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.
class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;
if (m == 0) {
return 0;
}
int n = obstacleGrid[0].length;
int[][] dp = new int[m][n];
dp[0][0] = obstacleGrid[0][0] == 1 ? 0 : 1;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i + j == 0) {
continue;
}
if (i == 0) {
dp[i][j] = obstacleGrid[i][j] == 1 ? 0 : dp[i][j - 1];
} else if (j == 0) {
dp[i][j] = obstacleGrid[i][j] == 1 ? 0 : dp[i - 1][j];
} else {
dp[i][j] = obstacleGrid[i][j] == 1 ? 0 : dp[i - 1][j] + dp[i][j - 1];
}
}
}
return dp[m - 1][n - 1];
}
}