LeetCode #2243 — EASY

Calculate Digit Sum of a String

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

Solve on LeetCode

The Problem

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:

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.
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.
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.

Step 01

Brute Force Baseline

Problem summary: 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: 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. 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. 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"11111222223"
3

Example 2

"00000000"
3

Core Insight

What unlocks the optimal approach

Try simulating the entire process to find the final answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2243: Calculate Digit Sum of a String
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;
    }
}

// Accepted solution for LeetCode #2243: Calculate Digit Sum of a String
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
}

# Accepted solution for LeetCode #2243: Calculate Digit Sum of a String
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

// Accepted solution for LeetCode #2243: Calculate Digit Sum of a String
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
    }
}

// Accepted solution for LeetCode #2243: Calculate Digit Sum of a String
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.