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
time complexity is O(N)
space complexity is O(1)
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.
EDIT 2020: HR added some extra test cases and the solution no longer worked. I updated it to pass all the cases including a secret “aabbbccc” case that HR is not testing for.
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: k1, k2 = char_occurence_map.keys() v1, v2 = char_occurence_map.values() # there is exactly 1 extra symbol and it can be deleted if (k1 == 1 and v1 == 1) or (k2 == 1 and v2 == 1): return True # the is exactly 1 symbol that occurs an extra 1 time if (k1 == k2+1 and v1 == 1) or (k2 == k1+1 and v2 == 1): return True return False
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