<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[0][0]) return -1;
        if (grid.size() == 1) return 1;
        queue<vector<int>> q; q.push({0,0});
        grid[0][0] = 1;
        unordered_set<int> visited;
        
        while(!q.empty()){
            int size = q.size();
            while(size){
                vector<int> tmp = q.front(); q.pop();
                int x = tmp[0], y = tmp[1];
                for (int i = -1; i <= 1; i++){
                    for (int j = -1; j <= 1; j++){
                        int nx = x + i; int ny = y + j;
                        if (nx >= 0 && ny >= 0 && nx < grid.size() && ny < grid[0].size() && !visited.count(nx * 1000 + ny)){
                            if (!grid[nx][ny]){
                                grid[nx][ny] = grid[x][y] + 1;
                                visited.insert(nx * 1000 + ny);
                                if (nx == grid.size()-1 && ny == grid[0].size()-1) {
                                    return grid[nx][ny];
                                }
                                q.push({nx, ny});
                            }
                        }
                    }
                }
                
                size--;
            }
        }
        return -1;
    }
};
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