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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.


class Solution {
    public String frequencySort(String s) {
        Map<Character, Integer> map = new HashMap<>();
        int maxFreq = 1;
        for (char c : s.toCharArray()) {
            map.put(c, map.getOrDefault(c, 0) + 1);
            maxFreq = Math.max(maxFreq, map.get(c));

        List<List<Character>> buckets = new ArrayList<>();
        for (int i = 0; i <= maxFreq; i++) {
            buckets.add(new ArrayList<Character>());
        for (Character key : map.keySet()) {
            int freq = map.get(key);

        StringBuilder sb = new StringBuilder();
        for (int count = buckets.size() - 1; count >= 1; count--) {
            for (Character c : buckets.get(count)) {
                for (int i = 0; i < count; i++) {
        return sb.toString();