Reverse String II

Reverse String II - Problem

Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.

Rules:

If there are fewer than k characters left, reverse all of them.
If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and leave the other as original.

Input & Output

Example 1 — Basic Case

$ Input: s = "abcdef", k = 2

› Output: "bacdfe"

💡 Note: First 2k=4 chars "abcd": reverse first k=2 to get "bacd". Last 2 chars "ef": reverse both to get "fe". Result: "bacdfe"

Example 2 — Exact Multiple

$ Input: s = "abcd", k = 2

› Output: "bacd"

💡 Note: String length equals 2k=4. Reverse first k=2 chars "ab" to "ba", keep "cd" as-is. Result: "bacd"

Example 3 — Single Character

$ Input: s = "a", k = 1

› Output: "a"

💡 Note: Less than k characters remaining, so reverse all of them. Single character "a" reversed is still "a"

Constraints

1 ≤ s.length ≤ 10⁴
s consists of only lowercase English letters
1 ≤ k ≤ 10⁴

Visualization

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Asked in

G Google 25 f Facebook 18 a Amazon 15

The key insight is to process the string in chunks of 2k characters and reverse only the first k characters of each chunk. Best approach uses two pointers for in-place reversal. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Character-by-Character Processing	O(n)	O(n)	Build result character by character, tracking position to decide reversal
Two Pointers In-Place Reversal	O(n)	O(n)	Use two pointers to reverse segments in-place within the string

Character-by-Character Processing — Algorithm Steps

Iterate through each character
Determine current segment and position within segment
If in reversible segment, collect and reverse characters
Add processed characters to result

Visualization

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Step-by-Step Walkthrough

Identify Chunks

Split string into 2k-sized chunks

Process Each Chunk

Reverse first k chars, keep rest as-is

Build Result

Combine all processed segments

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* solution(char* s, int k) {
    int n = strlen(s);
    char* result = (char*)malloc((n + 1) * sizeof(char));
    int pos = 0;
    int i = 0;
    
    while (i < n) {
        // Process chunk of size 2k
        int chunkEnd = (i + 2 * k < n) ? i + 2 * k : n;
        int reverseEnd = (i + k < n) ? i + k : n;
        
        // Reverse first k characters of this chunk
        for (int j = reverseEnd - 1; j >= i; j--) {
            result[pos++] = s[j];
        }
        
        // Add remaining characters as-is
        for (int j = reverseEnd; j < chunkEnd; j++) {
            result[pos++] = s[j];
        }
        
        i += 2 * k;
    }
    
    result[pos] = '\0';
    return result;
}

int main() {
    char s[10001];
    int k;
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    scanf("%d", &k);
    
    char* result = solution(s, k);
    printf("%s\n", result);
    free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through string with constant work per character

✓ Linear Growth

Space Complexity

O(n)

Extra space for result string and temporary character storage

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Character-by-Character Processing — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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