# LeetCode – Unique Paths II (Java)

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, there is one obstacle in the middle of a 3×3 grid as illustrated below,

```[
[0,0,0],
[0,1,0],
[0,0,0]
]
```

the total number of unique paths is 2.

Java Solution

```public int uniquePathsWithObstacles(int[][] obstacleGrid) { if(obstacleGrid==null||obstacleGrid.length==0) return 0;   int m = obstacleGrid.length; int n = obstacleGrid[0].length;   if(obstacleGrid[0][0]==1||obstacleGrid[m-1][n-1]==1) return 0;     int[][] dp = new int[m][n]; dp[0][0]=1;   //left column for(int i=1; i<m; i++){ if(obstacleGrid[i][0]==1){ dp[i][0] = 0; }else{ dp[i][0] = dp[i-1][0]; } }   //top row for(int i=1; i<n; i++){ if(obstacleGrid[0][i]==1){ dp[0][i] = 0; }else{ dp[0][i] = dp[0][i-1]; } }   //fill up cells inside for(int i=1; i<m; i++){ for(int j=1; j<n; j++){ if(obstacleGrid[i][j]==1){ dp[i][j]=0; }else{ dp[i][j]=dp[i-1][j]+dp[i][j-1]; }   } }   return dp[m-1][n-1]; }```

### 1 thought on “LeetCode – Unique Paths II (Java)”

1. Above mentioned solution will not work if the top-left cell had an obstacle.

dp[0][0] = (obstacleGrid[0][0] == 1)? 0 : 1;

``` public class Solution { public int uniquePathsWithObstacles(int[][] obstacleGrid) { if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0){ return -1; } int rows = obstacleGrid.length; int cols = obstacleGrid[0].length; int[][] dp = new int[rows][cols]; dp[0][0] = (obstacleGrid[0][0] == 1)? 0 : 1; for(int i = 1; i < cols; i++){ if(obstacleGrid[0][i] == 1){ dp[0][i] = 0; }else{ dp[0][i] = Math.min(dp[0][i-1], 1); } } for (int i = 1; i < rows; i++){ if(obstacleGrid[i][0] == 1){ dp[i][0] = 0; }else{ dp[i][0] = Math.min(dp[i-1][0], 1); } } for(int i = 1; i < rows; i++){ for(int j = 1; j < cols; j++){ if(obstacleGrid[i][j] == 1){ dp[i][j] = 0; }else{ dp[i][j] = dp[i][j-1] + dp[i-1][j]; } } } return dp[rows-1][cols-1]; } } ```