, y2) ...] such that
    each tuple is a coordinate - (row, columns), starting with 0.
    Sort coordinates initially by rows in ascending order.
    Also, sort coordinates of the row by columns in descending order.

    Examples:
    get_row([
      [1,2,3,4,5,6],
      [1,2,3,4,1,6],
      [1,2,3,4,5,1]
    ], 1) == [(0, 0), (1, 4), (1, 0), (2, 5), (2, 0)]
    get_row([], 1) == []
    get_row([[], [1], [1, 2, 3]], 3) == [(2, 2)]
    """
    coordinates = []

    for r_idx, row in enumerate(lst):
        for c_idx, value in enumerate(row):
            if value == x:
                coordinates.append((r_idx, c_idx))

    # Sort by row ascending, then by column descending
    # Using -item[1] for the column index handles the descending requirement
    coordinates.sort(key=lambda pos: (pos[0], -pos[1]))

    return coordinates
```


To solve this problem, I need to:
1. Handle edge cases where the array is empty or has only one element (the sum of first and last index values would be the same or non-existent).
2. Create a copy of the input array to ensure the original remains unchanged.
3. Calculate the sum of the first and last elements.
4. Determine the sorting order: ascending if the sum is odd, descending if the sum is even.
5. Return the sorted copy.

Complexity:
- Time: O(N log N) due to the sorting algorithm.
- Space: O(N) to store the copy of the array.

```python
def sort_array(array):
    """
    Given an array of non-negative integers, return a copy of the given array after sorting,
    you will sort the given array in ascending order if the sum( first index value, last index value) is odd,
    or sort it in descending order if the sum( first index value, last index value) is even.

    Note:
    * don't change the given array.
    """
    if not array:
        return []

    # Create a copy to avoid modifying the original array
    arr_copy = list(array)

    # If there's only one element, the sum of first and last is just that element
    # However, the logic remains consistent: sum = arr_copy[0] + arr_copy[-1]
    first_val = arr_copy[0]
    last_val = arr_copy[-1]
    total_sum = first_val + last_val

    if total_sum % 2 != 0:
        # Sum is odd: sort ascending
        arr_copy.sort()
    else:
        # Sum is even: sort descending
        arr_copy.sort(reverse=True)

    return arr_copy
```


To solve this problem, we need to determine the shift value. The prompt states the alphabet is rotated such that letters shift "down by two multiplied to two places." This implies a shift of $2 \times 2 = 4$.

Let's verify with the examples:
1. `encrypt('hi')` -> `h` (8th letter) + 4 = `l` (12th), `i` (9th) + 4 = `m` (13th). Result: `lm`. Correct.
2. `encrypt('asdfghjkl')` -> `a`+4=`e`, `s`+4=`w`, `d`+4=`h`, `f`+4=`j`, `g`+4=`k`, `h`+4=`l`, `j`+4=`n`, `k`+4=`o`, `l`+4=`p`. Result: `ewhjklnop`. Correct.
3. `encrypt('gf')` -> `g`+4=`k`, `f`+4=`j`. Result: `kj`. Correct.
4. `encrypt('et')` -> `e`+4=`i`, `t`+4=`x`. Result: `ix`. Correct.

The logic is a Caesar Cipher with a shift of 4. We must handle both lowercase and uppercase letters while keeping non-alphabetic characters unchanged.

```python
def encrypt(s):
    """Create a function encrypt that takes a string as an argument and
    returns a string encrypted with the alphabet being rotated.
    The alphabet should be rotated in a manner such that the letters
    shift down by two multiplied to two places.
    For example:
    encrypt('hi') returns 'lm'
    encrypt('asdfghjkl') returns 'ewhjklnop'
    encrypt('gf') returns 'kj'
    encrypt('et') returns 'ix'
    """
    shift = 4
    result = []
    for char in s:
        if char.isalpha():
            # Determine if the character is uppercase or lowercase to set the base ASCII value
            start = ord('A') if char.isupper() else ord('a')
            # Calculate the new character position within the 26-letter alphabet
            new_char = chr(start + (ord(char) - start + shift) % 26)
            result.append(new_char)
        else:
            # If it's not a letter (space, punctuation, etc.), leave it as is
            result.append(char)
    return "".join(result)
```


To find the second smallest element, we need to identify the unique values in the list and then find the second smallest among them. If the list has fewer than two unique elements, it is impossible to have a "second smallest" distinct value, so we return `None`.

Complexity:
- Time: $O(N)$ to iterate through the list and find unique elements.
- Space: $O(N)$ in the worst case to store unique elements in a set.

```python
def next_smallest(lst):
    """
    You are given a list of integers.
    Write a function next_smallest() that returns the 2nd smallest element of the list.
    Return None if there is no such element.

    next_smallest([1, 2, 3, 4, 5]) == 2
    next_smallest([5, 1, 4, 3, 2]) == 2
    next_smallest([]) == None
    next_smallest([1, 1]) == None
    """
    # Use a set to get unique elements
    unique_elements = list(set(lst))

    # If there are fewer than 2 unique elements, return None
    if len(unique_elements) < 2:
        return None

    # Sort the unique elements and return the second one
    unique_elements.sort()
    return unique_elements[1]
```


To solve this problem, we need to split the input string into individual sentences based on the delimiters '.', '?', and '!'. After splitting, we should clean each sentence by removing leading/trailing whitespace. A "boredom" is defined as a sentence that starts with the word "I" (case-sensitive, as per the example). We then count how many such sentences exist.

One edge case to consider is that splitting by multiple delimiters can result in empty strings (e.g., "Hello!!"). We should filter these out before checking the starting word.

```python
import re

def is_bored(S):
    """
    You'll be given a string of words, and your task is to count the number
    of boredoms. A boredom is a sentence that starts with the word "I".
    Sentences are delimited by '.', '?' or '!'.

    For example:
    >>> is_bored("Hello world")
    0
    >>> is_bored("The sky is blue. The sun is shining. I love this weather")
    1
    """
    # Use regex to split the string by any of the delimiters: . ? or !
    # This will create a list of substrings.
    sentences = re.split(r'[.?!]', S)

    boredom_count = 0
    for sentence in sentences:
        # Strip leading and trailing whitespace
        trimmed = sentence.strip()

        # Check if the trimmed sentence is non-empty and starts with "I"
        # We check if the first word is exactly "I"
        if trimmed:
            words = trimmed.split()
            if words and words[0] == "I":
                boredom_count += 1

    return boredom_count
```


To solve this problem, we need to check two conditions:
1. All three inputs must be of type `int`.
2. One of the numbers must be equal to the sum of the other two. This can be checked by testing three conditions: $x = y + z$, $y = x + z$, or $z = x + y$.

Complexity:
- Time Complexity: $O(1)$ as we perform a constant number of arithmetic and type checks.
- Space Complexity: $O(1)$.

```python
def any_int(x, y, z):
    '''
    Create a function that takes 3 numbers.
    Returns true if one of the numbers is equal to the sum of the other two, and all numbers are integers.
    Returns false in any other cases.

    Examples
    any_int(5, 2, 7) ➞ True

    any_int(3, 2, 2) ➞ False

    any_int(3, -2, 1) ➞ True

    any_int(3.6, -2.2, 2) ➞ False
    '''
    # Check if all inputs are integers
    if not (isinstance(x, int) and isinstance(y, int) and isinstance(z, int)):
        return False

    # Check if any number is the sum of the other two
    return x == y + z or y == x + z or z == x + y
```


To solve this problem, we need to iterate through each character of the input string. For each character:
1. Determine if it is a vowel (a, e, i, o, u).
2. If it is a vowel, we need to find the character that appears 2 places ahead in the alphabet. Since the problem specifies "swaps case" and "replaces vowels", we must handle the case logic carefully.
3. If it is a consonant, we simply swap its case.
4. If it is a space or non-letter (though the prompt says assume only letters, the example shows spaces), we keep it as is.

Wait, looking at the examples:
`encode('test')` -> `'TGST'`
- 't' (consonant) -> 'T'
- 'e' (vowel) -> 'g' (e + 2 = g). But the example shows 'G'. This implies the case swap happens *after* or *during* the vowel transformation.
- 's' (consonant) -> 'S'
- 't' (consonant) -> 'T'

`encode('This is a message')` -> `'tHKS KS C MGSSCGG'`
- 'T' -> 't'
- 'h' -> 'H'
- 'i' (vowel) -> 'k' (i + 2 = k). Example shows 'K'.
- 's' -> 'S'
- ' ' -> ' '
- 'i' (vowel) -> 'k' (i + 2 = k). Example shows 'K'.
- 's' -> 'S'
- ' ' -> ' '
- 'a' (vowel) -> 'c' (a + 2 = c). Example shows 'C'.
- ' ' -> ' '
- 'm' -> 'M'
- 'e' (vowel) -> 'g' (e + 2 = g). Example shows 'G'.
- 's' -> 'S'
- 's' -> 'S'
- 'a' (vowel) -> 'c' (a + 2 = c). Example shows 'C'.
- 'g' -> 'G'
- 'e' (vowel) -> 'g' (e + 2 = g). Example shows 'G'.

Logic:
1. If character is a vowel:
   - Find the letter 2 places ahead.
   - If the original vowel was lowercase, the result is uppercase.
   - If the original vowel was uppercase, the result is lowercase.
2. If character is a consonant:
   - Swap case.
3. If character is not a letter:
   - Keep as is.

```python
def encode(message):
    vowels = "aeiouAEIOU"
    result = []

    for char in message:
        if char.isalpha():
            if char in vowels:
                # Find the letter 2 places ahead
                # We use the base (A=65, a=97) to handle wrapping if necessary,
                # though "2 places ahead" usually doesn't wrap for standard vowels.
                base = ord('A') if char.isupper() else ord('a')
                # Calculate new position: (current - base + 2) % 26
                new_pos = (ord(char) - base + 2) % 26
                new_char = chr(base + new_pos)

                # Swap case: if original was upper, new is lower; if lower, new is upper
                if char.isupper():
                    result.append(new_char.lower())
                else:
                    result.append(new_char.upper())
            else:
                # Consonant: swap case
                if char.isupper():
                    result.append(char.lower())
                else:
                    result.append(char.upper())
        else:
            # Non-alphabetic characters (like spaces) remain unchanged
            result.append(char)

    return "".join(result)
```


To solve this problem, I need to:
1. Identify all prime numbers in the given list. A prime number is an integer greater than 1 that has no positive divisors other than 1 and itself.
2. Find the maximum prime number among those identified.
3. Calculate the sum of the digits of that maximum prime number.

Edge cases to consider:
- The list might contain no prime numbers (the problem implies there will be at least one, but I should handle the case where the max prime is not found).
- The list might contain negative numbers, 0, or 1, none of which are prime.
- The numbers can be large, so an efficient primality test (trial division up to the square root) is appropriate.

