<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:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        const int N = grid.size();
        if (grid[0][0] || grid[N-1][N-1]) return -1;

        static const pair<int,int> dirs[8] = {{-1,-1},{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1}};
        
        unordered_set<int> visited;
        auto get_neighbors = [&](int x, int y) -> vector<pair<int,int>>
        {
            vector<pair<int,int>> nbrs;
            for (const auto& [dx, dy] : dirs) {
                auto nx = x + dx, ny = y + dy;
                if (nx >= 0 && nx < N && ny >= 0 && ny < N && grid[nx][ny] == 0 && !visited.contains(N*nx + ny)) nbrs.push_back({nx,ny});
            }
            return nbrs;
        };

        queue<pair<int,int>> q;
        int distance = 1;
        q.push({0,0});
        visited.insert(0);
        while (!q.empty()) {
            for (int i = 0, qsize = q.size(); i < qsize; ++i) {
                auto [x,y] = q.front(); q.pop();
                if (x == N-1 && y == N-1) return distance;
                // visited.insert(N*x + y);
                for (auto& [nx,ny] : get_neighbors(x,y)) {
                    visited.insert(N*nx + ny);
                    q.push({nx,ny});
                }
            }
            ++distance;
        }
        return -1;
    }
};
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