Unique Paths in a Grid

Given a grid of size m * n, lets assume you are starting at (1,1) and your goal is to reach (m,n). At any instance, if you are on (x,y), you can either go to (x, y + 1) or (x + 1, y).

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.

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.

Solution:

Time/Space: O(mn)

public class Solution {
    public int uniquePathsWithObstacles(int[][] A) {
        // dp[i][j] = uniq path to i j
        // dp[i][j] = dp[i - 1][j] + dp[i][j - 1] if A[i][j] == 0
        // dp[0][0] = 1 - A[0][0]
        // dp[m - 1][n - 1]
        int m = A.length;
        int n = A[0].length;
        int[][] dp = new int[m][n];
        dp[0][0] = 1 - A[0][0];
        for (int j = 1; j < n; j ++) {
            if (A[0][j] == 0) {
                dp[0][j] = dp[0][j - 1];
            } 
        }
        for (int i = 1; i < m; i ++) {
            if (A[i][0] == 0) {
                dp[i][0] = dp[i - 1][0];
            }
        }
        for (int i = 1; i < m; i ++) {
            for (int j = 1; j < n; j ++) {
                if (A[i][j] == 0) {
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
                }
            }
        }
        return dp[m - 1][n - 1];
    }
}

Space optimized:

public class Solution {
    public int uniquePathsWithObstacles(int[][] A) {
        // dp[i][j] = uniq path to i j
        // dp[i][j] = dp[i - 1][j] + dp[i][j - 1] if A[i][j] == 0
        // dp[0][0] = 1 - A[0][0]
        // dp[m - 1][n - 1]
        int m = A.length;
        int n = A[0].length;
        int[] dp = new int[n];
        dp[0] = 1 - A[0][0];
        for (int j = 1; j < n; j ++) {
            if (A[0][j] == 0) {
                dp[j] = dp[j - 1];
            } 
        }
        for (int i = 1; i < m; i ++) {
            dp[0] = A[i][0] == 0 ? dp[0] : 0;
            for (int j = 1; j < n; j ++) {
                if (A[i][j] == 0) {
                    dp[j] = dp[j] + dp[j - 1];
                } else {
                    dp[j] = 0;
                }
            }
        }
        return dp[n - 1];
    }
}