Reverse String - Practice Coding Problems

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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Understanding the Visualization

Input Array

Character array: ["h","e","l","l","o"]

Reverse Process

Swap characters from both ends moving inward

Output Array

Reversed array: ["o","l","l","e","h"]

Key Takeaway

🎯 Key Insight: Two pointers from opposite ends swap characters while moving inward, achieving O(1) space reversal

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

Two Pointers - Optimal — Algorithm Steps

Initialize left pointer at start, right pointer at end
Swap characters at left and right positions
Move left forward, right backward until they meet

Visualization

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Step-by-Step Walkthrough

Initialize

Set left=0, right=n-1 pointers

Swap & Move

Swap characters, move pointers inward

Complete

Continue until pointers meet

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) {
        // Swap characters
        char* temp = s[left];
        s[left] = s[right];
        s[right] = temp;
        
        // Move pointers
        left++;
        right--;
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for demo - parse JSON array manually
    char* s[100];
    int size = 0;
    
    // Extract characters from JSON format
    char* token = strtok(line + 1, ",]"); // Skip opening bracket
    while (token != NULL && size < 100) {
        // Remove quotes and whitespace
        while (*token == ' ' || *token == '"') token++;
        char* end = token + strlen(token) - 1;
        while (end > token && (*end == ' ' || *end == '"' || *end == '\n')) *end-- = '\0';
        
        s[size] = malloc(2);
        s[size][0] = token[0];
        s[size][1] = '\0';
        size++;
        
        token = strtok(NULL, ",]");
    }
    
    solution(s, size);
    
    // Print result
    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 half the array to swap all characters

✓ Linear Growth

Space Complexity

O(1)

Only using two pointer variables, no extra data structures

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Two Pointers - Optimal — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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