Letter tile possibilities

We will generate all possible permutations using backtracking and then count the number of unique permutations.

class Solution:
    def numTilePossibilities(self, tiles: str) -> int:
        seen, res = set(), set()

        def backtrack(seen, res, curr):
            if curr != '' and curr not in res:
                res.add(curr)

            for index in range(len(tiles)):
                if index not in seen:
                    seen.add(index)
                    backtrack(seen, res, curr + tiles[index])
                    seen.remove(index)

        backtrack(seen, res, '')
        return len(res)

Time complexity: O(n!)
Space complexity: O(n)

We can use python's itertools.permutations to generate all possible permutations and then count the number of unique permutations.

class Solution:
    def numTilePossibilities(self, tiles: str) -> int:
        return len(set(itertools.permutations(tiles, len(tiles))))