Reverse Vowels of a String - Practice Coding Problems

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

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Understanding the Visualization

Input

Original string with vowels highlighted

Process

Identify and reverse only the vowels

Output

String with vowels in reversed positions

Key Takeaway

🎯 Key Insight: Two pointers efficiently swap vowels from both ends without needing extra storage for vowel collection

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

Approach	Time	Space	Notes
✓ Brute Force - Extract and Replace	O(n)	O(n)	Extract all vowels, then replace them in reverse order
Two Pointers - Optimal Solution	O(n)	O(n)	Use two pointers to swap vowels from both ends

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

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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

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Extract and Replace — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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