<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) {

        int n = grid.size();
        
        queue<vector<int>> q;
        vector<vector<bool>> visited(n, vector<bool>(n));

        if(grid[0][0] == 0) {
            q.push({0, 0, 1});
            visited[0][0] = true;
        }

        int d_x[] = {-1, -1, -1, 1, 1, 1, 0, 0};
        int d_y[] = {-1, 0, 1, -1, 0, 1, -1, 1};

        while(!q.empty()) {
            auto front = q.front();
            q.pop();

            if(front[0] == n-1 && front[1] == n-1) {
                return front[2];
            }

            for(int k=0; k<8; k++) {
                int n_i = d_x[k] + front[0];
                int n_j = d_y[k] + front[1];

                if(n_i >=0 && n_i <n && n_j >=0 && n_j <n && grid[n_i][n_j] == 0 && visited[n_i][n_j] == false) {
                    q.push({n_i, n_j, front[2] +1});
                    visited[n_i][n_j] = true;
                }
            }
        }

        return -1;
    }
};
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