Problem:

Given a string, sort it in decreasing order based on the frequency of characters.

Example 1:

Input:
“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:
“cccaaa”

Output:
“cccaaa”

Explanation:
Both ‘c’ and ‘a’ appear three times, so “aaaccc” is also a valid answer.
Note that “cacaca” is incorrect, as the same characters must be together.
Example 3:

Input:
“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.

Idea:

Counting sort

Solution 1: 

Sort the string based on element frequency

// Author: Huahua
// Running time: 289 ms
class Solution {
public:
    string frequencySort(string& s) {
        array<int, 128> f{0};
        for(char c:s) ++f[c];
        sort(s.begin(), s.end(), [&f] (char a, char b) {
            return f[a] > f[b] || (f[a] == f[b] && a > b);
        });
        return s;
    }
};

// Author: Huahua
// Running time: 139 ms
class Solution {
public:
    string frequencySort(string& s) {
        map<char, int> f;
        for(char c:s) ++f[c];
        sort(s.begin(), s.end(), [&f] (char a, char b) {
            return f[a] > f[b] || (f[a] == f[b] && a > b);
        });
        return s;
    }
};

// Author: Huahua
// Running time: 29 ms
class Solution {
public:
    string frequencySort(string& s) {
        array<int, 128> f{0};
        for(char c:s) ++f[c];
        sort(s.begin(), s.end(), [&f] (char a, char b) {
            return f[a] > f[b] || (f[a] == f[b] && a > b);
        });        
        return s;
    }
};

Solution 2:

Counting sort

// Author: Huahua
// Running time: 12 ms
class Solution {
public:
    string frequencySort(const string& s) {
        vector<int> f(128, 0);
        for(char c:s) ++f[c];
        vector<pair<int,char>> v;
        for (int i = 0; i < 128; ++i) {
            if (f[i] > 0) v.emplace_back(f[i], i);
        }
        sort(v.rbegin(), v.rend());
        string ans;
        for(const auto& kv : v) {
            ans.append(kv.first, kv.second);
        }
        return ans;
    }
};

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