Algorithm Puzzles: Unique Paths

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Algorithm Puzzles everyday every week sometimes: Unique Paths

Puzzle

Puzzle from leetcode:

There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The test cases are generated so that the answer will be less than or equal to 2 * 10^9.

Solution

it’s easy to resolve via dfs, but without lru_cache it might exceed time limit during check…

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from functools import lru_cache


class Solution:
    @lru_cache()
    def uniquePaths(self, m: int, n: int) -> int:
        ret = 0
        if m == 1 and n == 1:
            ret += 1
            return ret

        if n != 1:
            # move right
            ret += self.uniquePaths(m, n - 1)
        if m != 1:
            # move down
            ret += self.uniquePaths(m - 1, n)

        return ret

T.C. should be O(n^n)