# HackerRank ‘Sherlock and Valid String’ Solution

##### Short Problem Definition:

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 1 character at 1 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.

Sherlock and Valid String

##### Complexity:

time complexity is O(N)

space complexity is O(1)

##### Execution:

I optimized this solution to the minimal case that passes all tests on HackerRank. It seems that each character occurs 1 or 2 times. I did not pay the Hackos to verify the input :). The logic of the solution is as follows: count the character counts for each character.

• if they are all equal – it means that all characters occur exactly N times and there is no removal needed
• if 2 or more have less or more characters – there is no way to fix the string in just 1 removal
• if exactly 1 char has a different count than all other characters – remove this char completely and S is fixed.
##### Solution:
from collections import Counter

def isValid(S):
char_map = Counter(S)
char_occurence_map = Counter(char_map.values())

if len(char_occurence_map) == 1:
return True

if len(char_occurence_map) == 2:
for v in char_occurence_map.values():
if v == 1:
return True

return False

S = raw_input()
if isValid(S):
print "YES"
else:
print "NO"


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