# 451 Sort Characters By Frequency

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.

Return the sorted string. If there are multiple answers, return any of them.

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 * 105`
• `s` consists of uppercase and lowercase English letters and digits.
 `````` 1 2 3 4 5 6 7 8 9 10 `````` ``````class Solution: def frequencySort(self, s: str) -> str: counts = defaultdict(int) for c in s: counts[c] += 1 string_builder = [] for c, n in sorted(counts.items(), key=lambda x: -x[1]): string_builder.append(c * n) return ''.join(string_builder)``````