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.
Solutions
Solution 1: Memoization Search
We design a function $dfs(i, j)$ to represent the number of paths from the grid $(i, j)$ to the grid $(m - 1, n - 1)$, where $m$ and $n$ are the number of rows and columns of the grid, respectively.
The execution process of the function $dfs(i, j)$ is as follows:
If $i \ge m$ or $j \ge n$, or $obstacleGrid[i][j] = 1$, then the number of paths is $0$;
If $i = m - 1$ and $j = n - 1$, then the number of paths is $1$;
Otherwise, the number of paths is $dfs(i + 1, j) + dfs(i, j + 1)$.
To avoid repeated calculations, we can use the method of memoization search.
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns of the grid, respectively.
We define $f[i][j]$ as the number of paths to reach the grid $(i,j)$.
First, initialize all values in the first column and first row of $f$. Then, traverse other rows and columns, there are two cases:
If $obstacleGrid[i][j] = 1$, it means the number of paths is $0$, so $f[i][j] = 0$;
If $obstacleGrid[i][j] = 0$, then $f[i][j] = f[i - 1][j] + f[i][j - 1]$.
Finally, return $f[m - 1][n - 1]$.
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns of the grid, respectively.