LeetCode-in-Java

63. Unique Paths II

Medium

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.

Solution

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        // if start point has obstacle, there's no path
        if (obstacleGrid[0][0] == 1) {
            return 0;
        }
        obstacleGrid[0][0] = 1;
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        for (int i = 1; i < m; i++) {
            if (obstacleGrid[i][0] == 1) {
                obstacleGrid[i][0] = 0;
            } else {
                obstacleGrid[i][0] = obstacleGrid[i - 1][0];
            }
        }
        for (int j = 1; j < n; j++) {
            if (obstacleGrid[0][j] == 1) {
                obstacleGrid[0][j] = 0;
            } else {
                obstacleGrid[0][j] = obstacleGrid[0][j - 1];
            }
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    obstacleGrid[i][j] = 0;
                } else {
                    obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
                }
            }
        }
        return obstacleGrid[m - 1][n - 1];
    }
}

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