# 345. Reverse Vowels of a String - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions
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Visit original link: [345. Reverse Vowels of a String - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.blog/en/leetcode/345-reverse-vowels-of-a-string) for a better experience!
LeetCode link: [345. Reverse Vowels of a String](https://leetcode.com/problems/reverse-vowels-of-a-string), difficulty: **Easy**.
## LeetCode description of "345. Reverse Vowels of a String"
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.
### [Example 1]
**Input**: `s = "IceCreAm"`
**Output**: `"AceCreIm"`
**Explanation**:
The vowels in s are ['I', 'e', 'e', 'A']. On reversing the vowels, s becomes "AceCreIm".
### [Example 2]
**Input**: `"leetcode"`
**Output**: `"leotcede"`
### [Constraints]
- `1 <= s.length <= 3 * 10^5`
- `s` consist of **printable ASCII** characters.
## Intuition
Use two pointers to swap vowels in the string. The left pointer finds vowels from the start, the right pointer finds vowels from the end, and they swap until the pointers meet.
## Pattern of "left & right pointers"
The code framework is as follows:
```java
int r = target.length() - 1;
for (int l = 0; l < target.length(); l++) { // Important
// ...
if (l >= r) {
break;
}
// ...
r--;
}
```
## Step-by-Step Solution
1. Initialize a set of vowels
2. For some languages, convert the string to a character array (since strings are immutable)
3. Left pointer `l` starts at `0`, right pointer `r` starts at the end
4. Loop processing:
- Move `l` right until a vowel is found
- Move `r` left until a vowel is found
- Swap the two vowels when `l < r`
5. Return the processed string
## Complexity
> Space complexity is O(N) if converting to a character array
- Time complexity: `O(N)`.
- Space complexity: `O(1) or O(N)`.
## Python
```python
class Solution:
def reverseVowels(self, s: str) -> str:
vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
s = list(s)
r = len(s) - 1
for l in range(len(s)): # Important
if s[l] not in vowels:
continue
while r > 0 and s[r] not in vowels:
r -= 1
if l >= r:
break
s[l], s[r] = s[r], s[l]
r -= 1
return ''.join(s)
```
## Java
```java
class Solution {
public String reverseVowels(String s) {
var vowels = new HashSet<>(Arrays.asList('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'));
var chars = s.toCharArray();
int r = chars.length - 1;
for (int l = 0; l < chars.length; l++) { // important
if (!vowels.contains(chars[l])) {
continue;
}
while (r > 0 && !vowels.contains(chars[r])) {
r--;
}
if (l >= r) {
break;
}
// Swap the vowels
char temp = chars[l];
chars[l] = chars[r];
chars[r] = temp;
r--;
}
return new String(chars);
}
}
```
## C++
```cpp
class Solution {
public:
string reverseVowels(string s) {
unordered_set vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};
int r = s.size() - 1;
for (int l = 0; l < s.size(); l++) { // important
if (!vowels.contains(s[l])) {
continue;
}
while (r > 0 && !vowels.contains(s[r])) {
r--;
}
if (l >= r) {
break;
}
swap(s[l], s[r]);
r--;
}
return s;
}
};
```
## JavaScript
```javascript
var reverseVowels = function (s) {
const vowels = new Set(['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'])
const chars = s.split('')
let r = chars.length - 1
for (let l = 0; l < chars.length; l++) { // important
if (!vowels.has(chars[l])) {
continue
}
while (r > 0 && !vowels.has(chars[r])) {
r--
}
if (l >= r) {
break
}
[chars[l], chars[r]] = [chars[r], chars[l]]
r--
}
return chars.join('')
};
```
## Ruby
```ruby
def reverse_vowels(s)
vowels = Set.new(['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'])
r = s.size - 1
(0...s.size).each do |l|
unless vowels.include?(s[l])
next
end
while r > 0 && !vowels.include?(s[r])
r -= 1
end
if l >= r
break
end
s[l], s[r] = s[r], s[l]
r -= 1
end
s
end
```
## C#
```csharp
public class Solution
{
public string ReverseVowels(string s)
{
var vowels = new HashSet {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};
var chars = s.ToCharArray();
int r = chars.Length - 1;
// important
for (int l = 0; l < chars.Length; l++)
{
if (!vowels.Contains(chars[l]))
{
continue;
}
while (r > 0 && !vowels.Contains(chars[r]))
{
r--;
}
if (l >= r)
{
break;
}
(chars[l], chars[r]) = (chars[r], chars[l]);
r--;
}
return new string(chars);
}
}
```
## Go
```go
func reverseVowels(s string) string {
vowels := map[byte]bool{
'a': true, 'e': true, 'i': true, 'o': true, 'u': true,
'A': true, 'E': true, 'I': true, 'O': true, 'U': true,
}
chars := []byte(s)
r := len(chars) - 1
for l := 0; l < len(chars); l++ { // important
if !vowels[chars[l]] {
continue
}
for r > 0 && !vowels[chars[r]] {
r--
}
if l >= r {
break
}
chars[l], chars[r] = chars[r], chars[l]
r--
}
return string(chars)
}
```
## Other languages
```java
// Welcome to create a PR to complete the code of this language, thanks!
```
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---
Visit original link: [345. Reverse Vowels of a String - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.blog/en/leetcode/345-reverse-vowels-of-a-string) for a better experience!
GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).