<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:
    using T = std::tuple<int, int, int>;
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        if (grid[0][0] == 1) return -1;

        int m = grid.size();
        int n = grid[0].size();

        std::priority_queue<T, std::vector<T>, std::greater<T>> pq;
        pq.push({1, 0, 0});
        grid[0][0] = 1;

        while (!pq.empty()) {
            auto curr = pq.top(); pq.pop();
            if (std::get<1>(curr) == m-1 && std::get<2>(curr) == n-1) {
                return std::get<0>(curr);
            }

            for (const auto& d: dirs) {
                int i = d.first + std::get<1>(curr);
                int j = d.second + std::get<2>(curr);
                if (i >= 0 && i < m && j >= 0 && j < n && grid[i][j] != 1) {
                    grid[i][j] = 1;
                    pq.push({std::get<0>(curr) + 1, i, j});
                }
            }
        }

        return -1;
    }

private:
    std::vector<std::pair<int, int>> dirs{
        {0, 1},
        {0, -1},
        {1, 0},
        {-1, 0},
        {1, 1},
        {-1, -1},
        {-1, 1},
        {1, -1},
    };
};
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