63. Unique Paths II.cpp

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.








// TC -> O(m * n)
// SC -> O(m * n)
// Optimization: Instead of creating a new DP 2d vector
// obstacleGrid could have been itself used for storing results

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