<p>Given an <code>n x n</code> binary matrix <code>grid</code>, return <em>the length of the shortest <strong>clear path</strong> in the matrix</em>. If there is no clear path, return <code>-1</code>.</p>

<p>A <strong>clear path</strong> in a binary matrix is a path from the <strong>top-left</strong> cell (i.e., <code>(0, 0)</code>) to the <strong>bottom-right</strong> cell (i.e., <code>(n - 1, n - 1)</code>) such that:</p>

<ul>
	<li>All the visited cells of the path are <code>0</code>.</li>
	<li>All the adjacent cells of the path are <strong>8-directionally</strong> connected (i.e., they are different and they share an edge or a corner).</li>
</ul>

<p>The <strong>length of a clear path</strong> is the number of visited cells of this path.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/example1_1.png" style="width: 500px; height: 234px;" />
<pre>
<strong>Input:</strong> grid = [[0,1],[1,0]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/example2_1.png" style="height: 216px; width: 500px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0],[1,1,0],[1,1,0]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,0,0],[1,1,0],[1,1,0]]
<strong>Output:</strong> -1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>grid[i][j] is 0 or 1</code></li>
</ul>

class Solution {
public:
    struct Node
    {
        int r;
        int c;
        int dist;
    };
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        if (grid.size() == 0) {
            return -1;
        }

        queue<Node> q;
        q.push({0, 0, 1});

        vector<int> distances;
        while (!q.empty()) {
            auto node = q.front();
            q.pop();
            if (grid[node.r][node.c] == 0) {
                if (node.r == grid.size() - 1 && node.c == grid[0].size() - 1) {
                    distances.push_back(node.dist);
                } else {
                    grid[node.r][node.c] = 1;
                    if (node.r - 1 >= 0 && grid[node.r - 1][node.c] == 0) {
                        q.push({node.r - 1, node.c, node.dist + 1});
                        // grid[node.r - 1][node.c] = 1;
                    }
                    if (node.c - 1 >= 0 && grid[node.r][node.c - 1] == 0) {
                        q.push({node.r, node.c - 1, node.dist + 1});
                        // grid[node.r][node.c - 1] = 1;
                    }
                    if (node.r - 1 >= 0 && node.c - 1 >= 0 && grid[node.r - 1][node.c - 1] == 0) {
                        q.push({node.r - 1, node.c - 1, node.dist + 1});
                        // grid[node.r - 1][node.c - 1] = 1;
                    }
                    if (node.r + 1 < grid.size() && grid[node.r + 1][node.c] == 0) {
                        q.push({node.r + 1, node.c, node.dist + 1});
                        // grid[node.r + 1][node.c] = 1;
                    }
                    if (node.c + 1 < grid[0].size() && grid[node.r][node.c + 1] == 0) {
                        q.push({node.r, node.c + 1, node.dist + 1});
                        // grid[node.r][node.c + 1] = 1;
                    }
                    if (node.r + 1 < grid.size() && node.c + 1 < grid[0].size() && grid[node.r + 1][node.c + 1] == 0) {
                        q.push({node.r + 1, node.c + 1, node.dist + 1});
                        // grid[node.r + 1][node.c + 1] = 1;
                    }
                    if (node.r - 1 >= 0 && node.c + 1 < grid[0].size() && grid[node.r - 1][node.c + 1] == 0) {
                        q.push({node.r - 1, node.c + 1, node.dist + 1});
                        // grid[node.r - 1][node.c + 1] = 1;
                    }
                    if (node.r + 1 < grid.size() && node.c - 1 >= 0 && grid[node.r + 1][node.c - 1] == 0) {
                        q.push({node.r + 1, node.c - 1, node.dist + 1});
                        // grid[node.r + 1][node.c - 1] = 1;
                    }
                }
            }
        }
        int shortest = INT_MAX;
        for (auto dist : distances) {
            shortest = min(shortest, dist);
        }
        return shortest == INT_MAX ? -1 : shortest;
    }
};
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