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MEDIUM

Unique Paths II

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 go through any obstacles.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

Example

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

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Constraints

• m == obstacleGrid.length
• n == obstacleGrid[i].length
• 1 ≤ m, n ≤ 100
• obstacleGrid[i][j] is 0 or 1.

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Solution: Dynamic Programming

• Time Complexity: O(mn)
• Space Complexity: O(n)

C++

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        vector<int> dp(n, 0);
        dp[0] = (obstacleGrid[0][0] == 0) ? 1 : 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (obstacleGrid[i][j] == 1) {
                    dp[j] = 0;
                } else if (j > 0) {
                    dp[j] += dp[j - 1];
                }
            }
        }
        return dp[n - 1];
    }
};