---
title: '451. Sort Characters By Frequency'
description: Given a string s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string
sidebar:
  label: 'Sort Characters By Frequency'
  badge: 'Medium'
---

<Badge variant="accent">Hash Table</Badge>

### Example 1:
- Input: `s = "tree"`
- Output: `"eert"`
- Explanation: `'e'` appears twice while `'r'` and `'t'` both appear once. So `'e'` must appear before both `'r'` and `'t'`. Therefore `"eetr"` is also a valid answer.

### Example 2:
- Input: `s = "cccaaa"`
- Output: `"aaaccc"`
- Explanation: Both `'c'` and `'a'` appear three times, so both `"cccaaa"` and `"aaaccc"` are valid answers. Note that `"cacaca"` is incorrect, as the same characters must be together.

### Example 3:
- Input: `s = "Aabb"`
- Output: `"bbAa"`
- Explanation: `"bbaA"` is also a valid answer, but `"Aabb"` is incorrect. Note that `'A'` and `'a'` are treated as two different characters.

### Constraints:

- `1 <= s.length <= 5 * 10^5`
- `s` consists of uppercase and lowercase English letters and digits.

## Solution

```js
/**
 * @param {string} s
 * @return {string}
 */
var frequencySort = function (s) {
  // count the frequency of each character
  const freq = {};
  for (let c of s) freq[c] = (freq[c] || 0) + 1;

  // sort the characters by frequency
  return s
    .split("")
    .sort((a, b) => freq[b] - freq[a] || a.localeCompare(b))
    .join("");
};
```