Python 99th percentile


  • 1
    E

    This fails all cases immediately that don't have the same length or don't share the same letters. Follow on by looping through using the .count() approach to check if each has the same number of each type of letters, immediate failure if they don't match counts.

    def isAnagram(self, s, t):
            if len(s) != len(t) or len(list(set(s)^set(t))) != 0:
                return False
            else:
                for l in list(set(s)):
                    if s.count(l) != t.count(l):
                        return False
                        break
                else:
                    return True
    

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