<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, m;
    int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
    int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        if(grid[0][0] or grid.back().back())
            return -1;
        int x, y, len, x_, y_;
        queue<vector<int>> q;
        q.push({0, 0, 1});
        grid[0][0] = 1;
        n = grid.size();
        m = grid[0].size();
        
        while(!q.empty()) {
            x = q.front()[0];
            y = q.front()[1];
            len = q.front()[2];
            q.pop();
            if(x == n-1 and y == m-1)
                return len;
            for(int i = 0; i < 8; i++) {
                x_ = x + dx[i];
                y_ = y + dy[i];
                if(isValid(x_, y_) and !grid[x_][y_]) {
                    grid[x_][y_] = 1;
                    q.push({x_, y_, len+1});
                }
            }
        }
        return -1;
    }

    bool isValid(int x, int y) {
        return (x >= 0 and y >= 0 and x < n and y < m);
    }
};
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