Reverse Vowels of a String

Reverse Vowels of a String - Problem

Given a string s, reverse only all the vowels in the string and return it.

The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in both lower and upper cases, more than once.

Input & Output

Example 1 — Basic Case

$ Input: s = "hello"

› Output: "holle"

💡 Note: The vowels are 'e' and 'o'. After reversing them, we get 'o' and 'e', resulting in "holle"

Example 2 — Mixed Case Vowels

$ Input: s = "leetcode"

› Output: "leotcede"

💡 Note: The vowels are 'e', 'e', 'o', 'e'. After reversing: 'e', 'o', 'e', 'e', resulting in "leotcede"

Example 3 — Single Vowel

$ Input: s = "aA"

› Output: "Aa"

💡 Note: Two vowels 'a' and 'A'. After reversing positions, we get "Aa"

Constraints

1 ≤ s.length ≤ 3 × 10⁵
s consists of printable ASCII characters

Visualization

Tap to expand

Asked in

M Microsoft 35 a Amazon 25 f Facebook 20

The key insight is to use two pointers starting from both ends of the string, moving inward to find vowels and swap them in place. The optimal approach uses two pointers for O(n) time and O(n) space complexity.

Common Approaches

✓ Brute Force - Extract and Replace

⏱️ Time: O(n) Space: O(n)

First scan the string to collect all vowels in order. Then scan again and replace each vowel position with vowels from the collected list in reverse order.

Two Pointers - Optimal Solution

⏱️ Time: O(n) Space: O(n)

Place one pointer at the beginning and one at the end. Move pointers inward until both find vowels, then swap them. Continue until pointers meet.

Brute Force - Extract and Replace — Algorithm Steps

Step 1: Scan string and collect all vowels
Step 2: Scan again and replace vowels in reverse order

Visualization

Tap to expand

Step-by-Step Walkthrough

Extract

Collect all vowels in order

Reverse

Use vowels in reverse order

Replace

Build result string

Code -

solution.c — C

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

int isVowel(char c) {
    return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ||
           c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U';
}

char* solution(char* s) {
    int len = strlen(s);
    char* vowels = (char*)malloc(len + 1);
    int vowelCount = 0;
    
    // Collect all vowels
    for (int i = 0; i < len; i++) {
        if (isVowel(s[i])) {
            vowels[vowelCount++] = s[i];
        }
    }
    
    // Replace vowels in reverse order
    char* result = (char*)malloc(len + 1);
    int vowelIndex = vowelCount - 1;
    
    for (int i = 0; i < len; i++) {
        if (isVowel(s[i])) {
            result[i] = vowels[vowelIndex--];
        } else {
            result[i] = s[i];
        }
    }
    
    result[len] = '\0';
    free(vowels);
    return result;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    char* result = solution(s);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Two passes through the string of length n

✓ Linear Growth

Space Complexity

O(n)

Extra array to store vowels and result string

⚡ Linearithmic Space

85.0K Views

Medium Frequency

~15 min Avg. Time

2.2K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Extract and Replace — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler