Unique Paths II

Follow up for "Unique Paths":

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.

Analysis

From problem "unique path", this is a bottom-up DP problem

1. state: $$dp[i][j]$$: 从i,j到右下角的unique path数
2. transfer: $$dp[i][j]=dp[i][j+1]+dp[i+1][j]$$
3. initialize: all last row and last column 1
4. return $$dp[0][0]$$

Different from "unique path"

1. For all obstacles, dp = 0
2. Initialize difference: 一旦最后一行（列）有一个obstacle的话，它的左边（上面）全部为0， 右边和下面不受影响

Code (bottom to top)

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        if (obstacleGrid[m-1][n-1]==1) {return 0;}
        int[][] map = new int[m][n];

        //initialize last row
        for(int j=n-1; j>=0; j--){
            if(obstacleGrid[m-1][j]==1) {break;}
            else {map[m-1][j]=1;}
        }

        //initialize last colomn
        for(int i=m-2; i>=0;i--){
            if(obstacleGrid[i][n-1]==1) {break;}
            else {map[i][n-1]=1;}
        }

        for (int i=m-2; i>=0 ; i--){
            for(int j=n-2;j >=0 ;j--){
                if (obstacleGrid[i][j]==1) {map[i][j] =0;}
                else {map[i][j] = map[i][j+1] + map[i+1][j]; }
            }
        }
        return map[0][0];
    }
}

Code (top to bottom)

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid==null || obstacleGrid.length==0 || obstacleGrid[0].length==0) return 0;
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;

        int[][] dp = new int[m][n];

        for (int i=0; i<m; i++){
            if (obstacleGrid[i][0]==1) break;
            else dp[i][0]=1;
        }

        for (int j=0; j<n; j++){
            if (obstacleGrid[0][j]==1) break;
            else dp[0][j]=1;
        }

        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];
    }
}

Reference

Ref1