Reverse String

Reverse String - Problem

Write a function that reverses a string. The input string is given as an array of characters s.

You must do this by modifying the input array in-place with O(1) extra memory.

Input & Output

Example 1 — Basic String

$ Input: s = ["h","e","l","l","o"]

› Output: ["o","l","l","e","h"]

💡 Note: Reverse the array in-place: swap h↔o, e↔l, middle l stays. Result: ["o","l","l","e","h"]

Example 2 — Even Length

$ Input: s = ["H","a","n","n","a","h"]

› Output: ["h","a","n","n","a","H"]

💡 Note: Even length array: swap H↔h, a↔a, n↔n. Result: ["h","a","n","n","a","H"]

Example 3 — Single Character

$ Input: s = ["A"]

› Output: ["A"]

💡 Note: Single character array remains unchanged: ["A"]

Constraints

1 ≤ s.length ≤ 10⁵
s[i] is a printable ascii character

Visualization

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Asked in

G Google 45 a Amazon 38 M Microsoft 32 Apple 28

The key insight is to use two pointers from opposite ends, swapping characters while moving toward the center. The optimal two-pointer approach achieves in-place reversal with Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Extra Array	O(n)	O(n)	Create new array and copy characters in reverse order
Two Pointers - Optimal	O(n)	O(1)	Use two pointers from start and end, swap characters moving inward

Brute Force - Extra Array — Algorithm Steps

Create a new array of same size
Copy characters from end to start into new array
Copy new array back to original

Visualization

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Step-by-Step Walkthrough

Original Array

Start with input array

Copy Reversed

Create new array with characters in reverse

Copy Back

Copy temporary array back to original

Code -

solution.c — C

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

void solution(char* s[], int size) {
    int left = 0;
    int right = size - 1;
    
    while (left < right) {
        char* temp = s[left];
        s[left] = s[right];
        s[right] = temp;
        left++;
        right--;
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    char* s[100];
    int size = 0;
    
    // Remove brackets
    char* content = line + 1;
    char* end = strrchr(content, ']');
    if (end) *end = '\0';
    
    if (strlen(content) == 0) {
        printf("[]\n");
        return 0;
    }
    
    char* token = strtok(content, ",");
    while (token != NULL && size < 100) {
        // Remove quotes and whitespace
        while (*token == ' ' || *token == '"') token++;
        char* token_end = token + strlen(token) - 1;
        while (token_end > token && (*token_end == ' ' || *token_end == '"' || *token_end == '\n')) {
            *token_end = '\0';
            token_end--;
        }
        
        s[size] = malloc(2);
        s[size][0] = token[0];
        s[size][1] = '\0';
        size++;
        
        token = strtok(NULL, ",");
    }
    
    solution(s, size);
    
    printf("[");
    for (int i = 0; i < size; i++) {
        printf("\"%s\"", s[i]);
        if (i < size - 1) printf(",");
        free(s[i]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array to copy characters

✓ Linear Growth

Space Complexity

O(n)

Extra array of size n to store reversed characters

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Extra Array — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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