Given a string, your task is to reverse only the vowels of string.
Examples:
Input : hello
Output : holle
Input : hello world
Output : hollo werld
One simple solution is to store all the vowels while scanning the string and placing the vowels in the reverse order in another iteration of string.
Implementation:
C++
#include<bits/stdc++.h>
using namespace std;
bool isVowel(char c)
{
return (c=='a' || c=='A' || c=='e' ||
c=='E' || c=='i' || c=='I' ||
c=='o' || c=='O' || c=='u' ||
c=='U');
}
string reverseVowel(string str)
{
int j=0;
string vowel;
for (int i=0; str[i]!='\0'; i++)
if (isVowel(str[i]))
vowel[j++] = str[i];
for (int i=0; str[i]!='\0'; i++)
if (isVowel(str[i]))
str[i] = vowel[--j] ;
return str;
}
int main()
{
string str = "hello world";
cout << reverseVowel(str);
return 0;
}
|
Java
class GFG {
static boolean isVowel(char c) {
return (c == 'a' || c == 'A' || c == 'e'
|| c == 'E' || c == 'i' || c == 'I'
|| c == 'o' || c == 'O' || c == 'u'
|| c == 'U');
}
static String reverseVowel(String str1) {
int j = 0;
char[] str = str1.toCharArray();
String vowel = "";
for (int i = 0; i < str.length; i++) {
if (isVowel(str[i])) {
j++;
vowel += str[i];
}
}
for (int i = 0; i < str.length; i++) {
if (isVowel(str[i])) {
str[i] = vowel.charAt(--j);
}
}
return String.valueOf(str);
}
public static void main(String[] args) {
String str = "hello world";
System.out.println(reverseVowel(str));
}
}
|
Python3
def isVowel(c):
if (c == 'a' or c == 'A' or c == 'e' or
c == 'E' or c == 'i' or c == 'I' or
c == 'o' or c == 'O' or c == 'u' or c == 'U'):
return True
return False
def reverserVowel(string):
j = 0
vowel = [0] * len(string)
string = list(string)
for i in range(len(string)):
if isVowel(string[i]):
vowel[j] = string[i]
j += 1
for i in range(len(string)):
if isVowel(string[i]):
j -= 1
string[i] = vowel[j]
return ''.join(string)
if __name__ == "__main__":
string = "hello world"
print(reverserVowel(string))
|
C#
using System;
class GFG
{
static bool isVowel(char c)
{
return (c == 'a' || c == 'A' || c == 'e'
|| c == 'E' || c == 'i' || c == 'I'
|| c == 'o' || c == 'O' || c == 'u'
|| c == 'U');
}
static String reverseVowel(String str1)
{
int j = 0;
char[] str = str1.ToCharArray();
String vowel = "";
for (int i = 0; i < str.Length; i++)
{
if (isVowel(str[i]))
{
j++;
vowel += str[i];
}
}
for (int i = 0; i < str.Length; i++)
{
if (isVowel(str[i]))
{
str[i] = vowel[--j];
}
}
return String.Join("",str);
}
public static void Main(String[] args)
{
String str = "hello world";
Console.WriteLine(reverseVowel(str));
}
}
|
Javascript
<script>
function isVowel(c) {
return (c == 'a' || c == 'A' || c == 'e'
|| c == 'E' || c == 'i' || c == 'I'
|| c == 'o' || c == 'O' || c == 'u'
|| c == 'U');
}
function reverseVowel(str1) {
let j = 0;
let str = str1.split('');
let vowel = "";
for (let i = 0; i < str.length; i++) {
if (isVowel(str[i])) {
j++;
vowel += str[i];
}
}
for (let i = 0; i < str.length; i++) {
if (isVowel(str[i])) {
str[i] = vowel[--j];
}
}
return str.join("");
}
let str = "hello world";
document.write(reverseVowel(str));
</script>
|
Time complexity: O(n) where n = length of string
Auxiliary Space: O(v) where v = number of vowels in string
A better solution is to use two pointers scanning from beginning and end of the array respectively and manipulate vowels pointed by these pointers.
