# Sort Characters By Frequency Problem

## Description

LeetCode Problem 451.

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:

``````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:

``````1
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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:

``````1
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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 * 10^5
• s consists of uppercase and lowercase English letters and digits.

## Sample C++ Code

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class Solution {
public:
static bool comp(const pair<char, int>& a, const pair<char, int>& b) {
return a.second > b.second;
}
string frequencySort(string s) {
map<char, int> freq;
vector<pair<char, int>> freq_sorted;
for (int i = 0; i < s.size(); i ++) {
if (freq.find(s[i]) == freq.end())
freq[s[i]] = 0;
freq[s[i]] ++;
}
for (map<char, int>::iterator mit = freq.begin();
mit != freq.end(); mit ++) {
freq_sorted.push_back({mit->first, mit->second});
}
sort(freq_sorted.begin(), freq_sorted.end(), comp);

string sortedS = "";
for (vector<pair<char, int>>::iterator vit = freq_sorted.begin();
vit != freq_sorted.end(); vit ++) {
for (int i = 0; i < vit->second; i ++)
sortedS += vit->first;
}
return sortedS;
}
};
``````