## Problem

https://leetcode.com/problems/sort-vowels-in-a-string/description/

<br>

:::spoiler **Optimal Space & Time Complexity**

- Time complexity: O(nlogn) where n is string.length because of sorting

- Space complexity: O(n) where n is string.length

:::

<br>

<hr/>

## Solutions

:::spoiler 東

:::

<br>

:::spoiler Hao

:::

<br>

:::spoiler YC

:::

<br>

:::spoiler SOL

:::

---

## Supplement / Discussion

### 東

### Hao

[Explaination](https://leetcode.com/problems/sort-vowels-in-a-string/solutions/3803223/video-sorting-vowels-in-a-python-string-a-leetcode-challenge-solved/)

Basically there are several steps:

1. Extracting the vowels
- Checking if a char is vowel
- **string.prototype.indexOf()**: 'aeiouAEIOU'.indexOf(char) !== -1
- VOWELS.includes(char): besides declare an array, `new Set(['A', 'E', ...])` could also be employed

2. Sorting vowels
- string.prototype.charCodeAt(0)

3. Replacing the original vowels with sorted one
- Notice that you could check the index of vowel via
- checking if it is a vowel again

    s.split("")
      .map((char) => {
        if (VOWELS.includes(char)) return sortedVowels.pop();
        else return char;
      })
  .join("");

- use the recorded vowel indexs to iterate

    const vowelIndexs = [...];

    const newChars = s.split("");


    for (let i = 0; i < vowelIndexs.length; i += 1) {
      newChars[vowelIndexs[i]] = sortedVowels.pop();
    }

    return newChars.join("");

- DON'T: iterate recorded vowel indexs again when iterating the string which leads to time complexity as O(n^2)

    s.split('').map((char, i) => {
      if (vowelIdxs.includes(i)) return descSortedVowels.pop();
      else return char;
    }).join('');

### YC

### SOL