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"