<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) {
        if(grid.empty()) return -1;
        int n = grid.size(), m = grid[0].size();
        if(grid[0][0]) return -1;
        queue<pair<int,int> > q;
        vector<vector<bool> > vis(n, vector<bool>(m, false));
        q.push({0,0});
        int curPath = 0;
        vector<int> dx = {-1, 0, 1};
        vector<int> dy = {-1, 0, 1};
        while(!q.empty()){
            curPath++;
            int sz = q.size();
            for(int s = 0; s < sz; s++){
                auto cell = q.front(); q.pop();
                if(cell.first == n - 1 && cell.second == m - 1) return curPath;
                if(vis[cell.first][cell.second]) continue;
                vis[cell.first][cell.second] = true;
                for(auto i: dx){
                    for(auto j: dy){
                        if(i == 0 && j == 0) continue;
                        if(cell.first + i >= n || cell.first + i < 0 ||
                        cell.second + j >= m || cell.second + j < 0) continue;
                        if(grid[cell.first + i][cell.second + j] == 1) continue;
                        q.push({cell.first + i, cell.second + j});
                    }
                }
            }
        }
        return -1;
    }
};
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