<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:
    vector<vector<int>> directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    vector<vector<int>> grid;
    vector<vector<bool>> visited;
    int ways = 0;

    bool isValid(vector<int> node) {
        if (node[0] < 0 || node[0] >= grid.size()) return false;
        if (node[1] < 0 || node[1] >= grid[0].size()) return false;
        if (grid[node[0]][node[1]] == -1) return false;
        return true;
    }

    void dfs(vector<int> node, int remaining) {
        if (!isValid(node)) return;
        if (isValid(node) and grid[node[0]][node[1]] == 2 and remaining == 1) {
            ways++;
            return;
        }

        int result = 0;
        visited[node[0]][node[1]] = true;
        for (const auto direction: directions) {
            vector<int> neighbor = {node[0] + direction[0], node[1] + direction[1]};
            if (!isValid(neighbor) || visited[neighbor[0]][neighbor[1]] == true) continue;

            dfs(neighbor, remaining - 1);
        }

        visited[node[0]][node[1]] = false;
    }

    int uniquePathsIII(vector<vector<int>>& grid) {
        this->grid = grid;
        visited = vector<vector<bool>>(grid.size(), vector<bool>(grid[0].size(), false));
        int remaining = 0;
        int start_i, start_j = 0;

        for (int i = 0; i < grid.size(); i++) {
            for(int j = 0; j < grid[i].size(); j++) {
                if (grid[i][j] >= 0) {
                    remaining += 1;
                }
                if (grid[i][j] == 1) {
                    start_i = i;
                    start_j = j;
                }
            }
        }

        dfs({start_i, start_j}, remaining);
        return ways;
    }
};
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