<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:
    struct pair_hash{
        size_t operator()(const pair<int,int>& v) const{
            return v.first ^ v.second;
        }    
    };
    
    bool isValid(const int i, const int j, const int maxRow, const int maxCol){
        if (i<0 || j<0 || i>=maxRow || j>=maxCol) return false;
        return true;
    }
    bool dfs(vector<vector<int>>& grid, int i, int j, int goalRow, int goalCol, unordered_set<pair<int,int>, pair_hash>& canWalk){  
        
        // cout << i << " " << j << " " << canWalk.size() << endl;
        if (i==goalRow && j==goalCol){
            if (canWalk.size()==1){
                ++grid[i][j];
                return true;    
            } else{
                return false;
            }             
        }
        if (grid[i][j]<0) return false;
        vector<int> inc{0,1,0,-1,0};
        canWalk.erase(make_pair(i,j));
        for (int step=0; step<inc.size()-1; ++step){
            int newRow = i + inc[step];
            int newCol = j + inc[step+1];
            if (isValid(newRow, newCol, grid.size(), grid[0].size())){
                if (canWalk.find(make_pair(newRow, newCol)) != canWalk.end()){
                    dfs(grid, newRow, newCol, goalRow, goalCol, canWalk);
                }
                
            }
            
        }
        canWalk.insert(make_pair(i,j));
        return false;
    }
    int uniquePathsIII(vector<vector<int>>& grid) {
        pair<int, int> start, goal;
        unordered_set<pair<int,int>, pair_hash> canWalk;
        for (int i=0; i<grid.size(); ++i){
            for (int j=0; j<grid[i].size(); ++j){
                if (grid[i][j]==1){
                    start = make_pair(i, j);
                } else if (grid[i][j]==2){
                    goal = make_pair(i, j);
                    canWalk.insert(make_pair(i,j));
                } else if (grid[i][j]==0){
                    canWalk.insert(make_pair(i,j));
                }
            }
        }
    
        dfs(grid, start.first, start.second, goal.first, goal.second, canWalk);
        return grid[goal.first][goal.second]-2;
    }
};
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