'''
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.

solution:


'''


class Solution:
    def frequencySort(self, s: str) -> str:
        di = {}
        for i in s:
            if i in di.keys():
                di[i] += 1
            else:
                di[i] = 1
        a = sorted(di.values())
        c = ''
        d = []
        for i in a[::-1]:
            listOfKeys = [key for (key, value) in di.items() if value == i]
            for j in listOfKeys:
                if j not in d:
                    d.append(j)
                    c = c + (i * j)
        return c

