<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 startRow, startCol, endRow, endCol;
    int totalWalkable = 0; // Total cells that can be walked on (non-obstacle)
    int result = 0; // Number of valid unique paths
    
    void backtrack(vector<vector<int>>& grid, int row, int col, int remaining) {
        // Base case: If we reach the end and have visited all walkable cells
        if (row == endRow && col == endCol && remaining == 1) {
            result++; // Found a valid path
            return;
        }
        
        // Mark current cell as visited
        int temp = grid[row][col];
        grid[row][col] = -1; // Mark it as obstacle to indicate visited
        
        // Explore 4 possible directions: Down, Left, Right, Up
        vector<pair<int, int>> directions = {{1, 0}, {0, -1}, {0, 1}, {-1, 0}};
        for (auto [dr, dc] : directions) {
            int newRow = row + dr;
            int newCol = col + dc;
            
            // Check if the new position is within bounds and is walkable
            if (newRow >= 0 && newRow < grid.size() && newCol >= 0 && newCol < grid[0].size() && grid[newRow][newCol] != -1) {
                backtrack(grid, newRow, newCol, remaining - 1);
            }
        }
        
        // Unmark current cell (backtracking)
        grid[row][col] = temp;
    }
    
    int uniquePathsIII(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        
        // Step 1: Find start (1), end (2) and count walkable cells
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    startRow = i;
                    startCol = j;
                } else if (grid[i][j] == 2) {
                    endRow = i;
                    endCol = j;
                }
                if (grid[i][j] != -1) {
                    totalWalkable++; // Count non-obstacle cells
                }
            }
        }
        backtrack(grid, startRow, startCol, totalWalkable);        
        return result; 
    }
};
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