0451.Sort_Characters_By_Frequency.py

"""
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:

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:

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:

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.
"""

from collections import Counter
class Solution:
    def frequencySort(self, s: str) -> str:
        return ''.join(k * v for k, v in sorted(Counter(s).items(), key=lambda x:x[1], reverse = True))  
    
    #sorted not change orignial array. sort change array
    #sort the Counter base on frequency

    
class Solution:
    def frequencySort(self, s: str) -> str:
        
        ch_cnt = collections.Counter(s)
        max_freq = max(ch_cnt.values())
        
        buckets = [[] for _ in range(max_freq + 1)]
        for ch, freq in ch_cnt.items():
            buckets[freq].append(ch)
            
        string_builder = []
        for i in range(len(buckets) - 1, 0, -1):
            for ch in buckets[i]:
                string_builder.append(ch * i)
        return "".join(string_builder)