<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:
    void solve(vector<vector<int>>& grid,int sx,int sy,int fx,int fy,int k
    ,int& count,vector<vector<bool>>& visited){
        if(k==1 && sx==fx&&sy==fy){
            count++;
            return;
        }
        
        vector<vector<int>>direc = {{1,0},{-1,0},{0,-1},{0,1}};
        for(int i=0;i<4;i++){
            int x = sx + direc[i][0];
            int y = sy + direc[i][1];

            if(x>=0&&x<grid.size()&&y>=0&&y<grid[0].size()&&grid[x][y]!=-1&&!visited[x][y]){
                visited[x][y] = true;
                solve(grid,x,y,fx,fy,k-1,count,visited);
                visited[x][y] = false;
            }
        }
    }
    int uniquePathsIII(vector<vector<int>>& grid) {
        vector<int>curr;
        int m = grid.size();
        int n = grid[0].size();

        int obstacles = 0;
        int startX = 0;
        int startY = 0;
        int finishX = 0;
        int finishY = 0;
        
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==-1){
                    obstacles++;
                }
                if(grid[i][j]==1){
                    startX = i;
                    startY = j;
                }
                if(grid[i][j]==2){
                    finishX = i;
                    finishY = j;
                }
            }
        }
        int k = (m*n)-obstacles;
        int count = 0;
        vector<vector<bool>>visited(m,vector<bool>(n,false));
        visited[startX][startY] = true;
        solve(grid,startX,startY,finishX,finishY,k,count,visited);
        return count;
    }
};
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