An O(n) solution using stack, less complexity than two pointers.

class Solution {

  isVowel(s) {

    return ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'].includes(s);

  }



  reverseVowels(s) {

    const stack = [];



    for (const c of s) {

      if (this.isVowel(c)) {

        stack.push(c);

      }

    }



    let newS = '';

    for (let i = 0; i < s.length; i++) {

      const c = s.at(i);

      if (this.isVowel(c)) {

        const pop = stack.pop();

        newS += pop

      } else {

        newS += c

      }

    }





    return newS;

  }

}