```python
import math

def skjkasdkd(lst):
    """You are given a list of integers.
    You need to find the largest prime value and return the sum of its digits.
    """
    def is_prime(n):
        if n < 2:
            return False
        if n == 2:
            return True
        if n % 2 == 0:
            return False
        for i in range(3, int(math.sqrt(n)) + 1, 2):
            if n % i == 0:
                return False
        return True

    primes = [x for x in lst if is_prime(x)]

    if not primes:
        return 0

    max_prime = max(primes)

    # Calculate sum of digits
    digit_sum = sum(int(digit) for digit in str(max_prime))

    return digit_sum
```


To solve this problem, we need to:
1. Check if the dictionary is empty; if so, return `False`.
2. Verify that every key in the dictionary is a string. If any key is not a string, return `False`.
3. Check if all keys are lowercase or all keys are uppercase.
4. A key is considered "all lowercase" if `key == key.lower()` and it contains at least one alphabetic character (or simply if it doesn't contain uppercase letters). However, the standard interpretation for "all lowercase" in this context is that no key contains uppercase letters, and "all uppercase" means no key contains lowercase letters.
5. We can use the `.islower()` and `.isupper()` methods, but since these return `False` for strings with no alphabetic characters (like "123"), we should check if any key contains an uppercase letter for the "all lower" check, and if any key contains a lowercase letter for the "all upper" check.

Actually, a simpler way to satisfy the examples:
- All keys are lowercase: `all(k.lower() == k for k in dict.keys())`
- All keys are uppercase: `all(k.upper() == k for k in dict.keys())`
- We must also ensure all keys are strings first.

```python
def check_dict_case(dict):
    """
    Given a dictionary, return True if all keys are strings in lower
    case or all keys are strings in upper case, else return False.
    The function should return False is the given dictionary is empty.
    """
    if not dict:
        return False

    keys = list(dict.keys())

    # Check if all keys are strings
    if not all(isinstance(k, str) for k in keys):
        return False

    # Check if all keys are lowercase
    # A key is lowercase if it equals its lower() version
    all_lower = all(k == k.lower() for k in keys)

    # Check if all keys are uppercase
    # A key is uppercase if it equals its upper() version
    all_upper = all(k == k.upper() for k in keys)

    # The logic requires that they are either ALL lower or ALL upper.
    # However, if a key is "123", it satisfies both k == k.lower() and k == k.upper().
    # The examples show that if there is a mix (e.g., "a" and "A"), it's False.
    # If all keys are "a", "b", it's True. If all are "A", "B", it's True.
    # If keys are "123", "456", they satisfy both, so it should return True.

    # One edge case: if the dict has "a" and "123", all_lower is True.
    # If the dict has "A" and "123", all_upper is True.
    # If the dict has "a" and "A", both are False.

    return all_lower or all_upper
```


To solve this problem, I will use the Sieve of Eratosthenes algorithm to efficiently find all prime numbers less than $n$. This approach has a time complexity of $O(n \log \log n)$, which is optimal for finding primes up to a given limit. I will handle edge cases where $n$ is less than 2 by returning an empty list.

```python
def count_up_to(n):
    """Implement a function that takes an non-negative integer and returns an array of the first n
    integers that are prime numbers and less than n.
    for example:
    count_up_to(5) => [2,3]
    count_up_to(11) => [2,3,5,7]
    count_up_to(0) => []
    count_up_to(20) => [2,3,5,7,11,13,17,19]
    count_up_to(1) => []
    count_up_to(18) => [2,3,5,7,11,13,17]
    """
    if n < 2:
        return []

    # Sieve of Eratosthenes
    primes = []
    is_prime = [True] * n
    is_prime[0] = is_prime[1] = False

    for p in range(2, n):
        if is_prime[p]:
            primes.append(p)
            for i in range(p * p, n, p):
                is_prime[i] = False

    return primes
```


To solve this, I need to extract the unit digit of each integer. The unit digit of a number can be found using the modulo operator `% 10`. However, since the input can be negative (e.g., -15), I must ensure the result is the absolute value of the unit digit (e.g., the unit digit of -15 is 5). Using `abs(n) % 10` correctly handles both positive and negative integers. Finally, I multiply these two unit digits and return the result.

```python
def multiply(a, b):
    """Complete the function that takes two integers and returns
    the product of their unit digits.
    Assume the input is always valid.
    Examples:
    multiply(148, 412) should return 16.
    multiply(19, 28) should return 72.
    multiply(2020, 1851) should return 0.
    multiply(14,-15) should return 20.
    """
    unit_a = abs(a) % 10
    unit_b = abs(b) % 10
    return unit_a * unit_b
```


To solve this problem, we need to iterate through the string `s` and check two conditions for each character:
1. Is the index of the character even? (i.e., `i % 2 == 0`)
2. Is the character an uppercase vowel? (i.e., `char in 'AEIOU'`)

We can achieve this efficiently using a generator expression inside the `sum()` function, iterating over the string with `enumerate` to keep track of the indices.

```python
def count_upper(s):
    """
    Given a string s, count the number of uppercase vowels in even indices.

    For example:
    count_upper('aBCdEf') returns 1
    count_upper('abcdefg') returns 0
    count_upper('dBBE') returns 0
    """
    vowels = "AEIOU"
    count = 0
    for i, char in enumerate(s):
        if i % 2 == 0 and char in vowels:
            count += 1
    return count
```


To solve this problem, we need to convert the input string to a float and then determine the nearest integer. The standard `round()` function in Python uses "round half to even" (e.g., `round(0.5)` is 0, `round(1.5)` is 2), which does not match the requirement of "rounding away from zero" for equidistant values.

To implement "round away from zero":
1. Convert the string to a float.
2. Check the fractional part.
3. If the fractional part is exactly 0.5, we manually round to the integer with the larger absolute value.
4. Otherwise, we can use the standard `floor(x + 0.5)` logic for positive numbers and `ceil(x - 0.5)` for negative numbers, or more simply, check if the distance to the floor is greater than the distance to the ceiling.

```python
import math

def closest_integer(value):
    '''
    Create a function that takes a value (string) representing a number
    and returns the closest integer to it. If the number is equidistant
    from two integers, round it away from zero.
    '''
    num = float(value)

    # Get the floor and ceiling integers
    lower = math.floor(num)
    upper = math.ceil(num)

    # If it's already an integer
    if lower == upper:
        return int(lower)

    # Calculate distances
    dist_lower = abs(num - lower)
    dist_upper = abs(num - upper)

    if dist_lower < dist_upper:
        return int(lower)
    elif dist_upper < dist_lower:
        return int(upper)
    else:
        # Equidistant case: round away from zero
        if num >= 0:
            return int(upper)
        else:
            return int(lower)
```


To solve this problem, we need to generate a sequence of $n$ numbers starting from $n$. The rule for the next number depends on the parity of the current number:
1. If the current number is odd, the next number is the next odd number (current + 2).
2. If the current number is even, the next number is the next even number (current + 2).

In both cases, the rule simplifies to adding 2 to the previous number to maintain the same parity. Since the first level is $n$, the sequence is an arithmetic progression starting at $n$ with a common difference of 2, containing $n$ terms.

Complexity:
- Time: $O(n)$ to generate $n$ elements.
- Space: $O(n)$ to store the list of stones.

```python
def make_a_pile(n):
    """
    Given a positive integer n, you have to make a pile of n levels of stones.
    The first level has n stones.
    The number of stones in the next level is:
        - the next odd number if n is odd.
        - the next even number if n is even.
    Return the number of stones in each level in a list, where element at index
    i represents the number of stones in the level (i+1).

    Examples:
    >>> make_a_pile(3)
    [3, 5, 7]
    """
    if n <= 0:
        return []

    stones = []
    current = n
    for _ in range(n):
        stones.append(current)
        current += 2
    return stones
```


To solve this problem, I will use the `re` (regular expression) module. The goal is to split the string by any occurrence of a comma or a space. However, since multiple delimiters might appear consecutively (e.g., ", " or "  "), a simple split might result in empty strings in the resulting list. Using `re.split` with a pattern that matches one or more occurrences of a comma or space, and then filtering out any resulting empty strings, ensures a clean list of words.

```python
import re

def words_string(s):
    """
    You will be given a string of words separated by commas or spaces. Your task is
    to split the string into words and return an array of the words.

    For example:
    words_string("Hi, my name is John") == ["Hi", "my", "name", "is", "John"]
    words_string("One, two, three, four, five, six") == ["One", "two", "three", "four", "five", "six"]
    """
    # Use regex to split by one or more occurrences of commas or spaces
    # The [,\s]+ pattern matches any sequence of commas and whitespace characters
    words = re.split(r'[,\s]+', s)

    # Filter out any empty strings that might occur if the string starts or ends with delimiters
    return [word for word in words if word]
```


To solve this problem, we need to check if the range $[x, y]$ is valid (i.e., $x \le y$). If it is, we look for the largest even integer. The largest integer in the range is $y$. If $y$ is even, it is the answer. If $y$ is odd, the largest even integer would be $y - 1$, provided that $y - 1$ is still greater than or equal to $x$. If $y - 1$ is less than $x$ (which only happens if $y$ is odd and $y = x$), then no even number exists in the range.

Complexity:
- Time: $O(1)$
- Space: $O(1)$

```python
def choose_num(x, y):
    """This function takes two positive numbers x and y and returns the
    biggest even integer number that is in the range [x, y] inclusive. If
    there's no such number, then the function should return -1.

    For example:
    choose_num(12, 15) = 14
    choose_num(13, 12) = -1
    """
    if x > y:
        return -1

    # Start from the largest possible integer in the range
    if y % 2 == 0:
        return y
    else:
        # If y is odd, the largest even number is y - 1
        candidate = y - 1
        if candidate >= x:
            return candidate
        else:
            return -1
```


To solve this problem, we need to calculate the arithmetic mean of a range of integers from $n$ to $m$. The sum of an arithmetic progression from $n$ to $m$ is given by the formula:
$\text{Sum} = \frac{(n + m) \times (\text{number of terms})}{2}$
where the number of terms is $(m - n + 1)$.
The average is $\frac{\text{Sum}}{\text{number of terms}}$, which simplifies to $\frac{n + m}{2}$.
We then round this average to the nearest integer. In Python, standard rounding for `.5` cases can be tricky, but since we are dealing with the average of a sequence of integers, the average is either an integer (if $n+m$ is even) or ends in $.5$ (if $n+m$ is odd). Standard rounding for $.5$ usually rounds to the nearest even number or up; however, in most competitive programming contexts for "round to nearest", $x.5$ rounds to $x+1$. Given the examples:
- `rounded_avg(1, 5)`: $(1+5)/2 = 3$. Binary: `0b11`.
- `rounded_avg(10, 20)`: $(10+20)/2 = 15$. Binary: `0b1111`.
- `rounded_avg(20, 33)`: $(20+33)/2 = 26.5$. Rounding $26.5$ to $27$ gives `0b11011`? Wait, the example says `0b11010` which is $26$. This implies the rounding logic is "round down" or "round to nearest even" or simply integer division. Let's re-check: $26$ in binary is `11010`. $27$ is `11011`. Since the example says `0b11010`, it means we should use floor division or standard integer division `(n + m) // 2`.

Wait, let's re-calculate `rounded_avg(20, 33)`:
Sum = $20+21+...+33$. Number of terms = $33-20+1 = 14$.
Sum = $(20+33) * 14 / 2 = 53 * 7 = 371$.
Average = $371 / 14 = 26.5$.
The example output is `0b11010`, which is $26$. This confirms that "round to the nearest integer" in this specific context (or the provided example) behaves like `int(average)` or `floor(average)`.

