Unique Paths II - DFS approach - swift

Question

I am trying to solve the following question
https://leetcode.com/problems/unique-paths-ii/
Question is as follows:
An obstacle and space is marked as 1 and 0 respectively in the grid.
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:
Right -> Right -> Down -> Down
Down -> Down -> Right -> Right
My DFS Solution is as follows, but it returns 1 instead of 2. I am wondering what I am missing in the following approach?
class Solution {
func uniquePathsWithObstacles(_ obstacleGrid: [[Int]]) -> Int {
var column = obstacleGrid[0].count
var row = obstacleGrid.count
var visited = Array(repeating: Array(repeating: false, count:column), count: row)
return dfs(obstacleGrid, 0, 0, &visited)
}
func dfs(_ obstacleGrid: [[Int]], _ i: Int, _ j: Int, _ visited: inout [[Bool]]) -> Int {
if i < 0 || i >= obstacleGrid.count || j < 0 || j >= obstacleGrid[0].count || obstacleGrid[i][j] == 1 || visited[i][j] == true {
return 0
}
visited[i][j] = true
if i == obstacleGrid.count - 1 && j == obstacleGrid[0].count - 1 {
return 1
}
return dfs(obstacleGrid, i+1, 0, &visited) +
dfs(obstacleGrid, i-1, 0, &visited) +
dfs(obstacleGrid, i, j+1, &visited) +
dfs(obstacleGrid, i, j-1, &visited)
}
}

Answer 1

Returning 0 if visited[i][j] doesn't look correct. You should memorize the return value for visited cells, and return it again for the later calls on that cell. The reason is that you can reach that cell following different paths, and each of those should contribute to the result.
The fact that your code returns 1 makes sense, because the second time you reach the bottom-right cell, you will return 0 since that cell was 'visited' during the discovery of previous path.