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Leetcode 2243. Calculate Digit Sum of a String

Problem Statement

You are given a string s consisting of digits and an integer k. You need to transform the string into a new string according to the following process:

  1. Divide the string s into groups of size k.
  2. If the size of the last group is less than k, this is fine and you should proceed with summing the digits in the current groups to form a new string.
  3. Replace each group of s with the sum of the digits in the respective group.
  4. Repeat the process until the resultant string’s length is less than or equal to k.

Return the final string as the output.

Clarifying Questions

  1. What is the maximum length of the string s?
    • This will help in understanding if we need to handle very large strings.
  2. What should be done if s is already of length less than or equal to k?
    • It appears that we return s without any changes.
  3. Are there any constraints on the characters in the string s?
    • The problem specifies that s consists of digits only.

Code

#include <iostream>
#include <string>
#include <numeric>
#include <algorithm>

// Function to calculate the digit sum of a string
std::string digitSum(std::string s, int k) {
    while (s.length() > k) {
        std::string new_s = "";
        for (size_t i = 0; i < s.length(); i += k) {
            int sum = 0;
            for (size_t j = i; j < i + k && j < s.length(); ++j) {
                sum += s[j] - '0';  // Convert char digit to integer and add to sum
            }
            new_s += std::to_string(sum);  // Append the sum to the new string
        }
        s = new_s;
    }
    return s;
}

// Testing the function
int main() {
    std::string s = "1111122222";
    int k = 3;
    std::string result = digitSum(s, k);
    std::cout << "Result: " << result << std::endl;
    return 0;
}

Strategy

  1. Iterative Reduction:
    • Continuously reduce the length of the string by summing digits of segments of size k.
    • Iterate until the length of the resultant string becomes less than or equal to k.
  2. String Processing:
    • Use slices of size k to compute the sum for each segment.
    • Convert the sum back to a string and append to the new resultant string.
  3. Edge Case Handling:
    • If the string length is already less than or equal to k, return the string as is.

Time Complexity

The time complexity of this approach is O(n*log(n/k)):

Here, n is the length of the initial string s.

This ensures that the solution should work efficiently even for moderately large strings.

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