message: Added solution logic for Sherlock and valid strings
diff: +# Logic
  +# Create a frequency list for all alphabets,
  +# next compare the frequency of all alphabets from b to z with
  +# frquency of a, iteratively. If the frequencies don't match
  +# subtract the a's frequency from the current letter's frequency.
  +# Compare the frequency of the letter with a's frequency again, if they
  +# don't yet match increment diff_freqs. At the end of loop, if there's more
  +# than one letter with different frequnecy than a's, diff_freqs will
  +# be more than 1 and the string will be invalid!
  +
  +
  def isValid(s):
  freq = [0] * 26
  - # create frequency list of all letters
  for i in s:
  freq[ord(i) - ord('a')] += 1
  diff_freqs = 0
  