LeetCode Python/Java/C++/JS  >  String  >  345. Reverse Vowels of a String  >  Solved in Python, Java, C++, JavaScript, Ruby, C#, Go  >  GitHub or Repost

    LeetCode link: 345. Reverse Vowels of a String, difficulty: Easy.

    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
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    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:

    int r = target.length() - 1;

for (int l = 0; l < target.length(); l++) { // Important
    // ...
    if (l >= r) {
        break;
    }

    // ...
    r--;
}

    Step by Step Solutions

    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

    Time complexity

    O(N)

    Space complexity

    O(1) or O(N)

    Explanation

    Space complexity is O(N) if converting to a character array

    Solution in languages: Python Java C++ JavaScript Ruby C# Go

    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 # 

    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++ # 

    class Solution {
public:
    string reverseVowels(string s) {
        unordered_set<char> 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 # 

    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 # 

    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# # 

    public class Solution
{
    public string ReverseVowels(string s)
    {
        var vowels = new HashSet<char> {'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 # 

    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)
}
    Solution in languages: Python Java C++ JavaScript Ruby C# Go

    Other languages

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