<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>

#define ll long long
class Solution {
  public:
    int shortestPathBinaryMatrix (vector<vector<int>> &grid) {
        deque<array<int,3>>dq;
        int n = grid.size();
        vector<vector<int>>vis(n,vector<int>(n));
        if (grid[0][0] == 1 or grid.back().back() == 1) return -1;
        dq.push_back ({1, 0,0});
        while (!dq.empty() ) {
            auto [d,r,c] = dq.front();
            dq.pop_front();
            if (r == n - 1 and c == n - 1) return d;
            if(vis[r][c]) continue;
            vis[r][c]=1;
            vector<int> dx = { -1, 1, 0, 0, -1, -1, 1, 1};
            vector<int> dy = {0, 0, -1, 1, 1, -1, -1, 1};
            for (int k = 0; k < 8; k++) {
                int x = dx[k] + r;
                int y = dy[k] + c;

                if (x >= 0 and x<n and y >= 0 and y < n and grid[x][y] == 0) {
                   dq.push_back({d+1,x,y});
                }
            }
        }

        return -1;
    }
};

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