Leetcode POTD (11 September 2025) – Medium – 2785

2785. Sort Vowels in a String

Given a 0-indexed string s, permute s to get a new string t such that:

• All consonants remain in their original places. More formally, if there is an index i with 0 <= i < s.length such that s[i] is a consonant, then t[i] = s[i].
• The vowels must be sorted in the nondecreasing order of their ASCII values. More formally, for pairs of indices i, j with 0 <= i < j < s.length such that s[i] and s[j] are vowels, then t[i] must not have a higher ASCII value than t[j].

Return the resulting string.

The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in lowercase or uppercase. Consonants comprise all letters that are not vowels.

Example 1:

Input: s = "lEetcOde"
Output: "lEOtcede"
Explanation: 'E', 'O', and 'e' are the vowels in s; 'l', 't', 'c', and 'd' are all consonants. The vowels are sorted according to their ASCII values, and the consonants remain in the same places.

Example 2:

Input: s = "lYmpH"
Output: "lYmpH"
Explanation: There are no vowels in s (all characters in s are consonants), so we return "lYmpH".

Constraints:

• 1 <= s.length <= 105
• s consists only of letters of the English alphabet in uppercase and lowercase.

class Solution {
public:
    bool isVowel(char ch){
        ch = tolower(ch);
        return (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u');
    }
    string sortVowels(string s) {
        unordered_map<char, int> mp;
        for(char &chr:s){
            if(isVowel(chr)){
                mp[chr]++;
            }
        }
        string temp = "AEIOUaeiou";
        int j = 0;
        for(int i=0;i<s.length();i++){
            if(isVowel(s[i])){
                while(mp[temp[j]]== 0){
                    j++;
                }
                s[i] = temp[j];
                mp[temp[j]]--;
            }
        }
        return s;
    }
};

Complexity:

Space → O(n)

Time → O(1)

verdict :- we solved POTD!