2-line Python solution, AC with 350ms (some useful Python tricks)
\n\n        def anagrams(self, strs):\n            count = collections.Counter([tuple(sorted(s)) for s in strs])\n            return filter(lambda x: count[tupl
\n\n        def anagrams(self, strs):\n            count = collections.Counter([tuple(sorted(s)) for s in strs])\n            return filter(lambda x: count[tuple(sorted(x))]>1, strs)\n\n\n - collections.Counter creates a counter object. A counter object is like a specific kind of dictionary where it is build for counting  (objects that hashes to same value)\n - tuple(sorted(s)) is used here so that anagrams will be hashed to the same value. tuple is used because sorted returns a list which cannot be hashed but tuples can be hashed\n - filter: selects some elements of the list based on given function (first argument - a lambda function is given here)\n - lambda function defined here returns True if number of anagrams of that elements is greater than 1
['Python']