<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:
    //We will use recursion as we have to explore all paths in all 4 directions
    void f(vector<vector<int>> grid,int &count,vector<vector<int>> track,int x,int y){
        
        track[x][y]=1;

        //Base Case
        if(grid[x][y]==2){
            for(int i=0;i<track.size();i++){
                for(int j=0;j<track[0].size();j++){
                    if(track[i][j]==0) return; //that means we didnt visited a square
                }
            }
            //now we have reached end and every path is traced so inc count
            count++;
            return;
        }

        //Now we can go in 4 directions
        if(y+1<grid[0].size() and track[x][y+1]==0 and grid[x][y+1]!=-1) f(grid,count,track,x,y+1);
        if(x+1<grid.size() and track[x+1][y]==0 and grid[x+1][y]!=-1) f(grid,count,track,x+1,y);
        if(x-1>=0 and track[x-1][y]==0 and grid[x-1][y]!=-1) f(grid,count,track,x-1,y);
        if(y-1>=0 and track[x][y-1]==0 and grid[x][y-1]!=-1) f(grid,count,track,x,y-1); 
        
    }
    
    int uniquePathsIII(vector<vector<int>>& grid) {
        int count=0;
        vector<vector<int>> track(grid.size(),vector<int>(grid[0].size(),0)); //THis is used for tracking 
        
        //First find the start point
        int x; 
        int y;
        for(int i=0;i<grid.size();i++){
            for(int j=0;j<grid[0].size();j++){
                if(grid[i][j]==1){
                    x=i;
                    y=j;      
                }
                if(grid[i][j]==-1){
                    track[i][j]=-1; //so that at the base case we get correct ans
                }
            }
        }
        f(grid,count,track,x,y);
        return count;
    }
};
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