# 题目

Sherlock considers a string to be valid if all characters of the string appear the same number of times. It is also valid if he can remove just character at index in the string, and the remaining characters will occur the same number of times. Given a string , determine if it is valid. If so, return YES, otherwise return NO.

For example, if , it is a valid string because frequencies are . So is because we can remove one and have of each character in the remaining string. If however, the string is not valid as we can only remove occurrence of . That would leave character frequencies of .

Function Description

Complete the isValid function in the editor below. It should return either the string YES or the string NO.

isValid has the following parameter(s):

s: a string
Input Format

A single string .

Constraints

Each character
Output Format

Print YES if string is valid, otherwise, print NO.

Sample Input 0

aabbcd
Sample Output 0

NO
Explanation 0

Given , we would need to remove two characters, both c and d aabb or a and b abcd, to make it valid. We are limited to removing only one character, so is invalid.

Sample Input 1

aabbccddeefghi
Sample Output 1

NO
Explanation 1

Frequency counts for the letters are as follows:

{'a': 2, 'b': 2, 'c': 2, 'd': 2, 'e': 2, 'f': 1, 'g': 1, 'h': 1, 'i': 1}

There are two ways to make the valid string:

Remove characters with a frequency of : .
Remove characters of frequency : .
Neither of these is an option.

Sample Input 2

abcdefghhgfedecba
Sample Output 2

YES
Explanation 2

All characters occur twice except for which occurs times. We can delete one instance of to have a valid string

# 题目解析

hackerrank的题目首先的做阅读理解，搞清楚要满足什么情况，这个题目的核心是删除一个字母，一次。所以这个要理解到位，运用字典，再加上key和values的比对即可。

``````#!/bin/python3

import math
import os
import random
import re
import sys
from collections import Counter

# Complete the isValid function below.
def isValid(s):
s_c=Counter(s)
s_set=Counter(s_c.values())
s_keys=list(s_set.keys())
s_values=list(s_set.values())

if len(s_set)==1:
return 'YES'
if len(s_set)==2:
if (((s_keys[1]-s_keys[0])<=1) and (s_values[1]==1)):
return 'YES'
else:
return 'NO'
else:
return 'NO'
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')

s = input()

result = isValid(s)

fptr.write(result + '\n')

fptr.close()

``````