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

Sherlock and Valid String

Complexity:

time complexity is O(N)

space complexity is O(1)

Execution:

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.

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:
        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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