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/**
 * 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.
 */
 
public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid[0][0] == 1)
            return 0;
        int[][] paths = new int[obstacleGrid.length][obstacleGrid[0].length];
        // Initialization 
        paths[0][0] = 1;
        for(int i = 1; i < obstacleGrid.length; i++){
            if(obstacleGrid[i][0] == 0)
                paths[i][0] = paths[i - 1][0];
        }
        for(int i = 1; i < obstacleGrid[0].length; i++){
            if(obstacleGrid[0][i] == 0)
                paths[0][i] = paths[0][i - 1];
        }
        for(int i = 1; i < obstacleGrid.length; i++){
            for(int j = 1; j < obstacleGrid[0].length; j++)
                if(obstacleGrid[i][j] == 0)
                    paths[i][j] = paths[i - 1][j] + paths[i][j - 1];
        }
        return paths[obstacleGrid.length - 1][obstacleGrid[0].length - 1];
    }
}

/**
 * A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
 * The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
 * How many possible unique paths are there?
 * Note: m and n will be at most 100.
 */
 
public class Solution {
    public int uniquePaths(int m, int n) {
        int[][] paths = new int[m][n];
        // Initialization
        for(int i = 0; i < m; i++){
            paths[i][0] = 1;
        }
        for(int i = 0; i < n; i++){
            paths[0][i] = 1;
        }
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                paths[i][j] = paths[i - 1][j] + paths[i][j - 1];
            }
        }
        return paths[m - 1][n - 1];
    }
}