algoadvance

Problem Statement

Given a string s and an integer k, you need to divide the string into groups of size k. If the length of s is not a multiple of k, the final group should be filled with lowercase 'x' characters to make its length equal to k.

Return the list of groups formed.

Clarifying Questions

1. Input Constraints:
   • What is the maximum length of the string s?
   • Can k be greater than the length of s?
   • Are there any constraints on the characters in s (e.g. can it contain spaces, punctuation, numbers)?
2. Output Format:
   • Should the result be a list of strings?
   • If s is empty, should the result be an empty list?

Without further constraints, I’ll assume that:

• The input string s can be any non-empty string containing any characters.
• k is a positive integer.
• The result should be a list of strings.

Strategy

1. Initialize an empty list to store the resulting groups.
2. Traverse the string s in increments of k.
3. For each group:
   • If the group has less than k characters, append 'x' characters until the group’s length is k.
4. Append the final group to the result list.
5. Return the result list.

Code

Here’s the implementation of the described strategy:

def divideString(s: str, k: int, fill: str = 'x') -> list:
    result = []
    n = len(s)

    # Traverse the string in steps of k
    for i in range(0, n, k):
        group = s[i:i + k]
        # If the group is shorter than k, fill it with 'x'
        if len(group) < k:
            group += fill * (k - len(group))
        result.append(group)

    return result

# Example usage:
# s = "abcdefghi"
# k = 3
# Expected output: ["abc", "def", "ghi"]
print(divideString("abcdefghi", 3))

# s = "abcdefghij"
# k = 3
# Expected output: ["abc", "def", "ghi", "jxx"]
print(divideString("abcdefghij", 3))

Time Complexity

• Time Complexity: O(n) where n is the length of the input string s. We traverse the string in chunks of size k and perform O(1) operations for handling the last group.
• Space Complexity: O(n) for storing the result list that contains the groups.

Conclusion

This solution ensures that the string is divided into groups of size k, with the last group padded if necessary. The approach is efficient and handles edge cases effectively.

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