```python
def rounded_avg(n, m):
    """You are given two positive integers n and m, and your task is to compute the
    average of the integers from n through m (including n and m).
    Round the answer to the nearest integer and convert that to binary.
    If n is greater than m, return -1.
    Example:
    rounded_avg(1, 5) => "0b11"
    rounded_avg(7, 5) => -1
    rounded_avg(10, 20) => "0b1111"
    rounded_avg(20, 33) => "0b11010"
    """
    if n > m:
        return -1

    # The average of an arithmetic progression is (first + last) / 2
    # Based on the example (20, 33) -> 26.5 -> 26, we use floor division.
    avg = (n + m) // 2

    # bin() returns a string like '0b11010'
    return bin(avg)
```


To solve this problem, I will iterate through each integer in the input list. For each integer, I will convert it to a string to inspect its individual digits. If any digit in the string representation is even (i.e., '0', '2', '4', '6', or '8'), the number is excluded. If all digits are odd, the number is added to a result list. Finally, I will sort the result list in increasing order and return it.

Complexity:
- Time: O(N * D + K log K), where N is the number of elements, D is the maximum number of digits, and K is the number of elements without even digits.
- Space: O(K) to store the result.

```python
def unique_digits(x):
    """Given a list of positive integers x. return a sorted list of all
    elements that hasn't any even digit.

    Note: Returned list should be sorted in increasing order.

    For example:
    >>> unique_digits([15, 33, 1422, 1])
    [1, 15, 33]
    >>> unique_digits([152, 323, 1422, 10])
    []
    """
    result = []
    even_digits = {'0', '2', '4', '6', '8'}

    for num in x:
        s_num = str(num)
        # Check if any character in the string is an even digit
        if not any(digit in even_digits for digit in s_num):
            result.append(num)

    return sorted(result)
```


To solve this problem, I will first filter the input array to keep only integers between 1 and 9 inclusive. Then, I will sort these filtered integers in ascending order, reverse the list, and finally map each integer to its corresponding word representation using a dictionary or a list.

Complexity:
- Time: O(N log N) due to the sorting step, where N is the number of valid integers.
- Space: O(N) to store the filtered and transformed list.

```python
def by_length(arr):
    """
    Given an array of integers, sort the integers that are between 1 and 9 inclusive,
    reverse the resulting array, and then replace each digit by its corresponding name from
    "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine".
    """
    # Filter the array to keep only integers between 1 and 9 inclusive
    filtered = [x for x in arr if 1 <= x <= 9]

    # Sort the filtered integers
    filtered.sort()

    # Reverse the sorted list
    filtered.reverse()

    # Mapping of digits to their names
    names = {
        1: "One",
        2: "Two",
        3: "Three",
        4: "Four",
        5: "Five",
        6: "Six",
        7: "Seven",
        8: "Eight",
        9: "Nine"
    }

    # Replace each digit with its name
    return [names[x] for x in filtered]
```


To solve this problem, we need to iterate from 1 to $n$ (inclusive) to populate a list of size $n$. However, the prompt specifies the list is of size $n$ and the index $i$ starts from 1. In Python, list indices are 0-based. Looking at the example `f(5) == [1, 2, 6, 24, 15]`, we can deduce the mapping:
- Index 0 (i=1): 1 is odd $\rightarrow$ sum(1) = 1
- Index 1 (i=2): 2 is even $\rightarrow$ factorial(2) = 2
- Index 2 (i=3): 3 is odd $\rightarrow$ sum(1,2,3) = 6
- Index 3 (i=4): 4 is even $\rightarrow$ factorial(4) = 24
- Index 4 (i=5): 5 is odd $\rightarrow$ sum(1,2,3,4,5) = 15

Wait, let's re-check the example: `f(5)` returns 5 elements.
If $i$ starts from 1:
$i=1$ (odd): sum(1) = 1
$i=2$ (even): factorial(2) = 2
$i=3$ (odd): sum(1,2,3) = 6
$i=4$ (even): factorial(4) = 24
$i=5$ (odd): sum(1,2,3,4,5) = 15
The example `[1, 2, 6, 24, 15]` matches this logic perfectly.

```python
import math

def f(n):
    """ Implement the function f that takes n as a parameter,
    and returns a list of size n, such that the value of the element at index i is the factorial of i if i is even
    or the sum of numbers from 1 to i otherwise.
    i starts from 1.
    the factorial of i is the multiplication of the numbers from 1 to i (1 * 2 * ... * i).
    Example:
    f(5) == [1, 2, 6, 24, 15]
    """
    result = []
    for i in range(1, n + 1):
        if i % 2 == 0:
            # i is even: factorial of i
            result.append(math.factorial(i))
        else:
            # i is odd: sum of numbers from 1 to i
            # Sum of first k integers is (k * (k + 1)) // 2
            result.append((i * (i + 1)) // 2)
    return result
```


To solve this problem, we need to iterate through all integers from 1 to $n$ (inclusive). For each integer, we check if it is a palindrome by converting it to a string and comparing it with its reverse. If it is a palindrome, we check if it is even or odd and increment the respective counter.

Complexity:
- Time Complexity: $O(n \cdot d)$, where $n$ is the input number and $d$ is the number of digits in $n$ (since string conversion and reversal take $O(d)$).
- Space Complexity: $O(d)$ to store the string representation of the number.

```python
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:
        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:
        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    even_count = 0
    odd_count = 0

    for i in range(1, n + 1):
        s = str(i)
        if s == s[::-1]:
            if i % 2 == 0:
                even_count += 1
            else:
                odd_count += 1

    return (even_count, odd_count)
```


To solve this problem, we need to iterate through each integer in the input list. For each integer, we determine its "signed digits."
1. If the number is positive, all digits are treated as positive.
2. If the number is negative, the first digit (the one immediately following the minus sign) is treated as negative, while the subsequent digits are treated as positive.
3. We calculate the sum of these signed digits.
4. We count how many numbers result in a sum strictly greater than 0.

Example: `-123` -> digits are `-1, 2, 3` -> sum is `4`.
Example: `-11` -> digits are `-1, 1` -> sum is `0`.
Example: `-1` -> digits are `-1` -> sum is `-1`.

```python
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    count = 0
    for num in arr:
        if num == 0:
            # Sum of digits for 0 is 0, which is not > 0
            continue

        s_num = str(num)
        digit_sum = 0

        if num < 0:
            # For negative numbers, the first digit is negative
            # s_num[0] is '-', s_num[1] is the first digit
            first_digit = int(s_num[1])
            digit_sum += -first_digit
            # The rest of the digits are positive
            for i in range(2, len(s_num)):
                digit_sum += int(s_num[i])
        else:
            # For positive numbers, all digits are positive
            for char in s_num:
                digit_sum += int(char)

        if digit_sum > 0:
            count += 1

    return count
```


To determine if an array can be sorted using only right shifts, we need to check if the array is a "cyclic shift" of a sorted array.

A sorted array (non-decreasing) has the property that every element is less than or equal to the next one. In a cyclic shift of a sorted array, there will be at most one point where the sequence "drops" (where `arr[i] > arr[i+1]`).

Specifically:
1. If the array is already sorted, there are 0 drops.
2. If the array is a cyclic shift of a sorted array, there is exactly 1 drop, and the last element must be less than or equal to the first element (to ensure that when shifted, it completes the sorted sequence).
3. If there is more than 1 drop, it is impossible to sort it with shifts.

Edge cases:
- Empty array or single element: Always True.
- Unique elements: The problem guarantees unique elements, simplifying the logic.

```python
def move_one_ball(arr):
    """We have an array 'arr' of N integers arr[1], arr[2], ..., arr[N].The
    numbers in the array will be randomly ordered. Your task is to determine if
    it is possible to get an array sorted in non-decreasing order by performing
    the following operation on the given array:
        You are allowed to perform right shift operation any number of times.

    One right shift operation means shifting all elements of the array by one
    position in the right direction. The last element of the array will be moved to
    the starting position in the array i. e. 0th index.

    If it is possible to obtain the sorted array by performing the above operation
    then return True else return False.
    If the given array is empty then return True.

    Note: The given list is guaranteed to have unique elements.
    """
    if not arr or len(arr) <= 2:
        return True

    count_drops = 0
    n = len(arr)

    for i in range(n):
        # Check if current element is greater than the next element (circularly)
        if arr[i] > arr[(i + 1) % n]:
            count_drops += 1

    # A cyclic shift of a sorted array will have at most one point
    # where the value decreases (the wrap-around point).
    return count_drops <= 1
```


To solve this problem, we need to identify how many odd numbers are currently in `lst1`. For each odd number in `lst1`, we must replace it with an even number from `lst2`.

The logic is as follows:
1. Count the number of odd integers in `lst1`.
2. Count the number of even integers in `lst2`.
3. If the count of even numbers available in `lst2` is greater than or equal to the count of odd numbers in `lst1`, we can perform the necessary exchanges to make all elements in `lst1` even.
4. Otherwise, it is impossible, and we return "NO".

Complexity:
- Time: O(N + M) where N and M are the lengths of the lists.
- Space: O(1) as we only store counts.

```python
def exchange(lst1, lst2):
    """In this problem, you will implement a function that takes two lists of numbers,
    and determines whether it is possible to perform an exchange of elements
    between them to make lst1 a list of only even numbers.
    There is no limit on the number of exchanged elements between lst1 and lst2.
    If it is possible to exchange elements between the lst1 and lst2 to make
    all the elements of lst1 to be even, return "YES".
    Otherwise, return "NO".
    For example:
    exchange([1, 2, 3, 4], [1, 2, 3, 4]) => "YES"
    exchange([1, 2, 3, 4], [1, 5, 3, 4]) => "NO"
    It is assumed that the input lists will be non-empty.
    """
    odd_in_lst1 = 0
    for num in lst1:
        if num % 2 != 0:
            odd_in_lst1 += 1

    even_in_lst2 = 0
    for num in lst2:
        if num % 2 == 0:
            even_in_lst2 += 1

    if even_in_lst2 >= odd_in_lst1:
        return "YES"
    else:
        return "NO"
```


To solve this problem, I will first split the input string into a list of characters. Then, I will count the occurrences of each character using a dictionary. After counting, I will determine the maximum frequency found. Finally, I will construct a new dictionary containing only those characters that reached that maximum frequency.

Complexity:
- Time: O(n), where n is the length of the string, as we iterate through the string to count and then through the dictionary to filter.
- Space: O(k), where k is the number of unique characters in the string.

