<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 {
    int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        // bfs to visit cells until no more cells can be visited or has reached the dest
        int m = grid.size();
        int n = grid[0].size();
        if(grid[0][0] == 1 || grid[m-1][n-1] == 1) {
            return -1;
        }
        if(m == 1 && n == 1) {
            return 1;
        }
        queue<vector<int>> bfs; // x, y, depth
        vector<vector<bool>> visited(m , vector<bool>(n, false));
        bfs.push({0, 0, 1});
        while(bfs.size()) {
            auto front = bfs.front();
            int i = front[0];
            int j = front[1];
            int newDepth = front[2] + 1;
            bfs.pop();
            for(int k = 0; k < 8; k++) {
                int y = i + dir[k][0];
                int x = j + dir[k][1];
                if(y < 0 || y >= m || x < 0 || x >= n || visited[y][x] || grid[y][x]) {
                    continue;
                }
                if(y == m - 1 && x == n -1) {
                    return newDepth;
                }
                visited[y][x] = true;
                bfs.push({y, x, newDepth});
            }
        }
        return -1;
    }
};
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