# Solution to HackerRank Sherlock string problem

Solution to HackerRank Sherlock string problem
0

As promised at last Wednesdays’ Study Group Session, here is the solution to the Sherlock valid string problem, found at:

`````` my_list = list(s)
my_freq = {}

#build a dictionary
for item in my_list:
if (item in my_freq):
my_freq[item] += 1
else:
my_freq[item] = 1

#convert the dictionary into a list
freq_of_freq = list(my_freq.values())

#convert the list into a set
freq_of_freq_counts = set(freq_of_freq)

#if the set length is 1 the string is valid
if(len(freq_of_freq_counts)==1):
return "YES"
else:
# determine the special case.
# it's valid if there are only 2 counts,
# AND one of them is countOfOtherOne+1:1
my_freq = {}

#build a dictionary
for item in freq_of_freq:
if (item in my_freq):
my_freq[item] += 1
else:
my_freq[item] = 1
#convert the values and keys into lists
theValues = list(my_freq.values())
theKeys = list(my_freq.keys())

#if there are only 2 counts, check further, otherwise it's invalid
if(len(theValues) == 2):
#if you can remove 1 char from 1 index, it's met the special case
if( (theKeys[1] - theKeys[0] <= 1) and (theValues[1] == 1)):
return "YES"
else:
return "NO"
else:
return "NO"
``````