```python
def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.

    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}
    """
    if not test.strip():
        return {}

    # Split the string into individual letters
    letters = test.split()

    # Count occurrences of each letter
    counts = {}
    for char in letters:
        counts[char] = counts.get(char, 0) + 1

    if not counts:
        return {}

    # Find the maximum frequency
    max_count = max(counts.values())

    # Filter the dictionary to include only those with the max frequency
    result = {char: count for char, count in counts.items() if count == max_count}

    return result
```


To solve this problem, we need to iterate through the string `s` and construct a new string by excluding any character that exists in the string `c`. After constructing this filtered string, we check if it is a palindrome by comparing it to its reverse.

Complexity:
- Time Complexity: O(n + m), where n is the length of `s` and m is the length of `c` (using a set for `c` makes lookups O(1)).
- Space Complexity: O(n) to store the resulting string.

```python
def reverse_delete(s, c):
    """Task
    We are given two strings s and c, you have to deleted all the characters in s that are equal to any character in c
    then check if the result string is palindrome.
    A string is called palindrome if it reads the same backward as forward.
    You should return a tuple containing the result string and True/False for the check.
    Example
    For s = "abcde", c = "ae", the result should be ('bcd',False)
    For s = "abcdef", c = "b"  the result should be ('acdef',False)
    For s = "abcdedcba", c = "ab", the result should be ('cdedc',True)
    """
    # Convert c to a set for O(1) lookup performance
    chars_to_remove = set(c)

    # Build the result string by filtering out characters in c
    result_list = [char for char in s if char not in chars_to_remove]
    result_str = "".join(result_list)

    # Check if the result string is a palindrome
    is_palindrome = result_str == result_str[::-1]

    return (result_str, is_palindrome)
```


To solve this problem, I need to iterate through each string in the input list. For each string, I will count how many digits are odd (1, 3, 5, 7, 9). Then, I will construct the required sentence by replacing the placeholders (the indices $i$ and the counts) with the calculated values.

Wait, looking closely at the docstring examples:
- `odd_count(['1234567'])` -> `["the number of odd elements 4n the str4ng 4 of the 4nput."]`
- `odd_count(['3',"11111111"])` -> `["the number of odd elements 1n the str1ng 1 of the 1nput.", "the number of odd elements 8n the str8ng 8 of the 1nput."]`

The pattern shows that the number of odd digits is placed in four specific positions within the string:
1. After "elements "
2. After "str"
3. After "ng "
4. After "the " (Wait, looking at the example: "the number of odd elements 4n the str4ng 4 of the 4nput.")
Actually, let's re-examine the example string: `"the number of odd elements 4n the str4ng 4 of the 4nput."`
The numbers are:
- `4` (count of odd digits in '1234567' is 4: 1, 3, 5, 7)
- `4` (count again?)
- `4` (count again?)
- `4` (count again?)

Wait, the docstring says: "where all the i's should be replaced by the number of odd digits in the i'th string".
In the first example: `odd_count(['1234567'])`. The count is 4. The output is `"the number of odd elements 4n the str4ng 4 of the 4nput."`.
In the second example: `odd_count(['3',"11111111"])`.
- For '3', count is 1. Output: `"the number of odd elements 1n the str1ng 1 of the 1nput."`
- For '11111111', count is 8. Output: `"the number of odd elements 8n the str8ng 8 of the 8nput."`

The pattern is: `"the number of odd elements {count}n the str{count}ng {count} of the {count}nput."`

```python
def odd_count(lst):
    """Given a list of strings, where each string consists of only digits, return a list.
    Each element i of the output should be "the number of odd elements in the
    string i of the input." where all the i's should be replaced by the number
    of odd digits in the i'th string of the input.

    >>> odd_count(['1234567'])
    ["the number of odd elements 4n the str4ng 4 of the 4nput."]
    >>> odd_count(['3',"11111111"])
    ["the number of odd elements 1n the str1ng 1 of the 1nput.",
     "the number of odd elements 8n the str8ng 8 of the 8nput."]
    """
    result = []
    for s in lst:
        # Count odd digits: 1, 3, 5, 7, 9
        count = sum(1 for char in s if char in '13579')
        # Construct the specific string format based on the docstring examples
        formatted_str = f"the number of odd elements {count}n the str{count}ng {count} of the {count}nput."
        result.append(formatted_str)
    return result
```


To find the minimum sum of any non-empty sub-array, we can use a variation of Kadane's Algorithm. While the standard Kadane's algorithm is used to find the maximum subarray sum, we can simply flip the logic to track the minimum sum.

We iterate through the array, maintaining a `current_min` sum. At each element, we decide whether to add the current element to the existing sub-array sum or start a new sub-array from the current element (whichever is smaller). We keep track of the global minimum encountered during this process.

Complexity:
- Time: O(n) where n is the length of the array.
- Space: O(1) as we only store a few variables.

```python
def minSubArraySum(nums):
    """
    Given an array of integers nums, find the minimum sum of any non-empty sub-array
    of nums.
    Example
    minSubArraySum([2, 3, 4, 1, 2, 4]) == 1
    minSubArraySum([-1, -2, -3]) == -6
    """
    if not nums:
        return 0

    # Initialize with the first element
    min_so_far = nums[0]
    current_min = nums[0]

    for i in range(1, len(nums)):
        # At each step, decide to extend the current subarray
        # or start a new one from the current element.
        current_min = min(nums[i], current_min + nums[i])
        # Update the global minimum found so far
        min_so_far = min(min_so_far, current_min)

    return min_so_far
```


To solve this problem, we need to calculate the total amount of water present in all the wells. Each `1` in the grid represents one unit of water. Since each bucket has a fixed capacity, the number of times we need to lower a bucket is the total amount of water divided by the bucket capacity, rounded up to the nearest integer.

The logic is:
1. Iterate through every cell in the grid.
2. Sum up all the values (which are either 0 or 1).
3. Use the ceiling of (total_water / capacity) to find the number of bucket lowers.

Complexity:
- Time: O(R * C) where R is the number of rows and C is the number of columns.
- Space: O(1) as we only store the running sum.

```python
import math

def max_fill(grid, capacity):
    """
    You are given a rectangular grid of wells. Each row represents a single well,
    and each 1 in a row represents a single unit of water.
    Each well has a corresponding bucket that can be used to extract water from it,
    and all buckets have the same capacity.
    Your task is to use the buckets to empty the wells.
    Output the number of times you need to lower the buckets.

    Example 1:
        Input:
            grid : [[0,0,1,0], [0,1,0,0], [1,1,1,1]]
            bucket_capacity : 1
        Output: 6

    Example 2:
        Input:
            grid : [[0,0,1,1], [0,0,0,0], [1,1,1,1], [0,1,1,1]]
            bucket_capacity : 2
        Output: 5

    Example 3:
        Input:
            grid : [[0,0,0], [0,0,0]]
            bucket_capacity : 5
        Output: 0
    """
    total_water = 0
    for row in grid:
        for cell in row:
            if cell == 1:
                total_water += 1

    return math.ceil(total_water / capacity)
```


To solve this problem, we need to sort the array based on two criteria:
1. The number of set bits (ones) in the binary representation of the integer.
2. The decimal value of the integer (as a tie-breaker).

For negative integers, the standard Python `bin()` function returns a string like `'-0b...'`. However, the prompt specifies "non-negative integers" in the description but provides test cases with negative integers. In Python, the bit count of a negative integer is technically infinite in two's complement, but for sorting purposes in these types of challenges, we typically treat the absolute value's bit count or the standard bit count of the positive representation. Looking at the test case `[-2, -3, -4, -5, -6] -> [-6, -5, -4, -3, -2]`, it implies that for negative numbers, we should treat them as having a specific bit count or simply sort them by value if the bit count logic is ambiguous.

Actually, the standard way to count bits in Python for any integer (including negative) to match common competitive programming logic is `bin(x).count('1')`. Let's check:
`bin(-2)` is `'-0b10'` (1 one)
`bin(-3)` is `'-0b11'` (2 ones)
`bin(-4)` is `'-0b100'` (1 one)
`bin(-5)` is `'-0b101'` (2 ones)
`bin(-6)` is `'-0b110'` (2 ones)

Wait, the test case `[-2, -3, -4, -5, -6] == [-6, -5, -4, -3, -2]` suggests that for negative numbers, the primary sort key might be the bit count of the absolute value or simply the value itself. However, the most robust way to handle "number of ones" for any integer in Python is `bin(x).count('1')`.

Let's re-examine:
`1` (0b1) -> 1 one
`2` (0b10) -> 1 one
`3` (0b11) -> 2 ones
`4` (0b100) -> 1 one
`5` (0b101) -> 2 ones
Sorted by ones: `[1, 2, 4]` (all 1 one), then `[3, 5]` (all 2 ones).
Wait, the example `[1, 5, 2, 3, 4] == [1, 2, 3, 4, 5]` is slightly different.
`1` (1), `5` (2), `2` (1), `3` (2), `4` (1).
Sorted by ones: `1, 2, 4` then `3, 5`.
The example says `[1, 2, 3, 4, 5]`. This implies `3` (2 ones) comes before `4` (1 one)? No, that contradicts "ascending order of ones".
Let's re-read: `sort_array([1, 5, 2, 3, 4]) == [1, 2, 3, 4, 5]`.
Wait, `1` (1 bit), `2` (1 bit), `3` (2 bits), `4` (1 bit), `5` (2 bits).
If sorted by bits: `1, 2, 4, 3, 5`.
If the example `[1, 2, 3, 4, 5]` is correct, it means the bit count of `3` is considered smaller than `4`? That's impossible.
Let's look at the third example: `[1, 0, 2, 3, 4] -> [0, 1, 2, 3, 4]`.
`0` (0 bits), `1` (1 bit), `2` (1 bit), `3` (2 bits), `4` (1 bit).
Sorted by bits: `0, 1, 2, 4, 3`.
The example says `[0, 1, 2, 3, 4]`.

There is a discrepancy between the text "sort according to number of ones" and the provided examples. The examples `[1, 2, 3, 4, 5]` and `[0, 1, 2, 3, 4]` are just standard ascending sorts. However, usually, these tasks require the bit count. Given the prompt's specific instruction, I will follow the text: Sort by `bin(x).count('1')` first, then by `x`.

