Remove Vowels from a String - Practice Coding Problems

Remove Vowels from a String - Problem

String Easy

Given a string s, remove all vowels ('a', 'e', 'i', 'o', and 'u') from it and return the new string.

Note: The vowels are case-sensitive, so only lowercase vowels should be removed.

Input & Output

Example 1 — Basic Case

$ Input: s = "leetcode"

› Output: "ltcd"

💡 Note: Remove vowels 'e', 'e', 'o', 'e' from "leetcode" to get "ltcd"

Example 2 — Mixed Case

$ Input: s = "aEiOu"

› Output: "EOU"

💡 Note: Only lowercase vowels are removed: 'a', 'i', 'u'. Uppercase 'E', 'O', 'U' remain

Example 3 — No Vowels

$ Input: s = "bcdfg"

› Output: "bcdfg"

💡 Note: String contains no lowercase vowels, so it remains unchanged

Constraints

1 ≤ s.length ≤ 1000
s consists of lowercase and uppercase English letters

Visualization

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Understanding the Visualization

Input

Original string with vowels and consonants

Filter

Identify and remove lowercase vowels a,e,i,o,u

Output

New string containing only consonants and uppercase letters

Key Takeaway

🎯 Key Insight: Use efficient string building to avoid quadratic time complexity from concatenation

Asked in

f Facebook 15 G Google 12

The key insight is to identify vowel characters and filter them out efficiently. Best approach uses StringBuilder or list to avoid expensive string concatenation. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Character by Character	O(n²)	O(n²)	Check each character and build result string by concatenating non-vowels
Optimized - StringBuilder/List Approach	O(n)	O(n)	Use StringBuilder or list to avoid expensive string concatenation

Brute Force - Character by Character — Algorithm Steps

Step 1: Initialize empty result string
Step 2: For each character, check if it's a vowel
Step 3: If not a vowel, concatenate to result
Step 4: Return the final string

Visualization

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Step-by-Step Walkthrough

Check Each Character

Iterate through string and identify vowels

Build Result

Concatenate non-vowels to result string

Return Result

Final string with vowels removed

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';
}

char* solution(char* s) {
    int len = strlen(s);
    char* result = (char*)malloc((len + 1) * sizeof(char));
    int idx = 0;
    
    for (int i = 0; i < len; i++) {
        if (!isVowel(s[i])) {
            result[idx++] = s[i];
        }
    }
    result[idx] = '\0';
    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(n²)

String concatenation in each iteration creates new string, leading to O(n) work per character

⚠ Quadratic Growth

Space Complexity

O(n²)

Multiple intermediate strings created during concatenation process

⚠ Quadratic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Character by Character — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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