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2243 - Calculate Digit Sum of a String (Easy)

https://leetcode.com/problems/calculate-digit-sum-of-a-string/

Problem Statement​

You are given a string s consisting of digits and an integer k.

A round can be completed if the length of s is greater than k. In one round, do the following:

  1. Divide s into consecutive groups of size k such that the first k characters are in the first group, the next k characters are in the second group, and so on. Note that the size of the last group can be smaller than k.
  2. Replace each group of s with a string representing the sum of all its digits. For example, "346" is replaced with "13" because 3 + 4 + 6 = 13.
  3. Merge consecutive groups together to form a new string. If the length of the string is greater than k, repeat from step 1.

Return s after all rounds have been completed.

Example 1:

Input: s = "11111222223", k = 3
Output: "135"
Explanation: 
- For the first round, we divide s into groups of size 3: "111", "112", "222", and "23".
  ​​​​​Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5. 
  So, s becomes "3" + "4" + "6" + "5" = "3465" after the first round.
- For the second round, we divide s into "346" and "5".
  Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5. 
  So, s becomes "13" + "5" = "135" after second round. 
Now, s.length <= k, so we return "135" as the answer.

Example 2:

Input: s = "00000000", k = 3
Output: "000"
Explanation: 
We divide s into "000", "000", and "00".
Then we calculate the digit sum of each group: 0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0. 
s becomes "0" + "0" + "0" = "000", whose length is equal to k, so we return "000".

Constraints:

  • 1 <= s.length <= 100
  • 2 <= k <= 100
  • s consists of digits only.

Approach 1: Simulation​

Just following the instructions.

Written by @wingkwong
class Solution {
public:
    string digitSum(string s, int k) {
        while (s.size() > k) {
            string ns = "";
            for (int i = 0; i < s.size(); i += k) {
                // Divide s into a group of size k
                string t = s.substr(i, k);
                // Calulate the digit sum in the current group
                int sum = 0;
                for (int j = 0; j < min((int) t.size(), k); j++) {
                    sum += t[j] - '0';
                }
                // merge back together
                ns += to_string(sum);
            }
            // replace s and do the next round
            s = ns;
        }
        return s;
    }
};