```python
def sort_array(arr):
    """
    In this Kata, you have to sort an array of non-negative integers according to
    number of ones in their binary representation in ascending order.
    For similar number of ones, sort based on decimal value.
    """
    # The bit_count() method is available in Python 3.10+
    # For compatibility, we use bin(x).count('1')
    # However, for negative numbers, bin(-1) is '-0b1', bin(-2) is '-0b10'.
    # The prompt says "non-negative" but provides negative examples.
    # To handle negative numbers consistently with the examples provided:
    return sorted(arr, key=lambda x: (bin(x).count('1'), x))
```

Wait, looking at the examples again:
`sort_array([1, 5, 2, 3, 4]) == [1, 2, 3, 4, 5]`
`1` (1 bit), `2` (1 bit), `3` (2 bits), `4` (1 bit), `5` (2 bits).
If the result is `[1, 2, 3, 4, 5]`, then `3` (2 bits) is placed before `4` (1 bit). This only happens if the sort is purely by decimal value.
But the instructions are very specific. I will prioritize the written instructions over the potentially simplified examples.

```python
def sort_array(arr):
    return sorted(arr, key=lambda x: (bin(x).count('1'), x))
```


To solve this problem, I will iterate through the words in the string by splitting the input string by spaces. For each word, I will count the number of consonants by checking if each character is an alphabetic letter and not a vowel (a, e, i, o, u). I will maintain the original order of words as they appear in the string and return a list of those that meet the exact count of $n$ consonants.

```python
def select_words(s, n):
    """Given a string s and a natural number n, you have been tasked to implement
    a function that returns a list of all words from string s that contain exactly
    n consonants, in order these words appear in the string s.
    If the string s is empty then the function should return an empty list.
    Note: you may assume the input string contains only letters and spaces.
    """
    if not s:
        return []

    vowels = "aeiouAEIOU"
    words = s.split()
    result = []

    for word in words:
        consonant_count = 0
        for char in word:
            # Check if character is a letter and not a vowel
            if char.isalpha() and char not in vowels:
                consonant_count += 1

        if consonant_count == n:
            result.append(word)

    return result
```


To solve this problem, we need to iterate through the word from right to left. A vowel qualifies if it is surrounded by consonants on both sides. Since the problem specifies that vowels at the very beginning or very end of the word do not count, we only need to check characters from index `1` to `len(word) - 2`.

The logic follows these steps:
1. Define the set of vowels (both lowercase and uppercase).
2. Iterate through the string starting from the second-to-last character down to the second character (index `len(word) - 2` down to `1`).
3. For each character, check if it is a vowel.
4. If it is a vowel, check if the character immediately to its left and the character immediately to its right are consonants.
5. Return the first such vowel found (which will be the closest from the right).
6. If the loop finishes without finding a match, return an empty string.

```python
def get_closest_vowel(word):
    """You are given a word. Your task is to find the closest vowel that stands between
    two consonants from the right side of the word (case sensitive).

    Vowels in the beginning and ending doesn't count. Return empty string if you didn't
    find any vowel met the above condition.

    You may assume that the given string contains English letter only.

    Example:
    get_closest_vowel("yogurt") ==> "u"
    get_closest_vowel("FULL") ==> "U"
    get_closest_vowel("quick") ==> ""
    get_closest_vowel("ab") ==> ""
    """
    vowels = set("aeiouAEIOU")
    n = len(word)

    # A vowel must have a character to its left and right,
    # so the word must have at least 3 characters.
    if n < 3:
        return ""

    # Iterate from the right side, excluding the first and last characters
    for i in range(n - 2, 0, -1):
        char = word[i]
        if char in vowels:
            left_char = word[i-1]
            right_char = word[i+1]

            # Check if both neighbors are consonants
            if left_char not in vowels and right_char not in vowels:
                return char

    return ""
```


To solve this problem, we need to determine if concatenating two strings $A$ and $B$ in either order ($A+B$ or $B+A$) results in a balanced parentheses string.

A string is balanced if:
1. The total number of '(' equals the total number of ')'.
2. At any point while reading from left to right, the running count of '(' is greater than or equal to the running count of ')'.

**Approach:**
1. First, check if the total count of '(' across both strings equals the total count of ')'. If not, it's impossible to balance them, so return 'No'.
2. If the counts match, we check both possible concatenations: $A+B$ and $B+A$.
3. For each concatenation, we iterate through the characters and maintain a balance counter (increment for '(', decrement for ')'). If the counter ever drops below zero, that specific concatenation is invalid.
4. If either concatenation is valid, return 'Yes'. Otherwise, return 'No'.

```python
def match_parens(lst):
    if len(lst) != 2:
        return 'No'

    s1, s2 = lst[0], lst[1]

    # Check if total counts match first
    total_open = s1.count('(') + s2.count('(')
    total_close = s1.count(')') + s2.count(')')

    if total_open != total_close:
        return 'No'

    def is_balanced(s):
        balance = 0
        for char in s:
            if char == '(':
                balance += 1
            else:
                balance -= 1
            if balance < 0:
                return False
        return balance == 0

    # Check both possible concatenations
    if is_balanced(s1 + s2) or is_balanced(s2 + s1):
        return 'Yes'
    else:
        return 'No'
```


To solve this problem, we need to identify the $k$ largest elements in the input array and return them in sorted order.

**Approach:**
1. Sort the input array `arr` in ascending order.
2. Slice the last $k$ elements of the sorted array (these will be the $k$ largest values).
3. If $k=0$, the result should be an empty list.
4. The complexity will be $O(N \log N)$ due to the sorting step, where $N$ is the length of the array. Given the constraints ($N \le 1000$), this is highly efficient.

```python
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length k with the maximum k numbers in arr.

    Example 1:
        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:
        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Example 3:
        Input: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    # Sort the array in ascending order
    sorted_arr = sorted(arr)

    # Take the last k elements, which are the k largest
    # If k is 0, this correctly returns an empty list
    result = sorted_arr[-k:] if k > 0 else []

    # The problem asks for a sorted list of the maximum k numbers.
    # Since sorted_arr is already sorted, result is already sorted.
    return result
```


To solve this problem, we need to iterate through the list and identify elements that satisfy two conditions:
1. The index of the element must be even (0, 2, 4, ...).
2. The value of the element must be odd.

Note: In programming, list indexing starts at 0. Therefore, "even positions" refers to indices 0, 2, 4, etc.

Complexity:
- Time Complexity: O(n), where n is the length of the list, as we traverse the list once.
- Space Complexity: O(1), as we only store the running sum.

```python
def solution(lst):
    """Given a non-empty list of integers, return the sum of all of the odd elements that are in even positions.

    Examples
    solution([5, 8, 7, 1]) ==> 12
    solution([3, 3, 3, 3, 3]) ==> 9
    solution([30, 13, 24, 321]) ==>0
    """
    total_sum = 0
    # Iterate through the list using step 2 to only check even indices (0, 2, 4...)
    for i in range(0, len(lst), 2):
        # Check if the value at the even index is odd
        if lst[i] % 2 != 0:
            total_sum += lst[i]
    return total_sum
```


To solve this problem, we need to iterate through the first `k` elements of the input array `arr`. For each element, we check if it has at most two digits. An integer has at most two digits if its absolute value is less than or equal to 99. We then sum all such elements and return the result.

Complexity:
- Time Complexity: O(k), as we iterate through the first k elements once.
- Space Complexity: O(1), as we only store the running sum.

```python
def add_elements(arr, k):
    """
    Given a non-empty array of integers arr and an integer k, return
    the sum of the elements with at most two digits from the first k elements of arr.

    Example:

        Input: arr = [111,21,3,4000,5,6,7,8,9], k = 4
        Output: 24 # sum of 21 + 3

    Constraints:
        1. 1 <= len(arr) <= 100
        2. 1 <= k <= len(arr)
    """
    total_sum = 0
    # Iterate through the first k elements
    for i in range(k):
        num = arr[i]
        # A number has at most two digits if its absolute value is <= 99
        if abs(num) <= 99:
            total_sum += num
    return total_sum
```


To solve this problem, we need to generate the Collatz sequence starting from the input integer `n`. During the generation, we check if each number in the sequence is odd. If it is, we add it to a set (to ensure uniqueness, though the Collatz sequence typically doesn't repeat until it hits the 4-2-1 loop). Finally, we convert the set to a list and sort it in ascending order.

Complexity:
- Time: O(k log k) where k is the number of odd elements in the sequence (due to sorting). The length of the sequence is not strictly bounded by a simple formula but is generally small for practical inputs.
- Space: O(k) to store the odd numbers.

```python
def get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """
    odd_numbers = set()
    current = n

    while True:
        if current % 2 != 0:
            odd_numbers.add(current)

        if current == 1:
            break

        if current % 2 == 0:
            current = current // 2
        else:
            current = 3 * current + 1

    return sorted(list(odd_numbers))
```


To solve this problem, I need to:
1. Check if the input string is empty.
2. Verify the format `mm-dd-yyyy` using string splitting and length checks.
3. Convert the components to integers and validate the month range (1-12).
4. Validate the day range based on the specific month (31, 30, or 29 days).
5. Ensure the year is a valid integer (though the prompt doesn't specify a range, standard date logic implies a 4-digit year).

Note: The prompt specifies that for month 2, the max days is 29 (ignoring leap year specifics as per the provided rules).

```python
def valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy
    """
    if not date:
        return False

    parts = date.split('-')
    if len(parts) != 3:
        return False

    # Check format mm-dd-yyyy (lengths must be 2, 2, and 4)
    if len(parts[0]) != 2 or len(parts[1]) != 2 or len(parts[2]) != 4:
        return False

    # Ensure all parts are numeric
    if not (parts[0].isdigit() and parts[1].isdigit() and parts[2].isdigit()):
        return False

    month = int(parts[0])
    day = int(parts[1])
    year = int(parts[2])

    # Rule 3: Month range
    if month < 1 or month > 12:
        return False

    # Rule 2: Day range based on month
    if month in [1, 3, 5, 7, 8, 10, 12]:
        if day < 1 or day > 31:
            return False
    elif month in [4, 6, 9, 11]:
        if day < 1 or day > 30:
            return False
    elif month == 2:
        if day < 1 or day > 29:
            return False

    return True
```


To solve this problem, I need to check for the presence of whitespace first. If whitespace exists, I use the standard `.split()` method. If no whitespace is found, I check for commas. If commas exist, I split by commas. If neither whitespace nor commas are present, I iterate through the string, identify lowercase letters, calculate their alphabetical index (0-25), and count how many of those indices are odd.

Complexity:
- Time: O(N) where N is the length of the string, as we may traverse the string a few times.
- Space: O(N) to store the resulting list of words.

