##### 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`

.

##### Link

##### Complexity:

time complexity is O(N)

space complexity is O(N)

##### Execution:

This is one of the easier medium problems. Create a character occurrence map. Then create an occurrence-occurrence map. If all the occurrences are the same (size is 1) the string is valid. If there is exactly one character that occurs exactly once, it is also valid. Otherwise invalid

##### Solution:

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

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