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  1. Prepare
  2. Algorithms
  3. Strings
  4. Sherlock and Anagrams
  5. Discussions

Sherlock and Anagrams

  • coder1033
     + 9 comments

    yes you're right, we don't need to list all combinations to count pairs. using your formula - 

    from collections import Counter
def sherlockAndAnagrams(s):
    count = 0
    for i in range(1,len(s)+1):
        a = ["".join(sorted(s[j:j+i])) for j in range(len(s)-i+1)]
        b = Counter(a)
        for j in b:
            count+=b[j]*(b[j]-1)/2
    return int(count)