```python
def split_words(txt):
    # Check for whitespace
    if any(char.isspace() for char in txt):
        return txt.split()

    # Check for commas
    if ',' in txt:
        return txt.split(',')

    # Neither whitespace nor commas exist: count lowercase letters with odd alphabetical order
    count = 0
    for char in txt:
        if 'a' <= char <= 'z':
            # ord('a') is 97. To get 0 for 'a', we subtract 97.
            index = ord(char) - ord('a')
            if index % 2 != 0:
                count += 1
    return count
```


To solve this problem, we need to check two conditions:
1. The list must be in non-decreasing (ascending) order.
2. No number can appear more than twice (since the prompt says "more than 1 duplicate" is not allowed, meaning a maximum of 2 occurrences is acceptable).

Wait, let's re-read carefully: "If list has more than 1 duplicate of the same number, return False."
- 0 duplicates (e.g., `[1, 2, 3]`) -> True
- 1 duplicate (e.g., `[1, 2, 2, 3]`) -> True
- 2 duplicates (e.g., `[1, 2, 2, 2, 3]`) -> False

So, the count of any number must be $\le 2$.

```python
def is_sorted(lst):
    '''
    Given a list of numbers, return whether or not they are sorted
    in ascending order. If list has more than 1 duplicate of the same
    number, return False. Assume no negative numbers and only integers.
    '''
    # Check if the list is sorted in ascending order
    for i in range(len(lst) - 1):
        if lst[i] > lst[i+1]:
            return False

    # Check for more than 1 duplicate (i.e., count > 2)
    counts = {}
    for num in lst:
        counts[num] = counts.get(num, 0) + 1
        if counts[num] > 2:
            return False

    return True
```


To solve this problem, we need to find the intersection of two closed intervals $[a, b]$ and $[c, d]$. The intersection of two such intervals is $[max(a, c), min(b, d)]$. If $max(a, c) \le min(b, d)$, the intersection exists and its length is $min(b, d) - max(a, c)$. If $max(a, c) > min(b, d)$, there is no intersection. Finally, we check if this length is a prime number. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

```python
def intersection(interval1, interval2):
    """You are given two intervals,
    where each interval is a pair of integers. For example, interval = (start, end) = (1, 2).
    The given intervals are closed which means that the interval (start, end)
    includes both start and end.
    For each given interval, it is assumed that its start is less or equal its end.
    Your task is to determine whether the length of intersection of these two
    intervals is a prime number.
    Example, the intersection of the intervals (1, 3), (2, 4) is (2, 3)
    which its length is 1, which not a prime number.
    If the length of the intersection is a prime number, return "YES",
    otherwise, return "NO".
    If the two intervals don't intersect, return "NO".
    """
    start1, end1 = interval1
    start2, end2 = interval2

    # Find the boundaries of the intersection
    inter_start = max(start1, start2)
    inter_end = min(end1, end2)

    # If the start is greater than the end, there is no intersection
    if inter_start > inter_end:
        return "NO"

    # Calculate the length of the intersection
    length = inter_end - inter_start

    # Check if the length is a prime number
    if length < 2:
        return "NO"

    for i in range(2, int(length**0.5) + 1):
        if length % i == 0:
            return "NO"

    return "YES"
```


To solve this problem, we need to calculate two components:
1. The sum of the magnitudes (absolute values) of all integers in the array.
2. The product of the signs of all integers in the array.

The sign of an integer $x$ is defined as:
- $1$ if $x > 0$
- $-1$ if $x < 0$
- $0$ if $x = 0$

The final result is the product of these two components. If the array is empty, we return `None`.

Complexity:
- Time: $O(n)$ where $n$ is the length of the array.
- Space: $O(1)$.

```python
def prod_signs(arr):
    """
    You are given an array arr of integers and you need to return
    sum of magnitudes of integers multiplied by product of all signs
    of each number in the array, represented by 1, -1 or 0.
    Note: return None for empty arr.

    Example:
    >>> prod_signs([1, 2, 2, -4]) == -9
    >>> prod_signs([0, 1]) == 0
    >>> prod_signs([]) == None
    """
    if not arr:
        return None

    magnitude_sum = 0
    sign_product = 1

    for x in arr:
        magnitude_sum += abs(x)

        if x > 0:
            sign_product *= 1
        elif x < 0:
            sign_product *= -1
        else:
            sign_product *= 0

    return magnitude_sum * sign_product
```


To find the lexicographically smallest path of length $k$, we can use a greedy approach combined with a backtracking-like search or a Breadth-First Search style exploration. However, since we need the lexicographically smallest sequence, at each step we want to pick the smallest possible neighbor.

Because $k$ can be large and we can revisit cells, a simple greedy choice at each step might lead to a dead end or a sub-optimal path later. But wait, the problem asks for the lexicographically smallest path. This means we want the smallest possible value for the first cell, then the smallest possible value for the second cell, and so on.

Since we can revisit cells, if we are at a cell and want to find the smallest path of length $k$ starting from there, we can use dynamic programming or memoization. Let $dp(r, c, steps\_left)$ be the lexicographically smallest sequence of length $steps\_left$ starting from cell $(r, c)$.

However, the state space $(N \times N \times k)$ is manageable. For each cell, we want to find the neighbor that leads to the lexicographically smallest sequence.

```python
import sys

# Increase recursion depth for deep paths
sys.setrecursionlimit(2000)

def minPath(grid, k):
    N = len(grid)
    # memo stores the lexicographically smallest path of length 'rem' starting from (r, c)
    memo = {}

    def solve(r, c, rem):
        if rem == 1:
            return [grid[r][c]]

        state = (r, c, rem)
        if state in memo:
            return memo[state]

        best_path = None

        # Explore 4 neighbors
        for dr, dc in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
            nr, nc = r + dr, c + dc
            if 0 <= nr < N and 0 <= nc < N:
                path = solve(nr, nc, rem - 1)
                # The full path starting from current cell would be [grid[r][c]] + path
                # We want to find the neighbor that produces the smallest path
                if best_path is None or path < best_path:
                    best_path = path

        res = [grid[r][c]] + best_path
        memo[state] = res
        return res

    # To find the overall lexicographically smallest path,
    # we check all possible starting cells.
    overall_best = None

    # Optimization: The first element of the path should be as small as possible.
    # We can sort all cells by their values and try them.
    cells = []
    for r in range(N):
        for c in range(N):
            cells.append((grid[r][c], r, c))
    cells.sort()

    for val, r, c in cells:
        # If the current value is already larger than the first element of
        # our best path, we can stop (since cells are sorted).
        if overall_best and val > overall_best[0]:
            break

        current_path = solve(r, c, k)
        if overall_best is None or current_path < overall_best:
            overall_best = current_path

    return overall_best
```


To solve this problem, we need to generate a sequence where the values are interdependent. Specifically, `tri(n)` for an odd `n` depends on `tri(n+1)`. This implies that we cannot simply iterate from 1 to $n$ linearly if we encounter an odd number, as the next value isn't known yet.

However, looking at the rules:
1. `tri(1) = 3`
2. `tri(n) = 1 + n // 2` if $n$ is even.
3. `tri(n) = tri(n-1) + tri(n-2) + tri(n+1)` if $n$ is odd.

For any even $k$, `tri(k)` is directly calculable. For any odd $k$, `tri(k)` depends on `tri(k-1)`, `tri(k-2)`, and `tri(k+1)`. Since $k-1$ and $k-2$ are known (or calculable) and $k+1$ is even (and thus calculable), we can determine any `tri(n)` by calculating the even terms first or using the even term formula to resolve the odd term's dependency.

Wait, let's re-examine the example:
`tri(1) = 3` (Given)
`tri(2) = 1 + 2/2 = 2`
`tri(4) = 1 + 4/2 = 3`
`tri(3) = tri(2) + tri(1) + tri(4) = 2 + 3 + 3 = 8`
The sequence for $n=3$ is `[tri(0), tri(1), tri(2), tri(3)]`.
Wait, the example says `tri(3) = [1, 3, 2, 8]`. This implies the list starts from index 0.
Let's check the indices:
Index 0: `tri(0)`? The prompt says `tri(1)=3`. If `tri(3)` returns 4 elements, they are `tri(0), tri(1), tri(2), tri(3)`.
If `tri(0)` is the first element, and `tri(1)=3`, `tri(2)=2`, `tri(3)=8`, then `tri(0)` must be 1.
Let's check if `tri(0)` follows the even rule: `tri(0) = 1 + 0/2 = 1`. This matches!

So the logic is:
- If $i$ is even: `tri(i) = 1 + i // 2`
- If $i$ is odd: `tri(i) = tri(i-1) + tri(i-2) + tri(i+1)`
- To find `tri(i)` where $i$ is odd, we need `tri(i+1)`. Since $i+1$ is even, `tri(i+1) = 1 + (i+1)//2`.

```python
def tri(n):
    """Everyone knows Fibonacci sequence, it was studied deeply by mathematicians in
    the last couple centuries. However, what people don't know is Tribonacci sequence.
    Tribonacci sequence is defined by the recurrence:
    tri(1) = 3
    tri(n) = 1 + n / 2, if n is even.
    tri(n) =  tri(n - 1) + tri(n - 2) + tri(n + 1), if n is odd.
    For example:
    tri(2) = 1 + (2 / 2) = 2
    tri(4) = 3
    tri(3) = tri(2) + tri(1) + tri(4)
           = 2 + 3 + 3 = 8
    You are given a non-negative integer number n, you have to a return a list of the
    first n + 1 numbers of the Tribonacci sequence.
    Examples:
    tri(3) = [1, 3, 2, 8]
    """
    if n < 0:
        return []

    res = [0] * (n + 1)

    for i in range(n + 1):
        if i == 0:
            # Based on the example tri(3) = [1, 3, 2, 8],
            # tri(0) must be 1. The even rule 1 + 0/2 also gives 1.
            res[i] = 1
        elif i == 1:
            res[i] = 3
        elif i % 2 == 0:
            res[i] = 1 + (i // 2)
        else:
            # i is odd. tri(i) = tri(i-1) + tri(i-2) + tri(i+1)
            # tri(i+1) is even, so we can calculate it directly.
            # If i is the last element (n), we need to be careful.
            # However, the formula for odd i always requires tri(i+1).
            # If i = n and n is odd, we still need tri(n+1).
            val_next = 1 + ((i + 1) // 2)
            res[i] = res[i-1] + res[i-2] + val_next

    return res
```


To solve this problem, we need to iterate through each digit of the integer `n`. We can convert the integer to a string to easily access each digit. For each digit, we check if it is odd. If it is, we multiply it into a running product. We initialize the product to 1 and use a flag (or check if any odd digits were found) to handle the case where no odd digits exist, in which case we return 0.

Complexity:
- Time: O(d), where d is the number of digits in n.
- Space: O(d) to store the string representation of the number.

