2785. Sort Vowels in a String

Problem

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.

Solution (Java)

class Solution {
    public String sortVowels(String s) {
        char ch[]=s.toCharArray();
        StringBuilder sb=new StringBuilder();
        char temp[]=new char[ch.length];
        for(int i=0;i<ch.length;i++){
            if(ch[i]=='a'||ch[i]=='e'||ch[i]=='i'||ch[i]=='o'||ch[i]=='u'){
               sb.append(ch[i]);
            }
            else if(ch[i]=='A'||ch[i]=='E'||ch[i]=='I'||ch[i]=='O'||ch[i]=='U'){
               sb.append(ch[i]);

            }
            else{
                temp[i]=ch[i];
            }
        }

        char c[]=sb.toString().toCharArray();
        Arrays.sort(c);
        int j=0;
        for(int i=0;i<ch.length;i++){
            if(ch[i]=='a'||ch[i]=='e'||ch[i]=='i'||ch[i]=='o'||ch[i]=='u'){
                temp[i]=c[j];
                j++;
            }
            else if(ch[i]=='A'||ch[i]=='E'||ch[i]=='I'||ch[i]=='O'||ch[i]=='U'){
                temp[i]=c[j];
                j++;
            }
        }
        String ans="";
        for(int i=0;i<temp.length;i++){
            ans+=temp[i];
        }
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).