2785. Sort Vowels in a String.cpp

2785. Sort Vowels in a String
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Medium
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Hint
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:
    // Returns true if the character is a vowel.
    bool isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'o'|| c == 'u'|| c == 'i' 
            || c == 'A' || c == 'E' || c == 'O'|| c == 'U'|| c == 'I';
    }
    
    string sortVowels(string s) {
        unordered_map<char, int> count;

        // Store the frequencies for each character.
        for (char c : s) {
            if (isVowel(c)) {
                count[c]++;
            }
        }

        // Sorted string having all the vowels.
        string sortedVowel = "AEIOUaeiou";
        string ans;
        int j = 0;
        for (int i = 0; i < s.size(); i++) {
            if (!isVowel(s[i])) {
                ans += s[i];
            } else {
                // Skip to the character which is having remaining count.
                while (count[sortedVowel[j]] == 0) {
                    j++;
                }

                ans += sortedVowel[j];
                count[sortedVowel[j]]--;
            }
        }
        return ans;
    }
};