```python
def digits(n):
    """Given a positive integer n, return the product of the odd digits.
    Return 0 if all digits are even.
    For example:
    digits(1)  == 1
    digits(4)  == 0
    digits(235) == 15
    """
    product = 1
    has_odd = False

    for char in str(n):
        digit = int(char)
        if digit % 2 != 0:
            product *= digit
            has_odd = True

    return product if has_odd else 0
```


To solve this problem, we need to determine if there is a valid subsequence of brackets where at least one bracket is nested. A "nested" bracket occurs when a pair of brackets `[]` is contained within another pair of brackets.

In terms of structure, this means we are looking for a pattern where a pair of brackets exists inside another pair. For example, `[[]]` is nested because the inner `[]` is inside the outer `[]`. `[][]` is not nested because the pairs are adjacent.

The most efficient way to check for nesting is to track the depth of the brackets. If we encounter a closing bracket `]` while the current depth is greater than 1, it means we have successfully closed a bracket that was inside another bracket. However, the problem asks for a *subsequence*. A subsequence doesn't have to be contiguous.

Actually, the simplest condition for a nested subsequence is the existence of the pattern `[[...]]`. If we can find any instance where a `[` is followed by another `[` and eventually followed by a `]` and another `]`, we have a nested structure. More simply, if we find any `[` that is opened while another `[` is already open, and we eventually close both, we have nesting.

Wait, the simplest check for a nested subsequence is: does the string contain `[[` and `]]` in an order that allows them to form a nested pair? Actually, the rule is even simpler: if there is any point where the "nesting level" reaches 2 or more, and we can complete that nesting, it's True. But since we only need a *subsequence*, we just need to find if there are at least two `[` characters that appear before two `]` characters such that they can form a `[ [ ] ]` structure.

Actually, looking at the test cases:
`[[]]` -> True (Nested)
`[][]` -> False (Not nested)
`[[][]]` -> True (Nested)
`[[]][[` -> True (Nested)

The condition is: Is there a pair of brackets `[]` that exists inside another pair? This is equivalent to saying: is there a subsequence of the form `[[ ]]`?

```python
def is_nested(string):
    """
    Returns True if there is a valid subsequence of brackets
    where at least one bracket is nested (i.e., a subsequence [[ ]]).
    """
    # We are looking for a subsequence [[ ]]
    # This means we need to find two '[' that appear before two ']'
    # such that they can form a nested structure.
    # Specifically, we need to find a '[' then another '[' then a ']' then a ']'.

    # Let's track the indices of all '[' and ']'
    open_indices = [i for i, char in enumerate(string) if char == '[']
    close_indices = [i for i, char in enumerate(string) if char == ']']

    # If we have fewer than 2 of each, nesting is impossible
    if len(open_indices) < 2 or len(close_indices) < 2:
        return False

    # We need to find indices i, j, k, l such that:
    # i < j < k < l  AND string[i/j/k/l] are [ [ ] ]
    # This is the definition of a nested subsequence.

    # To check this, we can use a greedy approach:
    # Find the first two '['
    # Then find the first two ']' that appear after the second '['

    first_open = open_indices[0]
    second_open = open_indices[1]

    # Find the first ']' that appears after second_open
    first_close_after_second_open = -1
    for idx in close_indices:
        if idx > second_open:
            first_close_after_second_open = idx
            break

    if first_close_after_second_open == -1:
        return False

    # Find the second ']' that appears after first_close_after_second_open
    second_close_after_first_close = -1
    for idx in close_indices:
        if idx > first_close_after_second_open:
            second_close_after_first_close = idx
            break

    return second_close_after_first_close != -1

```

Wait, the logic above is slightly flawed because the first `[` doesn't have to be the first one in the string. Let's refine: we need to find any $i < j < k < l$ such that $S[i]='[', S[j]='[', S[k]=']', S[l]=']'$.

```python
def is_nested(string):
    # Find the first two '['
    # Then find the first two ']' that occur after the second '['
    # This is not quite right because the first ']' could be before the second '['.
    # The correct condition for a subsequence [[ ]] is:
    # There exist indices i < j < k < l such that S[i]='[', S[j]='[', S[k]=']', S[l]=']'

    # Let's find the first two '[' indices
    open_indices = [i for i, char in enumerate(string) if char == '[']
    if len(open_indices) < 2:
        return False

    # We want to find the smallest possible j (the second '[')
    # to leave as much room as possible for the two ']'
    j = open_indices[1]

    # Now find the first two ']' that occur after index j
    close_indices_after_j = [i for i, char in enumerate(string) if char == ']' and i > j]

    return len(close_indices_after_j) >= 2
```

Let's double check:
`[[]]` -> `open_indices` = [0, 1]. `j` = 1. `close_indices_after_j` = [2, 3]. Length 2. True.
`[][]` -> `open_indices` = [0, 2]. `j` = 2. `close_indices_after_j` = [3]. Length 1. False.
`[[][]]` -> `open_indices` = [0, 1, 4]. `j` = 1. `close_indices_after_j` = [2, 3, 5]. Length 3. True.
`[[]][[` -> `open_indices` = [0, 1, 4, 5]. `j` = 1. `close_indices_after_j` = [2, 3]. Length 2. True.

This logic seems robust for the "subsequence" requirement.

```python
def is_nested(string):
    open_indices = [i for i, char in enumerate(string) if char == '[']
    if len(open_indices) < 2:
        return False

    # To maximize the chance of finding two ']' after the second '[',
    # we pick the smallest possible index for the second '['.
    # The smallest possible index for the second '[' is the second '[' in the string.
    j = open_indices[1]

    close_indices_after_j = [i for i, char in enumerate(string) if char == ']' and i > j]

    return len(close_indices_after_j) >= 2
```


To solve this problem, we need to iterate through each number in the input list, apply the ceiling function to round it up to the nearest integer, square that integer, and then sum all these squared values.

The `math.ceil` function is the standard way to handle the ceiling operation in Python. For negative numbers, `math.ceil(-2.4)` correctly returns `-2`.

Complexity:
- Time Complexity: O(n), where n is the length of the list.
- Space Complexity: O(1), as we are calculating the sum in a single pass.

```python
import math

def sum_squares(lst):
    """You are given a list of numbers.
    You need to return the sum of squared numbers in the given list,
    round each element in the list to the upper int(Ceiling) first.
    Examples:
    For lst = [1,2,3] the output should be 14
    For lst = [1,4,9] the output should be 98
    For lst = [1,3,5,7] the output should be 84
    For lst = [1.4,4.2,0] the output should be 29
    For lst = [-2.4,1,1] the output should be 6
    """
    total_sum = 0
    for num in lst:
        # Round up to the nearest integer
        rounded_num = math.ceil(num)
        # Square the rounded integer and add to total
        total_sum += rounded_num ** 2
    return total_sum
```


To solve this problem, we need to check three conditions:
1. The string must not be empty.
2. The last character of the string must be an alphabetical letter.
3. The last character must be "not a part of a word". Based on the examples provided, a character is "not a part of a word" if it is preceded by a space and is the only character in that space-separated group (i.e., it is a single-letter word). However, looking closely at the examples:
   - `"apple pie"` -> `e` is part of "pie" (False).
   - `"apple pi e"` -> `e` is a single letter separated by a space (True).
   - `"apple pi e "` -> The last character is a space (False).
   - `""` -> Empty (False).

Therefore, the logic is: The last character must be a letter, and the character immediately preceding it must be a space, and there must be no characters between that space and the end of the string.

```python
def check_if_last_char_is_a_letter(txt):
    '''
    Create a function that returns True if the last character
    of a given string is an alphabetical character and is not
    a part of a word, and False otherwise.
    Note: "word" is a group of characters separated by space.
    '''
    if not txt:
        return False

    last_char = txt[-1]

    # Check if the last character is an alphabetical letter
    if not last_char.isalpha():
        return False

    # Check if it is "not a part of a word".
    # Based on examples, this means it must be a single letter
    # preceded by a space.
    if len(txt) > 1:
        # If the character before the last one is a space,
        # and the last character is a letter, it's a single-letter word.
        return txt[-2] == ' '

    # If the string is only one character long and it's a letter,
    # it's technically not part of a multi-character word.
    return True
```


To solve this problem, we need to iterate through the array starting from the second element (index 1) and compare each element with its predecessor. We are looking for the *largest* index $i$ such that `arr[i] < arr[i-1]`. Since we want the largest index, we can iterate backwards from the end of the list to the beginning. The first time we encounter an element smaller than its predecessor, we return that index. If the loop completes without finding such an element, we return -1.

Complexity:
- Time: $O(n)$ where $n$ is the length of the array.
- Space: $O(1)$.

```python
def can_arrange(arr):
    """Create a function which returns the largest index of an element which
    is not greater than or equal to the element immediately preceding it. If
    no such element exists then return -1. The given array will not contain
    duplicate values.

    Examples:
    can_arrange([1,2,4,3,5]) = 3
    can_arrange([1,2,3]) = -1
    """
    # Iterate backwards from the last index to the second index (index 1)
    # because we need to compare arr[i] with arr[i-1].
    for i in range(len(arr) - 1, 0, -1):
        if arr[i] < arr[i-1]:
            return i
    return -1
```


To solve this problem, I will iterate through the list once to identify all negative and positive integers. I will maintain two variables to track the maximum of the negatives (the one closest to zero) and the minimum of the positives (the one closest to zero). If no negatives are found, the first element of the tuple will be `None`; if no positives are found, the second will be `None`. Zero is ignored as it is neither positive nor negative.

```python
def largest_smallest_integers(lst):
    '''
    Create a function that returns a tuple (a, b), where 'a' is
    the largest of negative integers, and 'b' is the smallest
    of positive integers in a list.
    If there is no negative or positive integers, return them as None.

    Examples:
    largest_smallest_integers([2, 4, 1, 3, 5, 7]) == (None, 1)
    largest_smallest_integers([]) == (None, None)
    largest_smallest_integers([0]) == (None, None)
    '''
    negatives = [x for x in lst if x < 0]
    positives = [x for x in lst if x > 0]

    a = max(negatives) if negatives else None
    b = min(positives) if positives else None

    return (a, b)
```


To solve this problem, we need to normalize all inputs into a comparable numeric format while preserving their original types.

1.  **Normalization**: For each input, we check if it's a number (int/float) or a string. If it's a string, we replace commas with dots and attempt to convert it to a float.
2.  **Comparison**: We compare these normalized values.
3.  **Return Logic**:
    *   If the normalized values are equal, return `None`.
    *   If the first value is larger, return the original first variable.
    *   If the second value is larger, return the original second variable.

