> For the complete documentation index, see [llms.txt](https://jimmylin1991.gitbook.io/practice-of-algorithm-problems/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://jimmylin1991.gitbook.io/practice-of-algorithm-problems/backtracking/980.-unique-paths-iii.md).

# 980. Unique Paths III

On a 2-dimensional `grid`, there are 4 types of squares:

* `1` represents the starting square.  There is exactly one starting square.
* `2` represents the ending square.  There is exactly one ending square.
* `0` represents empty squares we can walk over.
* `-1` represents obstacles that we cannot walk over.

Return the number of 4-directional walks from the starting square to the ending square, that **walk over every non-obstacle square exactly once**.

**Example 1:**

```
Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
Output: 2
Explanation: We have the following two paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)
```

**Example 2:**

```
Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
Output: 4
Explanation: We have the following four paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)
```

**Example 3:**

```
Input: [[0,1],[2,0]]
Output: 0
Explanation: 
There is no path that walks over every empty square exactly once.
Note that the starting and ending square can be anywhere in the grid.
```

**Note:**

1. `1 <= grid.length * grid[0].length <= 20`

```cpp
// Recursive DFS
void dfs(vector<vector<int> >& grid, vector<vector<bool> >& visited, int x, int y, int end_x, int end_y, int& zeros, int& res) {
    if (x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] < 0 || visited[x][y]) return;
    if (x == end_x && y == end_y) {
        if (zeros == 0) ++res;
        return;
    }
    visited[x][y] = true;
    --zeros;
    dfs(grid, visited, x + 1, y, end_x, end_y, zeros, res);
    dfs(grid, visited, x - 1, y, end_x, end_y, zeros, res);
    dfs(grid, visited, x, y + 1, end_x, end_y, zeros, res);
    dfs(grid, visited, x, y - 1, end_x, end_y, zeros, res);
    ++zeros;
    visited[x][y] = false;
}
int uniquePathsIII(vector<vector<int>>& grid) { // time: O(4^(mn)); space: O(mn)
    int m = grid.size(), n = grid[0].size();
    int res = 0, zeros = 0, start_x = -1, start_y = -1, end_x = -1, end_y = -1;
    vector<vector<bool> > visited(m, vector<bool> (n, false));
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            if (grid[i][j] == 0) ++zeros;
            else if (grid[i][j] == 1) {
                ++zeros;
                start_x = i;
                start_y = j;
            } else if (grid[i][j] == 2) {
                end_x = i;
                end_y = j;
            }
        }
    }
    dfs(grid, visited, start_x, start_y, end_x, end_y, zeros, res);
    return res;
}
```

```cpp
// Brute Force Recursion
bool check(vector<vector<int>>& grid, int x, int y) {
    int m = grid.size(), n = grid[0].size();
    return 0 <= x && x < m && 0 <= y && y < n && grid[x][y] >= 0;
}
void dfs(vector<vector<int>>& grid, int x, int y, int end_x, int end_y, int& zeros, int& res) {
    if (!check(grid, x, y)) return;
    if (x == end_x && y == end_y) {
        if (zeros == 0) ++res;
        return;
    }
    grid[x][y] = -1;
    --zeros;
    dfs(grid, x + 1, y, end_x, end_y, zeros, res);
    dfs(grid, x - 1, y, end_x, end_y, zeros, res);
    dfs(grid, x, y + 1, end_x, end_y, zeros, res);
    dfs(grid, x, y - 1, end_x, end_y, zeros, res);
    ++zeros;
    grid[x][y] = 0;
}
int uniquePathsIII(vector<vector<int>>& grid) { // time: O(4^(mn)); space: O(mn)
    int res = 0, zeros = 1, start_x = -1, start_y = -1, end_x = -1, end_y = -1;
    int m = grid.size(), n = grid[0].size();
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            if (grid[i][j] == 0) ++zeros;
            else if (grid[i][j] == 1) {
                start_x = i;
                start_y = j;
            } else if (grid[i][j] == 2) {
                end_x = i;
                end_y = j;
            }
        }
    }
    dfs(grid, start_x, start_y, end_x, end_y, zeros, res);
    return res;
}
```


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