source: https://leetcode.com/problems/unique-paths-ii/
Unique Paths II
Table of Contents
Description
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 3×3 grid above.
There are two ways to reach the bottom-right corner:
- Right -> Right -> Down -> Down
- 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.
Solution
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class Solution { public int uniquePathsWithObstacles(int[][] obstacleGrid) { int[][] dp = new int[obstacleGrid.length][obstacleGrid[0].length]; for(int i = 0 ;i<dp.length;i++) { for(int j = 0;j<dp[0].length;j++) { if(i == 0 && j == 0 && obstacleGrid[i][j] !=1) { //number of ways to reach the cell (0,0) is 1 dp[i][j] = 1; } //if there is an obstacle at the current location else if(obstacleGrid[i][j] == 1) { dp[i][j] =0; //if there is a space at the current location } else { //if you are moving down to reach current cell if(i!=0) dp[i][j] += dp[i-1][j]; //if you are moving right to reach the current cell if(j!=0) dp[i][j] += dp[i][j-1]; } } } return dp[dp.length-1][dp[0].length-1]; } } |