[Mock] Google 3

1079. Letter Tile Possibilities

You have n tiles, where each tile has one letter tiles[i] printed on it.

Return the number of possible non-empty sequences of letters you can make using the letters printed on those tiles.

My Answer 1: Accepted (Runtime: 116 ms - 43.08% / Memory Usage: 23.6 MB - 12.79%)

class Solution:
    def numTilePossibilities(self, tiles: str) -> int:
        def backtrack(path, tiles):
            seqlist.add(path)
            
            if not tiles:
                return path
            
            for i in range(len(tiles)):
                backtrack(path + tiles[i], tiles[:i]+tiles[i+1:])
        
        seqlist = set()
        backtrack("", tiles)
        
        return len(seqlist)-1

전에 backtrack 으로 조합찾던게 생각나서 그걸로 풀었읍니다

근데 좀 엉성..

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1042. Flower Planting With No Adjacent

You have n gardens, labeled from 1 to n, and an array paths where paths[i] = [xi, yi] describes a bidirectional path between garden xi to garden yi. In each garden, you want to plant one of 4 types of flowers.

All gardens have at most 3 paths coming into or leaving it.

Your task is to choose a flower type for each garden such that, for any two gardens connected by a path, they have different types of flowers.

Return any such a choice as an array answer, where answer[i] is the type of flower planted in the (i+1)th garden. The flower types are denoted 1, 2, 3, or 4. It is guaranteed an answer exists.

My Answer 1: Wrong Answer (3 / 51 test cases passed.)

class Solution:
    def gardenNoAdj(self, n: int, paths: List[List[int]]) -> List[int]:
        result = []
        pathmap = {}
        
        for i in range(len(paths)):
            if paths[i][0] not in pathmap:
                pathmap[paths[i][0]] = [paths[i][1]]
            else:
                pathmap[paths[i][0]].append(paths[i][1])
        
        if n == 0:
            return []
        
        start = paths[0][0]
        result.append(start)
        
        for i in range(n-1):
            if result[-1] in pathmap and pathmap[result[-1]]:
                result.append(pathmap[result[-1]][0])
            else:
                result.append(start)
        
        return result

문제 이해를 잘못해서 망~

이거 뭔가 풀 수 있을 거 같은데.. 안되네요