UniquePaths-2.java

/**
 * You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.
 *
 * An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.
 *
 * Return the number of possible unique paths that the robot can take to reach the bottom-right corner.
 *
 * The testcases are generated so that the answer will be less than or equal to 2 * 109.
 *
 *
 *
 * Example 1:
 *
 *
 * Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
 * Output: 2
 * Explanation: There is one obstacle in the middle of the 3x3 grid above.
 * There are two ways to reach the bottom-right corner:
 * 1. Right -> Right -> Down -> Down
 * 2. Down -> Down -> Right -> Right
 * Example 2:
 *
 *
 * Input: obstacleGrid = [[0,1],[0,0]]
 * Output: 1
 *
 *
 * Constraints:
 *
 * m == obstacleGrid.length
 * n == obstacleGrid[i].length
 * 1 <= m, n <= 100
 * obstacleGrid[i][j] is 0 or 1.
 */

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