There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid). 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 * 109.
Input: m = 3, n = 7 Output: 28
Input: m = 3, n = 2 Output: 3 Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner: 1. Right -> Down -> Down 2. Down -> Down -> Right 3. Down -> Right -> Down
- 1 <= m, n <= 100
I solved this problem like this.
If we use dp, we can solve this problem.
class Solution: def uniquePaths(self, m: int, n: int) -> int: dp = [[0 for x in range(n)] for _ in range(m)] dp = 1 for r in range(m): for c in range(n): if r == 0 and c == 0: continue up = dp[r-1][c] if r - 1 >= 0 else 0 left = dp[r][c-1] if c - 1 >= 0 else 0 dp[r][c] = up + left return dp[m-1][n-1]
The way to corner is always same right and down count. So we can get answer combination.
class Solution: def uniquePaths(self, m: int, n: int) -> int: # nCk k = n - 1 n = m + n - 2 ans = 1 k = min(k, n - k) for x in range(n, n-k, -1): ans *= x for x in range(1, k+1): ans /= x return int(ans)