<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<pair<int,int>>bfsQ;
        vector<vector<int>>visited(n,vector<int>(n,0));
        vector<vector<int>>dis(n,vector<int>(n,-1));
        if(grid[0][0])
            return -1;
        bfsQ.push({0,0});
        // visited[0][0]=1;
        dis[0][0]=1;
        while(!bfsQ.empty())
        {
            auto frontNode = bfsQ.front();
            bfsQ.pop();
            if(visited[frontNode.first][frontNode.second])
            {
                continue;
            }
            visited[frontNode.first][frontNode.second] = 1;
            // printf("(%d,%d) -> %d\n",frontNode.first, frontNode.second, dis[frontNode.first][frontNode.second]);
            if(frontNode.first == n-1 && frontNode.second == n-1)
            {
                break;
            }
            vector<int>drow = {0,-1,-1,-1,0,1,1,1};
            vector<int>dcol = {-1,-1,0,1,1,1,0,-1};
            for(int i = 0;i<drow.size();i++)
            {
                int nrow = frontNode.first + drow[i];
                int ncol = frontNode.second + dcol[i];
                if(nrow<0 || nrow>=n || ncol<0 || ncol>=n || visited[nrow][ncol]!=0 || grid[nrow][ncol])
                {
                    continue;
                }
                // printf("pushing: (%d,%d)\n",nrow, ncol);

                if(dis[nrow][ncol]==-1)
                {
                    dis[nrow][ncol] = dis[frontNode.first][frontNode.second] + 1;
                }
                bfsQ.push({nrow,ncol});
            }
        }
        return dis[n-1][n-1];

    }
};
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