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Divide a String Into Groups of Size k

Difficulty: Easy

Problem Description

Given a string s, the size of each group k, and a character fill, partition the string into groups of size k. If the last group does not have k characters, use the fill character to complete the group. Return an array representing each group.

Key Insights

  • The string is partitioned into groups of size k.
  • If the last group is smaller than k, fill with the specified character.
  • The output should maintain the original order of characters.

Space and Time Complexity

Time Complexity: O(n), where n is the length of the string, as we iterate through the string to create groups. Space Complexity: O(n/k), which is equivalent to O(n) in terms of the output array size, since we are creating a new array of groups.

Solution

To solve this problem, we will:

  1. Initialize an empty list to hold the groups.
  2. Use a loop to iterate through the string in increments of k.
  3. For each group, check if there are enough characters to fill it; if not, append the fill character until the group reaches size k.
  4. Finally, return the list of groups.

The primary data structure used here is a list (or array) to store the resulting groups.

Code Solutions

def divideString(s: str, k: int, fill: str) -> List[str]:
    groups = []
    for i in range(0, len(s), k):
        group = s[i:i+k]
        if len(group) < k:
            group += fill * (k - len(group))  # Fill the group if it's smaller than k
        groups.append(group)
    return groups
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