<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 ans=0;
    int m,n;



    void recursion(vector<vector<int>>& grid, vector<vector<bool>> past,int i, int j,int total )
    {
        past[i][j]=1;
        total--;

        if(grid[i][j]==2)
        {
            if(total==0)
                ans++;

            return;
        }

        //go up
        if(i-1>=0 && grid[i-1][j]!=-1 && past[i-1][j]!=1 )
            recursion(grid,past,i-1,j,total);

        //go down
        if(i+1<m && grid[i+1][j]!=-1 && past[i+1][j]!=1 )
            recursion(grid,past,i+1,j,total);   

        //go right
        if(j+1<n && grid[i][j+1]!=-1 && past[i][j+1]!=1)
            recursion(grid,past,i,j+1,total);

        //go left
        if(j-1>=0 && grid[i][j-1]!=-1 && past[i][j-1]!=1)
            recursion(grid,past,i,j-1,total);        

    }

    int uniquePathsIII(vector<vector<int>>& grid) {
        //row size
        m=grid.size();

        //column size
        n=grid[0].size();

        //making a 2d boolean array to store visited positions
        vector<vector<bool>> past(m,vector<bool>(n,0));
        
        int starti;
        int startj;

        //finding starting postiton
        int count=0;

        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
                {
                    if(grid[i][j]==1)
                        {
                            starti=i;
                            startj=j;
                        }

                    if(grid[i][j]==-1)
                        count++;  
                }
        }
        int total=m*n - count;

        //going through all possible paths and trying to reach end
        recursion(grid,past,starti,startj,total);

        return ans;
    }
};
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