leetcode1647. Minimum Deletions to Make Character Frequencies Unique

Posted 2022-02-03 seyjs

题目如下：

A string s is called good if there are no two different characters in s that have the same frequency.

Given a string s, return the minimum number of characters you need to delete to make s good.

The frequency of a character in a string is the number of times it appears in the string. For example, in the string "aab", the frequency of \'a\' is 2, while the frequency of \'b\' is 1. 

Example 1:

Input: s = "aab"
Output: 0
Explanation: s is already good.

Example 2:

Input: s = "aaabbbcc"
Output: 2
Explanation: You can delete two \'b\'s resulting in the good string "aaabcc".
Another way it to delete one \'b\' and one \'c\' resulting in the good string "aaabbc".

Example 3:

Input: s = "ceabaacb"
Output: 2
Explanation: You can delete both \'c\'s resulting in the good string "eabaab".
Note that we only care about characters that are still in the string at the end (i.e. frequency of 0 is ignored).

Constraints:

• 1 <= s.length <= 105
• s contains only lowercase English letters.

解题思路：先统计出每个字符出现的次数，然后从最大的出现次数开始，若不唯一，则将次数减一，再判断是否重复，循环直到不重复为止。

代码如下：

class Solution(object):
    def minDeletions(self, s):
        """
        :type s: str
        :rtype: int
        """
        res = 0
        dic = 
        for i in s:
            dic[i] = dic.setdefault(i,0)+1

        times = sorted(dic.itervalues())

        for val in dic.itervalues():
            if times.count(val) <= 1:continue
            tmp = val
            while tmp >= 0:
                if tmp not in times:break
                tmp -= 1
            times.pop(times.index(val))
            if tmp != 0: times.append(tmp)
            res += (val - tmp)
        return res