题解 | #删除字符串中出现次数最少的字符#

import sys

# aabcddd
oldStr = sys.stdin.readline()
strLength = len(oldStr)
oldStrList = list(oldStr)[0 : strLength - 1]
minCount = 20
n = strLength - 1
dict1 = {}

for i in range(n):
    tempCount = 0
    for j in range(n):

        if oldStrList[j] == oldStrList[i]:
            tempCount += 1
    # 字典中，键是唯一且不可重复的
    # updata()函数只能增加键值不一样的
    # 但字母是不一样的，可做键
    dict1.update({oldStrList[i]: tempCount})

minCount = min(dict1.values())
allKeys = list(dict1.keys())

for word in allKeys:
	#通过if判断，找到最小的次数对应的字母
    if dict1[word] == minCount:
	#需要根据次数删除所有的符合条件的字母
        times = dict1[word]
        while int(times) > 0:
            oldStrList.remove(word)
            times -= 1

print("".join(oldStrList))