Most Easy and Simplest solution
\n# Complexity\n- Time complexity: O(N*Log N) corrected \n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(N)\n Add your space complexity her
\n# Complexity\n- Time complexity: O(N*Log N) corrected \n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\nUpvote if you like my approach\n# Code\n```\nclass Solution {\npublic:\n    string sortVowels(string s) {\n        string vowel = "";\n        set<char> st{\'a\',\'e\',\'i\',\'o\',\'u\',\'A\',\'E\',\'I\',\'O\',\'U\'};\n        for(int i = 0; i<s.size(); i++){\n            if(st.find(s[i]) != st.end())\n                vowel += s[i];\n        }\n        sort(vowel.begin(),vowel.end());\n        \n        int j = 0;\n        for(int i = 0; i<s.size(); i++){\n            if(st.find(s[i]) != st.end())\n                s[i] = vowel[j++];\n        }\n        \n        return s;\n    }\n};\n```
['C']