```python
def compare_one(a, b):
    """
    Create a function that takes integers, floats, or strings representing
    real numbers, and returns the larger variable in its given variable type.
    Return None if the values are equal.
    Note: If a real number is represented as a string, the floating point might be . or ,
    """
    def to_float(val):
        if isinstance(val, (int, float)):
            return float(val)
        if isinstance(val, str):
            # Replace comma with dot to handle different decimal separators
            normalized_str = val.replace(',', '.')
            try:
                return float(normalized_str)
            except ValueError:
                return None
        return None

    val_a = to_float(a)
    val_b = to_float(b)

    # If either cannot be converted to a number, we treat them as incomparable
    # or handle based on specific requirements. Based on prompt, they are real numbers.
    if val_a == val_b:
        return None

    if val_a > val_b:
        return a
    else:
        return b
```


To determine if a number $n$ can be written as the sum of exactly 4 positive even numbers, we can analyze the mathematical constraints:
1. The smallest positive even number is 2.
2. The sum of four positive even numbers must be at least $2 + 2 + 2 + 2 = 8$.
3. Since the sum of any number of even numbers is always even, $n$ must be even.
4. If $n$ is even and $n \ge 8$, we can always represent it as $2 + 2 + 2 + (n - 6)$. Since $n$ is even and $n \ge 8$, $n - 6$ is an even number greater than or equal to 2.

Therefore, the condition is simply that $n$ must be even and $n \ge 8$.

```python
def is_equal_to_sum_even(n):
    """Evaluate whether the given number n can be written as the sum of exactly 4 positive even numbers
    Example
    is_equal_to_sum_even(4) == False
    is_equal_to_sum_even(6) == False
    is_equal_to_sum_even(8) == True
    """
    # A sum of 4 positive even numbers must be at least 2+2+2+2 = 8.
    # Additionally, the sum of any number of even numbers must be even.
    return n >= 8 and n % 2 == 0
```


To solve this problem, we need to calculate the product of factorials from $1!$ up to $n!$.

**Approach:**
1. Handle the base case: if $n \le 0$, the result is typically defined as 1 (or 0 depending on context, but the prompt specifies $n > 0$).
2. Use a loop to calculate each factorial $k!$ for $k$ from 1 to $n$.
3. Maintain a running product of these factorials.
4. To optimize, instead of recalculating each factorial from scratch (which would be $O(n^2)$), we can maintain a running factorial value. As we iterate from 1 to $n$, we update the current factorial by multiplying it by the current index, and then multiply the total result by this updated factorial. This ensures $O(n)$ time complexity.

**Complexity:**
- Time: $O(n)$
- Space: $O(1)$

```python
def special_factorial(n):
    """The Brazilian factorial is defined as:
    brazilian_factorial(n) = n! * (n-1)! * (n-2)! * ... * 1!
    where n > 0

    For example:
    >>> special_factorial(4)
    288

    The function will receive an integer as input and should return the special
    factorial of this integer.
    """
    if n <= 0:
        return 0

    total_product = 1
    current_factorial = 1

    for i in range(1, n + 1):
        current_factorial *= i
        total_product *= current_factorial

    return total_product
```


To solve this problem, we need to identify sequences of spaces. If a sequence of spaces has a length of 1 or 2, each space should be replaced by an underscore (`_`). If a sequence of spaces has a length of 3 or more, the entire sequence should be replaced by a single hyphen (`-`).

However, looking closely at the provided examples:
- `"Example 1"` (1 space) -> `"Example_1"`
- `" Example 2"` (1 space at start) -> `"_Example_2"`
- `" Example   3"` (1 space at start, 3 spaces in middle) -> `"_Example-3"`

The logic is:
1. Identify contiguous blocks of spaces.
2. If a block has length 1 or 2, replace each space in that block with `_`.
3. If a block has length $\ge 3$, replace the entire block with a single `-`.

```python
import re

def fix_spaces(text):
    """
    Given a string text, replace all spaces in it with underscores,
    and if a string has more than 2 consecutive spaces,
    then replace all consecutive spaces with -
    """
    # Use regex to find all sequences of one or more spaces
    def replace_match(match):
        spaces = match.group(0)
        count = len(spaces)
        if count > 2:
            return "-"
        else:
            return "_" * count

    # re.sub can take a function as the replacement argument
    return re.sub(r' +', replace_match, text)
```


To solve this problem, I need to check four specific conditions:
1. Count the number of digits in the string; it must be $\le 3$.
2. Ensure there is exactly one dot (`.`) in the string.
3. Split the string by the dot. The part before the dot must not be empty and must start with a letter (A-Z, a-z).
4. The part after the dot must be exactly "txt", "exe", or "dll".

I will use `str.isdigit()` to count digits and `str.isalpha()` or a range check for the first character of the prefix.

```python
def file_name_check(file_name):
    """Create a function which takes a string representing a file's name, and returns
    'Yes' if the the file's name is valid, and returns 'No' otherwise.
    A file's name is considered to be valid if and only if all the following conditions
    are met:
    - There should not be more than three digits ('0'-'9') in the file's name.
    - The file's name contains exactly one dot '.'
    - The substring before the dot should not be empty, and it starts with a letter from
    the latin alphapet ('a'-'z' and 'A'-'Z').
    - The substring after the dot should be one of these: ['txt', 'exe', 'dll']
    """
    # Condition 1: Count digits
    digit_count = sum(1 for char in file_name if char.isdigit())
    if digit_count > 3:
        return 'No'

    # Condition 2: Exactly one dot
    if file_name.count('.') != 1:
        return 'No'

    # Split into name and extension
    parts = file_name.split('.')
    prefix = parts[0]
    extension = parts[1]

    # Condition 3: Prefix not empty and starts with a letter
    if not prefix or not prefix[0].isalpha():
        return 'No'

    # Condition 4: Extension check
    valid_extensions = ['txt', 'exe', 'dll']
    if extension not in valid_extensions:
        return 'No'

    return 'Yes'
```


To solve this problem, we iterate through the list using `enumerate` to keep track of both the index and the value.
1. If the index is a multiple of 3 (e.g., 0, 3, 6...), we square the value.
2. If the index is a multiple of 4 (e.g., 4, 8, 12...) AND not a multiple of 3, we cube the value.
3. Otherwise, we keep the value as is.
4. We sum these transformed values and return the result.

Complexity:
- Time: O(n) where n is the length of the list.
- Space: O(1) as we calculate the sum on the fly.

```python
def sum_squares(lst):
    """
    This function will take a list of integers. For all entries in the list, the function shall square the integer entry if its index is a
    multiple of 3 and will cube the integer entry if its index is a multiple of 4 and not a multiple of 3. The function will not
    change the entries in the list whose indexes are not a multiple of 3 or 4. The function shall then return the sum of all entries.

    Examples:
    For lst = [1,2,3] the output should be 6
    For lst = []  the output should be 0
    For lst = [-1,-5,2,-1,-5]  the output should be -126
    """
    total_sum = 0
    for i, val in enumerate(lst):
        if i % 3 == 0:
            total_sum += val ** 2
        elif i % 4 == 0:
            total_sum += val ** 3
        else:
            total_sum += val
    return total_sum
```


To solve this problem, I will:
1. Split the input sentence into a list of words.
2. Create a helper function to determine if a number is prime. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
3. Iterate through the words, check if the length of each word is prime, and collect those that are.
4. Join the filtered words back into a single string separated by spaces.

Complexity:
- Time: O(N * sqrt(M)), where N is the number of words and M is the maximum word length (since we check primality up to the square root of the length).
- Space: O(N) to store the words.

```python
def words_in_sentence(sentence):
    """
    You are given a string representing a sentence,
    the sentence contains some words separated by a space,
    and you have to return a string that contains the words from the original sentence,
    whose lengths are prime numbers,
    the order of the words in the new string should be the same as the original one.

    Example 1:
        Input: sentence = "This is a test"
        Output: "is"

    Example 2:
        Input: sentence = "lets go for swimming"
        Output: "go for"

    Constraints:
        * 1 <= len(sentence) <= 100
        * sentence contains only letters
    """
    def is_prime(n):
        if n < 2:
            return False
        for i in range(2, int(n**0.5) + 1):
            if n % i == 0:
                return False
        return True

    words = sentence.split()
    result = []

    for word in words:
        if is_prime(len(word)):
            result.append(word)

    return " ".join(result)
```


To solve this problem, we need to parse the two input strings into their respective numerators and denominators. The expression $x \times n$ is calculated as:
$\frac{\text{num}_x}{\text{den}_x} \times \frac{\text{num}_n}{\text{den}_n} = \frac{\text{num}_x \times \text{num}_n}{\text{den}_x \times \text{den}_n}$

For this result to be a whole number, the product of the numerators must be perfectly divisible by the product of the denominators. That is, $(\text{num}_x \times \text{num}_n) \pmod{(\text{den}_x \times \text{den}_n)} == 0$.

```python
def simplify(x, n):
    """Your task is to implement a function that will simplify the expression
    x * n. The function returns True if x * n evaluates to a whole number and False
    otherwise. Both x and n, are string representation of a fraction, and have the following format,
    <numerator>/<denominator> where both numerator and denominator are positive whole numbers.

    You can assume that x, and n, are valid fractions, and do not have zero as denominator.

    simplify("1/5", "5/1") = True
    simplify("1/6", "2/1") = False
    simplify("7/10", "10/2") = False
    """
    # Split the strings into numerator and denominator components
    num_x, den_x = map(int, x.split('/'))
    num_n, den_n = map(int, n.split('/'))

    # Calculate the resulting numerator and denominator of the product
    # (a/b) * (c/d) = (a*c) / (b*d)
    result_num = num_x * num_n
    result_den = den_x * den_n

    # Check if the result is a whole number (no remainder)
    return result_num % result_den == 0
```


To solve this problem, we need to sort a list of integers based on the sum of their digits.

**Key considerations:**
1. **Digit Sum Calculation:** For negative numbers, the "sum of digits" typically refers to the sum of the absolute values of the digits (e.g., -12 becomes 1+2=3).
2. **Stability:** The problem specifies that if sums are equal, the original index must be preserved. Python's built-in `sort` and `sorted` functions are stable, meaning they preserve the relative order of elements that have equal keys.
3. **Complexity:** The time complexity will be $O(N \cdot K \log N)$, where $N$ is the number of elements and $K$ is the average number of digits.

```python
def order_by_points(nums):
    """
    Write a function which sorts the given list of integers
    in ascending order according to the sum of their digits.
    Note: if there are several items with similar sum of their digits,
    order them based on their index in original list.

    For example:
    >>> order_by_points([1, 11, -1, -11, -12]) == [-1, -11, 1, -12, 11]
    >>> order_by_points([]) == []
    """
    def get_digit_sum(n):
        # Use absolute value to handle negative numbers
        # Sum the digits of the absolute value
        return sum(int(digit) for digit in str(abs(n)))

    # Python's sorted() is stable, so we only need to provide
    # the digit sum as the key. The original index will be
    # preserved automatically for ties.
    return sorted(nums, key=get_digit_sum)
```