<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();
        if (grid[0][0] == 1 || grid[n-1][n-1] == 1) return -1;

        queue<vector<int>> q;  // vector: [i,j,level]
        q.push({0,0,1});
        grid[0][0] = 1;

        while (!q.empty()) {
            vector<int>& vec = q.front();
            int i = vec[0];
            int j = vec[1];
            int lvl = vec[2];
            cout << "visiting: " << i << "," << j << endl;
            q.pop();
            if (i == n-1 && j == n-1) return lvl;

            vector<int> options {-1, 0, 1};

            for (int x : options) {
                for (int y : options) {
                    //cout << "neighbor: " << i+x << "," << j+y;
                    if (i+x >= 0 && i+x <= n-1 && j+y >= 0 && j+y <= n-1 && grid[i+x][j+y] == 0) {
                        q.push({i+x,j+y,lvl+1});
                        grid[i+x][j+y] = 1;
                    }
                }
            } 
        }

        return -1;
    }
};

// bfs from topleft cell
// count levels
// check if arrived (n-1,n-1)
41404