Question

一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为 “Start” ）。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为 “Finish” ）。

问总共有多少条不同的路径？

输入：m = 3, n = 2
输出：3

-> m
[0][ ][ ]
[ ][ ][x]

Solution

a. DFS

My first thought was DFS to solve this problem.

• Every step has two directions: x or y.
• To reach the destination, the robot's x should be less than m and y should be less than n.
• When x is equal to m and y is equal to n, the robot reaches the destination.

Here is the code:

def uniquePaths(m: int, n: int) -> int:    x, y = 1, 1    ans = [0]    def dfs(x, y, m, n):        if x == m and y == n:            ans[0] += 1            return 0        if x < m:            dfs(x+1, y, m, n)        if y < n:            dfs(x, y+1, m, n)    dfs(x, y, m, n)    return ans[0]

Not surprisingly, this method will time out. We need another algorithm to scale down the Time complexity.

b. DP (Dynamic Programming)

DP is Divide-and-Conque (D&C) with Memorization

Take a closer look to is problem, I found that divide-and-conque (D&C) algorithm is really suitable. The problem f(m, n) could be divided into two parts: f(m, n-1) and f(m-1, n)

Additional, there is only 1 way if m is equal to 1 or n is equal to 1.

So we have the following Recurrence:

f(m,n) = 1, if m = 1 or n = 1

f(m,n) = f(m-1, n) + f(m, n-1)

Implementing the algorithm with python3.

def uniquePaths(m: int, n: int) -> int:    dp = [[1]*n] + [[1]+[0]*(n-1) for _ in range(m-1)]  # init lookup table    for i in range(1, m):        for j in range(1, n):            dp[i][j] = dp[i-1][j] + dp[i][j-1]    return dp[m-1][n-1]