Reverse Vowels of a String - Practice Coding Problems

Reverse Vowels of a String - Problem

Reverse Vowels of a String

Given a string s, your task is to reverse only the vowels in the string while keeping all other characters in their original positions. Return the modified string.

The vowels are: 'a', 'e', 'i', 'o', 'u' and they can appear in both lowercase and uppercase, and may appear multiple times in the string.

Example:
Input: "hello"
Output: "holle"
Explanation: The vowels 'e' and 'o' are swapped while 'h', 'l', 'l' stay in place.

Goal: Implement an efficient algorithm to swap vowels while preserving the original structure of the string.

Input & Output

example_1.py — Basic Example

$ Input: s = "hello"

› Output: "holle"

💡 Note: The vowels in 'hello' are 'e' and 'o'. When reversed, 'e' goes to position 4 and 'o' goes to position 1, resulting in 'holle'.

example_2.py — Mixed Case

$ Input: s = "leetcode"

› Output: "leotcede"

💡 Note: The vowels are 'e', 'e', 'o', 'e'. When reversed: 'e', 'o', 'e', 'e'. Result: 'leotcede'.

example_3.py — Edge Case

$ Input: s = "aA"

› Output: "Aa"

💡 Note: Both characters are vowels ('a' and 'A'). When reversed, 'A' comes first and 'a' comes second.

Constraints

1 ≤ s.length ≤ 3 × 10⁵
s consists of printable ASCII characters
The vowels are 'a', 'e', 'i', 'o', 'u' in both lowercase and uppercase

Visualization

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Understanding the Visualization

Setup

Place two organizers at opposite ends of the row

Find

Each organizer walks toward center until they find a red chair

Swap

The people in the red chairs swap positions

Continue

Organizers move past swapped chairs and repeat

Key Takeaway

🎯 Key Insight: Two pointers efficiently swap elements from opposite ends without needing extra space, making this the optimal O(n) time, O(1) space solution.

Asked in

G Google 23 a Amazon 18 ⊞ Microsoft 15 Apple 12

The optimal solution uses the two pointers technique with O(n) time and O(1) space complexity. Initialize pointers at both ends, move them inward to find vowels, then swap them. This approach is elegant, efficient, and handles all edge cases naturally.

Common Approaches

Approach	Time	Space	Notes
✓ Two Pointers (Optimal)	O(n)	O(1)	Use two pointers from both ends to swap vowels efficiently
Brute Force (Find All Vowels First)	O(n)	O(n)	Extract all vowels, reverse them, then rebuild the string

Two Pointers (Optimal) — Algorithm Steps

Convert string to character array for in-place modification
Initialize left pointer at start (0) and right pointer at end (n-1)
Move left pointer right until it finds a vowel or meets right pointer
Move right pointer left until it finds a vowel or meets left pointer
If both pointers found vowels, swap them
Continue until left >= right
Convert character array back to string

Visualization

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Step-by-Step Walkthrough

Initialize Pointers

Left pointer at start, right pointer at end

Find Vowels

Move pointers to locate vowels from both ends

Swap Vowels

Swap the vowels and move pointers inward

Continue

Repeat until pointers meet in the middle

Code -

solution.c — C

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

bool 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* reverseVowels(char* s) {
    int left = 0, right = strlen(s) - 1;
    
    while (left < right) {
        // Move left pointer to next vowel
        while (left < right && !isVowel(s[left])) {
            left++;
        }
        
        // Move right pointer to previous vowel
        while (left < right && !isVowel(s[right])) {
            right--;
        }
        
        // Swap vowels if both pointers found vowels
        if (left < right) {
            char temp = s[left];
            s[left] = s[right];
            s[right] = temp;
            left++;
            right--;
        }
    }
    
    return s;
}

int main() {
    char s[1000];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0; // Remove newline
    printf("%s\n", reverseVowels(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through string, each character visited at most once

✓ Linear Growth

Space Complexity

O(1)

Only using two pointer variables and vowel check, plus O(n) for array conversion

✓ Linear Space

142.4K Views

Medium Frequency

~15 min Avg. Time

2.8K Likes

Ln 1, Col 1

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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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