<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 delRow[4] = {-1, 0, 1, 0};
    int delCol[4] = {0, 1, 0, -1};

    void dfs(int i, int j, int n, int m, int& tot, vector<vector<int>>& grid, vector<vector<int>> vis, pair<int, int> end, int count, int& ans){
        if(i==end.first && j==end.second){
            if(count==tot) ans = ans + 1;
        }
        vis[i][j] = 1;
        for(int k=0; k<4; k++){
            int ni = i + delRow[k];
            int nj = j + delCol[k];
            if(ni>=0 && ni<n && nj>=0 && nj<m && !vis[ni][nj] && (grid[ni][nj]==0 || grid[ni][nj]==2)){
                dfs(ni, nj, n, m, tot, grid, vis, end, count+1, ans);
            }
        }
    }

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

        pair<int, int> strt, end;
        int count = 0;
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(grid[i][j]==0 || grid[i][j]==2) count++;
                if(grid[i][j]==1) strt = {i, j};
                if(grid[i][j]==2) end = {i, j};
            }
        }
        int cnt = 0, ans = 0;
        vector<vector<int>> vis(n, vector<int>(m, 0));
        dfs(strt.first, strt.second, n, m, count, grid, vis, end, cnt, ans);
        return ans;
    }
};
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