Unique Paths II

Description

You are given an m x n integer array grid. There is a robot 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.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 109.

Example 1:

Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]] Output: 2 Explanation: There is one obstacle in the middle of the 3x3 grid above. There are two ways to reach the bottom-right corner:

1. Right → Right → Down → Down
2. Down → Down → Right → Right

Example 2:

Input: obstacleGrid = [[0,1],[0,0]] Output: 1

Constraints:

• m == obstacleGrid.length
• n == obstacleGrid[i].length
• 1 <= m, n <= 100
• obstacleGrid[i][j] is 0 or 1.

Code

Time Complexity: $O(n^2)$, Space Complexity: $O(n^2)$

只是 Unique Paths 加了障礙物，所以 DP 的 state 要排除障礙物，其他都一樣。

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int n = obstacleGrid.size();
        int m = obstacleGrid[0].size();
        vector<vector<int>> DP(n + 1, vector<int>(m + 1, 0));
        DP[0][0] = 1;
        DP[0][1] = 1;
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                if (!obstacleGrid[i - 1][j - 1]) DP[i][j] = DP[i-1][j] + DP[i][j-1];
            }
        }
        return DP[n][m];
    }
};

Source

• Unique Paths II - LeetCode