Given a string, remove the vowels from the string and print the string without vowels.
Examples:
Input : welcome to geeksforgeeks
Output : wlcm t gksfrgks
Input : what is your name ?
Output : wht s yr nm ?
A loop is designed that goes through a list composed of the characters of that string, removes the vowels and then joins them.
Implementation:
C++14
#include <bits/stdc++.h>
using namespace std;
string remVowel(string str)
{
vector< char > vowels = { 'a' , 'e' , 'i' , 'o' , 'u' ,
'A' , 'E' , 'I' , 'O' , 'U' };
for ( int i = 0; i < str.length(); i++)
{
if (find(vowels.begin(), vowels.end(),
str[i]) != vowels.end())
{
str = str.replace(i, 1, "" );
i -= 1;
}
}
return str;
}
int main()
{
string str = "GeeeksforGeeks - A Computer"
" Science Portal for Geeks" ;
cout << remVowel(str) << endl;
return 0;
}
|
Java
import java.util.Scanner;
public class Practice {
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
String s = sc.nextLine();
for ( int i = 0 ; i < s.length(); i++) {
if (s.charAt(i) == 'a' || s.charAt(i) == 'e'
|| s.charAt(i) == 'i' || s.charAt(i) == 'o'
|| s.charAt(i) == 'u' || s.charAt(i) == 'A'
|| s.charAt(i) == 'E' || s.charAt(i) == 'I'
|| s.charAt(i) == 'O'
|| s.charAt(i) == 'U' ) {
continue ;
}
else {
System.out.print(s.charAt(i));
}
}
}
}
|
Python3
def rem_vowel(string):
vowels = [ 'a' , 'e' , 'i' , 'o' , 'u' ]
result = [letter for letter in string if letter.lower() not in vowels]
result = ''.join(result)
print (result)
string = "GeeksforGeeks - A Computer Science Portal for Geeks"
rem_vowel(string)
string = "Loving Python LOL"
rem_vowel(string)
|
C#
using System;
using System.Text;
using System.Linq;
using System.Collections.Generic;
public class Test
{
static String remVowel(String str)
{
char []vowels = { 'a' , 'e' , 'i' , 'o' , 'u' , 'A' , 'E' , 'I' , 'O' , 'U' };
List< char > al = vowels.OfType< char >().ToList();;
StringBuilder sb = new StringBuilder(str);
for ( int i = 0; i < sb.Length; i++) {
if (al.Contains(sb[i])){
sb.Replace(sb[i].ToString(), "" ) ;
i--;
}
}
return sb.ToString();
}
public static void Main()
{
String str = "GeeeksforGeeks - A Computer Science Portal for Geeks" ;
Console.Write(remVowel(str));
}
}
|
Javascript
<script>
function remVowel(str)
{
let al = [ 'a' , 'e' , 'i' , 'o' , 'u' ,
'A' , 'E' , 'I' , 'O' , 'U' ];
let result = "" ;
for (let i = 0; i < str.length; i++)
{
if (!al.includes(str[i]))
{
result += str[i];
}
}
return result;
}
let str = "GeeeksforGeeks - A Computer Science " +
"Portal for Geeks" ;
document.write(remVowel(str));
</script>
|
OutputGksfrGks - Cmptr Scnc Prtl fr Gks
Time Complexity: O(n), where n is the length of the string
Auxiliary Space: O(1)
We can improve the above solution by using Regular Expressions.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
string remVowel(string str)
{
regex r( "[aeiouAEIOU]" );
return regex_replace(str, r, "" );
}
int main()
{
string str = "GeeeksforGeeks - A Computer Science Portal for Geeks" ;
cout << (remVowel(str));
return 0;
}
|
Java
import java.util.Arrays;
import java.util.List;
class GFG
{
static String remVowel(String str)
{
return str.replaceAll( "[aeiouAEIOU]" , "" );
}
public static void main(String[] args)
{
String str = "GeeeksforGeeks - A Computer Science Portal for Geeks" ;
System.out.println(remVowel(str));
}
}
|
Python3
import re
def rem_vowel(string):
return (re.sub( "[aeiouAEIOU]" ,"",string))
string = "GeeksforGeeks - A Computer Science Portal for Geeks"
print (rem_vowel(string))
|
C#
using System;
using System.Text.RegularExpressions;
class GFG
{
static String remVowel(String str)
{
str = Regex.Replace(str, "[aeiouAEIOU]" , "" );
return str;
}
public static void Main()
{
String str = "GeeeksforGeeks - A Computer Science Portal for Geeks" ;
Console.WriteLine(remVowel(str));
}
}
|
Javascript
<script>
function remVowel(str)
{
let r= new RegExp( "[aeiouAEIOU]" , 'g' )
return str.replace(r, "" );
}
let str = "GeeeksforGeeks - A Computer Science Portal for Geeks" ;
document.write(remVowel(str));
</script>
|
OutputGksfrGks - Cmptr Scnc Prtl fr Gks
Time Complexity: O(n), where n is the length of the string
Auxiliary Space: O(1)
This article is contributed by Pramod Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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