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2243. Calculate Digit Sum of a String

Description

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.

Solutions

  • class Solution { public String digitSum(String s, int k) { while (s.length() > k) { int n = s.length(); StringBuilder t = new StringBuilder(); for (int i = 0; i < n; i += k) { int x = 0; for (int j = i; j < Math.min(i + k, n); ++j) { x += s.charAt(j) - '0'; } t.append(x); } s = t.toString(); } return s; } } 
  • class Solution { public: string digitSum(string s, int k) { while (s.size() > k) { string t; int n = s.size(); for (int i = 0; i < n; i += k) { int x = 0; for (int j = i; j < min(i + k, n); ++j) { x += s[j] - '0'; } t += to_string(x); } s = t; } return s; } }; 
  • class Solution: def digitSum(self, s: str, k: int) -> str: while len(s) > k: t = [] n = len(s) for i in range(0, n, k): x = 0 for j in range(i, min(i + k, n)): x += int(s[j]) t.append(str(x)) s = "".join(t) return s 
  • func digitSum(s string, k int) string { for len(s) > k { t := &strings.Builder{} n := len(s) for i := 0; i < n; i += k { x := 0 for j := i; j < i+k && j < n; j++ { x += int(s[j] - '0') } t.WriteString(strconv.Itoa(x)) } s = t.String() } return s } 
  • function digitSum(s: string, k: number): string { let ans = []; while (s.length > k) { for (let i = 0; i < s.length; i += k) { let cur = s.slice(i, i + k); ans.push(cur.split('').reduce((a, c) => a + parseInt(c), 0)); } s = ans.join(''); ans = []; } return s; } 
  • /** * @param {string} s * @param {number} k * @return {string} */ var digitSum = function (s, k) { while (s.length > k) { const t = []; for (let i = 0; i < s.length; i += k) { const x = s .slice(i, i + k) .split('') .reduce((a, b) => a + +b, 0); t.push(x); } s = t.join(''); } return s; }; 
  • impl Solution { pub fn digit_sum(s: String, k: i32) -> String { let mut s = s; let k = k as usize; while s.len() > k { let mut t = Vec::new(); for chunk in s.as_bytes().chunks(k) { let sum: i32 = chunk.iter().map(|&c| (c - b'0') as i32).sum(); t.push(sum.to_string()); } s = t.join(""); } s } }

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