2.11 Unique Paths

Description

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] ]

Method

it is a dynamic problem. if we want to know how many paths it can have if we want to reach(i, j), we need to know the paths for (i - 1, j) and (i, j - 1) since only this two locations can reach (i, j) so we use another matrix to store the paths for each location(i, j) and then we calculate from (0, 0) to (m - 1, n -1)

Time and Space Complexity

o(m*n)

Code

public class Solution {

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

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

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

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

}