<p>You are given an <code>m x n</code> integer array <code>grid</code> where <code>grid[i][j]</code> could be:</p>

<ul>
	<li><code>1</code> representing the starting square. There is exactly one starting square.</li>
	<li><code>2</code> representing the ending square. There is exactly one ending square.</li>
	<li><code>0</code> representing empty squares we can walk over.</li>
	<li><code>-1</code> representing obstacles that we cannot walk over.</li>
</ul>

<p>Return <em>the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/02/lc-unique1.jpg" style="width: 324px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 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)
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/02/lc-unique2.jpg" style="width: 324px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> 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)
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/02/lc-unique3-.jpg" style="width: 164px; height: 165px;" />
<pre>
<strong>Input:</strong> grid = [[0,1],[2,0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 20</code></li>
	<li><code>1 &lt;= m * n &lt;= 20</code></li>
	<li><code>-1 &lt;= grid[i][j] &lt;= 2</code></li>
	<li>There is exactly one starting cell and one ending cell.</li>
</ul>

class Solution {
public:
    int m, n;
    vector<vector<int>> grid;
    int count;

    bool isValid(int i, int j) { return i >= 0 && j >= 0 && i < m && j < n; }

    vector<vector<int>> dir = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};

    int dfs(int x, int y, int destx, int desty, vector<vector<bool>>& vis,
            set<pair<int, int>>& st) {
        if (x == destx && y == desty && st.size() == count) {
            return 1;
        }
        if (x == destx && y == desty) {
            return 0;
        }
        int res = 0;
        vis[x][y] = true;
        for (auto d : dir) {
            int nx = x + d[0];
            int ny = y + d[1];
            if (isValid(nx, ny) && !vis[nx][ny] && grid[nx][ny] != -1) {
                st.insert({nx, ny});
                res += dfs(nx, ny, destx, desty, vis, st);
                st.erase({nx, ny});  // Backtrack
            }
        }
        vis[x][y] = false;  // Backtrack
        return res;
    }

    int uniquePathsIII(vector<vector<int>>& grid) {
        this->grid = grid;
        m = grid.size();
        n = grid[0].size();
        count = 0;

        int startx, starty, destx, desty;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    startx = i;
                    starty = j;
                }
                if (grid[i][j] == 2) {
                    destx = i;
                    desty = j;
                }
                if (grid[i][j] == 0) {
                    count++;
                }
            }
        }
        count = count + 2;
        vector<vector<bool>> vis(m, vector<bool>(n, false));
        set<pair<int, int>> st;
        st.insert({startx, starty});
        return dfs(startx, starty, destx, desty, vis, st);
    }
};

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