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:
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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:
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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:
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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.
classSolution { public: string frequencySort(string s){ map<int, int> m; for (char c : s) { m[c - 'a']++; } priority_queue<vector<int>, vector<vector<int>>, less<vector<int>>> pq; for (auto it : m) { int c = it.first; int f = it.second; pq.push({f, c}); } string res; while (!pq.empty()) { auto top = pq.top(); pq.pop(); char c = top[1] + 'a'; int f = top[0]; for (int i = 0; i < f; i++) { res += c; } } return res; } };