README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2200-2299/2243.Calculate%20Digit%20Sum%20of%20a%20String/README_EN.md	1301	Weekly Contest 289 Q1	String Simulation

2243. Calculate Digit Sum of a String

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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

Solution 1: Simulation

According to the problem statement, we can simulate the operations described in the problem until the length of the string is less than or equal to $k$. Finally, return the string.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $s$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

function digitSum(s: string, k: number): string {
    while (s.length > k) {
        const t: number[] = [];
        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;
}

Rust

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
    }
}

JavaScript

/**
 * @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;
};