Solution 1: Two Pointers

We can use two pointers $i$ and $j$ , initially pointing to the start and end of the string respectively.

In each loop, we check whether the character at $i$ is a vowel. If it's not, we move $i$ forward. Similarly, we check whether the character at $j$ is a vowel. If it's not, we move $j$ backward. If $i < j$ at this point, then both characters at $i$ and $j$ are vowels, so we swap these two characters. Then, we move $i$ forward and $j$ backward. We continue the above operations until $i \geq j$ .

The time complexity is $O (n)$ , where $n$ is the length of the string. The space complexity is $O (∣Σ∣)$ , where $Σ$ is the size of the character set.

Python3

class Solution:
    def reverseVowels(self, s: str) -> str:
        vowels = "aeiouAEIOU"
        i, j = 0, len(s) - 1
        cs = list(s)
        while i < j:
            while i < j and cs[i] not in vowels:
                i += 1
            while i < j and cs[j] not in vowels:
                j -= 1
            if i < j:
                cs[i], cs[j] = cs[j], cs[i]
                i, j = i + 1, j - 1
        return "".join(cs)

Java

class Solution {
    public String reverseVowels(String s) {
        boolean[] vowels = new boolean[128];
        for (char c : "aeiouAEIOU".toCharArray()) {
            vowels[c] = true;
        }
        char[] cs = s.toCharArray();
        int i = 0, j = cs.length - 1;
        while (i < j) {
            while (i < j && !vowels[cs[i]]) {
                ++i;
            }
            while (i < j && !vowels[cs[j]]) {
                --j;
            }
            if (i < j) {
                char t = cs[i];
                cs[i] = cs[j];
                cs[j] = t;
                ++i;
                --j;
            }
        }
        return String.valueOf(cs);
    }
}

C++

class Solution {
public:
    string reverseVowels(string s) {
        bool vowels[128];
        memset(vowels, false, sizeof(vowels));
        for (char c : "aeiouAEIOU") {
            vowels[c] = true;
        }
        int i = 0, j = s.size() - 1;
        while (i < j) {
            while (i < j && !vowels[s[i]]) {
                ++i;
            }
            while (i < j && !vowels[s[j]]) {
                --j;
            }
            if (i < j) {
                swap(s[i++], s[j--]);
            }
        }
        return s;
    }
};

Go

func reverseVowels(s string) string {
    vowels := [128]bool{}
    for _, c := range "aeiouAEIOU" {
        vowels[c] = true
    }
    cs := []byte(s)
    i, j := 0, len(cs)-1
    for i < j {
        for i < j && !vowels[cs[i]] {
            i++
        }
        for i < j && !vowels[cs[j]] {
            j--
        }
        if i < j {
            cs[i], cs[j] = cs[j], cs[i]
            i, j = i+1, j-1
        }
    }
    return string(cs)
}

TypeScript

function reverseVowels(s: string): string {
    const vowels = new Set(['a', 'e', 'i', 'o', 'u']);
    const cs = s.split('');
    for (let i = 0, j = cs.length - 1; i < j; ++i, --j) {
        while (i < j && !vowels.has(cs[i].toLowerCase())) {
            ++i;
        }
        while (i < j && !vowels.has(cs[j].toLowerCase())) {
            --j;
        }
        [cs[i], cs[j]] = [cs[j], cs[i]];
    }
    return cs.join('');
}

Rust

impl Solution {
    pub fn reverse_vowels(s: String) -> String {
        let vowel = String::from("aeiouAEIOU");
        let mut data: Vec<char> = s.chars().collect();
        let (mut i, mut j) = (0, s.len() - 1);
        while i < j {
            while i < j && !vowel.contains(data[i]) {
                i += 1;
            }
            while i < j && !vowel.contains(data[j]) {
                j -= 1;
            }
            if i < j {
                data.swap(i, j);
                i += 1;
                j -= 1;
            }
        }
        data.iter().collect()
    }
}

Problem