##### 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(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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