<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 {
private:
    typedef pair<int, int> pr;
    bool isValid(vector<vector<int>>& grid, pr& idx) {
        int i = idx.first, j = idx.second, r = grid.size(), c = grid[0].size();
        if (i >= 0 && i < r && j >= 0 && j < c && grid[i][j] != -1) {
            return true;
        }
        return false;
    }
 int paths(vector<vector<int>>& grid, pr currIdx, int currCount, int &countEmpty, vector<vector<bool>>& vis){
    cout<<"x: "<<currIdx.first<<" y: "<<currIdx.second<<" "<<"value: "<<grid[currIdx.first][currIdx.second]<<" ";
    cout<<"count = "<<currCount<<endl;;
    if(grid[currIdx.first][currIdx.second] == -1){
        return 0;
    }
    if(grid[currIdx.first][currIdx.second] == 2 && currCount == countEmpty+1){
        return 1;
    }
    int n_paths=0;
    vector<int> dirX = {1,-1,0,0};
    vector<int> dirY = {0,0,1,-1};
    vis[currIdx.first][currIdx.second] = true;
    for(int i=0;i<4;i++){
        pr idx = {currIdx.first + dirX[i], currIdx.second + dirY[i]};
        if(isValid(grid,idx) && !vis[idx.first][idx.second]){
            n_paths = n_paths + paths(grid, idx, currCount+1, countEmpty, vis);
        }
    }
    vis[currIdx.first][currIdx.second] = false;
    return n_paths;
 }
public:
    int uniquePathsIII(vector<vector<int>>& grid) {
        if (grid.empty() || grid[0].empty()) {
            return 0;
        }
        int nPaths = 0;
        int rows = grid.size();
        int cols = grid[0].size();
        vector<vector<bool>> vis(rows, vector<bool>(cols, false));
        int countEmpty = 0;
        pair<int, int> startIdx, endIdx;
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == 0) {
                    countEmpty++;
                } else if (grid[i][j] == 1) {
                    startIdx = {i, j};
                } else if (grid[i][j] == 2) {
                    endIdx = {i, j};
                }
            }
        }
        cout<<"Total 0's = "<<countEmpty<<endl;
        return paths(grid,startIdx,0,countEmpty,vis);
    }
};
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