<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 N;
    bool in_bounds(int r, int c) {
        return 0 <= r && r < N && 0 <= c && c < N;
    }
    vector<pair<int, int>> neighborhood(int i, int j) {
        return {{i - 1, j}, {i + 1, j}, {i, j - 1}, {i, j + 1}, {i - 1, j - 1}, {i - 1, j + 1}, {i + 1, j - 1}, {i + 1, j + 1}};
    }
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        if (grid[0][0]) return -1;
        N = grid.size();
        queue<pair<int, int>> q;
        q.emplace(0, 0);
        set<pair<int, int>> vis = {{0, 0}};
        int ans = 0;
        while (!q.empty()) {
            ans++;
            int cnt = q.size();
            while (cnt--) {
                auto [r, c] = q.front();
                q.pop();
                if (r == N - 1 && c == N - 1) return ans;
                for (auto [nr, nc] : neighborhood(r, c)) {
                    if (!in_bounds(nr, nc)) continue;
                    if (vis.find({nr, nc}) != vis.end()) continue;
                    if (grid[nr][nc]) continue;
                    vis.insert({nr, nc});
                    q.emplace(nr, nc);
                }
            }
        }
        return -1;
    }
};
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