63. Unique Paths II

Problem:

---

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.

Note: m and n will be at most 100.

Analysis:

Follow up of Unique Path. When O[i][j] == 1, dp[i][j] == 0. Implement this problem with 1D array is a little tricky. Since dp[0] can not be updated by dp[j] += d[j - 1].  We can set the initial value of dp[0] to 1. If O[i][0] == 1, set dp[0] to 0.

Solution:

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[] dp = new int[n];
        dp[0] = 1;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (obstacleGrid[i][j] == 1) dp[j] = 0;
                else if (j > 0) dp[j] += dp[j - 1];
        }
        }
        return dp[n - 1];
    }
}