LeetCode – Unique Paths II (Java)

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.

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];
}
Category >> Algorithms >> Interview  
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  1. mg on 2017-6-5

    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];
    }
    }

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