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.
class Solution:
     def frequencySort(self, s):
         """
         :type s: str
         :rtype: str
         """
         charToFreq = {}
         freqToChar = {}
         for c in s:
             if c not in charToFreq:
                 charToFreq[c] = 0
             charToFreq[c] += 1
         print(charToFreq)
         for key, value in list(charToFreq.items()):
             if value not in freqToChar:
                 freqToChar[value] = []
             freqToChar[value].append(key)
         print(freqToChar)
         result = []
         for key in sorted(freqToChar, reverse = True):
             for char in freqToChar[key]:
                 result += [char] * key
         return "".join(result)

class Solution:
     def frequencySort(self, s):
         """
         :type s: str
         :rtype: str
         """
         charToFreq = {}
         freqToChar = {}
         for c in s:
             if c not in charToFreq:
                 charToFreq[c] = 0
             charToFreq[c] += 1
         print(charToFreq)
         for key, value in list(charToFreq.items()):
             if value not in freqToChar:
                 freqToChar[value] = []
             freqToChar[value].append(key)
         print(freqToChar)
         result = []
         for key in sorted(freqToChar, reverse = True):
             for char in freqToChar[key]:
                 result += [char] * key
         return "".join(result)
