Remove Vowels from a String - Practice Coding Problems

Remove Vowels from a String - Problem

String Easy

Given a string s, your task is to remove all vowels from it and return the resulting string. The vowels to remove are the five standard English vowels: 'a', 'e', 'i', 'o', and 'u' (both lowercase and uppercase).

This is a fundamental string manipulation problem that tests your understanding of character filtering and string building techniques. It's commonly used as a warm-up question in coding interviews to assess basic programming skills.

Example:

Input: "hello world"
Output: "hll wrld"
Explanation: We remove 'e', 'o', 'o' from the original string

Input & Output

example_1.py — Basic String

$ Input: s = "hello"

› Output: "hll"

💡 Note: We remove the vowels 'e' and 'o' from "hello", leaving us with "hll"

example_2.py — Mixed Case

$ Input: s = "AEIOU"

› Output: ""

💡 Note: All characters are vowels (uppercase), so the result is an empty string

example_3.py — No Vowels

$ Input: s = "bcdfg"

› Output: "bcdfg"

💡 Note: No vowels present in the input, so the string remains unchanged

Constraints

0 ≤ s.length ≤ 1000
s consists of only lowercase and uppercase English letters
Vowels are case-sensitive: both 'a' and 'A' should be removed

Visualization

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Understanding the Visualization

Create the Filter

Set up a hash set with all vowels for instant recognition

Process Each Character

Check each character against our filter in constant time

Build Result

Collect all non-vowel characters efficiently

Return Clean String

Return the filtered string with all vowels removed

Key Takeaway

🎯 Key Insight: Using a hash set for vowel lookup transforms multiple comparisons per character into a single O(1) operation, making the algorithm both faster and more maintainable.

Asked in

G Google 25 a Amazon 18 f Meta 12 ⊞ Microsoft 8

The optimal approach uses a hash set to store vowels for O(1) lookup time. We iterate through the string once, checking each character against the vowel set and building our result. This achieves O(n) time complexity with clean, readable code that's easily extensible for different vowel sets.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Character-by-Character Check)	O(n)	O(n)	Check each character against every vowel individually
One-Pass with Set (Optimal)	O(n)	O(n)	Use a set for O(1) vowel lookup and filter in one pass

Brute Force (Character-by-Character Check) — Algorithm Steps

Initialize an empty result string
Iterate through each character in the input string
For each character, check if it equals any of the 10 vowels
If it's not a vowel, append it to the result
Return the result string

Visualization

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Step-by-Step Walkthrough

Start with input

Begin with the input string 'hello'

Check 'h'

Compare 'h' against all 10 vowels - not a match, keep it

Check 'e'

Compare 'e' against vowels - matches 'e', skip it

Continue process

Repeat for remaining characters

Code -

solution.c — C

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

char* removeVowels(char* s) {
    int len = strlen(s);
    char* result = malloc((len + 1) * sizeof(char));
    int j = 0;
    
    for (int i = 0; i < len; i++) {
        char c = s[i];
        if (c != 'a' && c != 'e' && c != 'i' && c != 'o' && c != 'u' &&
            c != 'A' && c != 'E' && c != 'I' && c != 'O' && c != 'U') {
            result[j++] = c;
        }
    }
    result[j] = '\0';
    return result;
}

int main() {
    char s[1000];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0; // Remove newline
    if (s[0] == '"') memmove(s, s+1, strlen(s)); // Remove quotes
    if (s[strlen(s)-1] == '"') s[strlen(s)-1] = '\0';
    printf("%s\n", removeVowels(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We visit each character once, but each character requires up to 10 comparisons with vowels

✓ Linear Growth

Space Complexity

O(n)

Space needed for the result string in worst case (no vowels to remove)

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Character-by-Character Check) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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