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UniquePathsII.java
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UniquePathsII.java
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package Leetcode;
/**
* @author kalpak
*
* 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).
*
* Now consider if some obstacles are added to the grids. How many unique paths would there be?
*
* An obstacle and space is marked as 1 and 0 respectively in the grid.
*
*
* 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.
*/
public class UniquePathsII {
public static int uniquePathsWithObstacles(int[][] obstacleGrid) {
if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0)
return 0;
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[][] dp = new int [m][n];
// First Column
for(int i = 0; i < m; i++) {
if(obstacleGrid[i][0] != 1)
dp[i][0] = 1;
else
break; // Important: Since there is an obstacle, cant proceed further
}
// First Row
for(int i = 0; i < n; i++) {
if(obstacleGrid[0][i] != 1)
dp[0][i] = 1;
else
break; // Important: Since there is an obstacle, cant proceed further
}
for(int i = 1; i < m; i++) {
for(int j = 1; j < n; j++) {
if(obstacleGrid[i][j] != 1) {
dp[i][j] = dp[i-1][j] + dp[i][j-1];
}
}
}
return dp[m-1][n-1];
}
public static void main(String[] args) {
int[][] obstacleGrid = new int[][]{{0,0,0},{0,1,0},{0,0,0}};
System.out.println(uniquePathsWithObstacles(obstacleGrid));
}
}