Implementation:
C++
#include<bits/stdc++.h>
using namespace std;
bool isVowel(char c)
{
return (c=='a' || c=='A' || c=='e' ||
c=='E' || c=='i' || c=='I' ||
c=='o' || c=='O' || c=='u' ||
c=='U');
}
string reverseVowel(string str)
{
int i = 0;
int j = str.length()-1;
while (i < j)
{
if (!isVowel(str[i]))
{
i++;
continue;
}
if (!isVowel(str[j]))
{
j--;
continue;
}
swap(str[i], str[j]);
i++;
j--;
}
return str;
}
int main()
{
string str = "hello world";
cout << reverseVowel(str);
return 0;
}
|
Java
class GFG {
static boolean isVowel(char c) {
return (c == 'a' || c == 'A' || c == 'e'
|| c == 'E' || c == 'i' || c == 'I'
|| c == 'o' || c == 'O' || c == 'u'
|| c == 'U');
}
static String reverseVowel(String str) {
int i = 0;
int j = str.length()-1;
char[] str1 = str.toCharArray();
while (i < j)
{
if (!isVowel(str1[i]))
{
i++;
continue;
}
if (!isVowel(str1[j]))
{
j--;
continue;
}
char t = str1[i];
str1[i]= str1[j];
str1[j]= t;
i++;
j--;
}
String str2 = String.copyValueOf(str1);
return str2;
}
public static void main(String[] args) {
String str = "hello world";
System.out.println(reverseVowel(str));
}
}
|
Python3
def isVowel(c):
return (c=='a' or c=='A' or c=='e' or
c=='E' or c=='i' or c=='I' or
c=='o' or c=='O' or c=='u' or
c=='U')
def reverseVowel(str):
i = 0
j = len(str) - 1
while (i < j):
if not isVowel(str[i]):
i += 1
continue
if (not isVowel(str[j])):
j -= 1
continue
str[i], str[j] = str[j], str[i]
i += 1
j -= 1
return str
if __name__ == "__main__":
str = "hello world"
print(*reverseVowel(list(str)), sep = "")
|
C#
using System;
class GFG
{
static Boolean isVowel(char c)
{
return (c == 'a' || c == 'A' || c == 'e'
|| c == 'E' || c == 'i' || c == 'I'
|| c == 'o' || c == 'O' || c == 'u'
|| c == 'U');
}
static String reverseVowel(String str)
{
int i = 0;
int j = str.Length-1;
char[] str1 = str.ToCharArray();
while (i < j)
{
if (!isVowel(str1[i]))
{
i++;
continue;
}
if (!isVowel(str1[j]))
{
j--;
continue;
}
char t = str1[i];
str1[i]= str1[j];
str1[j]= t;
i++;
j--;
}
String str2 = String.Join("",str1);
return str2;
}
public static void Main(String[] args)
{
String str = "hello world";
Console.WriteLine(reverseVowel(str));
}
}
|
Javascript
<script>
function isVowel(c)
{
return (c == 'a' || c == 'A' || c == 'e'
|| c == 'E' || c == 'i' || c == 'I'
|| c == 'o' || c == 'O' || c == 'u'
|| c == 'U');
}
function reverseVowel(str)
{
let i = 0;
let j = str.length-1;
let str1 = str.split("");
while (i < j)
{
if (!isVowel(str1[i]))
{
i++;
continue;
}
if (!isVowel(str1[j]))
{
j--;
continue;
}
let t = str1[i];
str1[i]= str1[j];
str1[j]= t;
i++;
j--;
}
let str2 = (str1).join("");
return str2;
}
let str = "hello world";
document.write(reverseVowel(str));
</script>
|
Time complexity : O(n) where n = length of string
Auxiliary Space : O(1)
This article is contributed by Gaurav Ahirwar and Gaurav Miglani. 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.