# 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 3x3 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
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• mg

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];
}
}