<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 <vector<int>>q;
        if(grid[0][0])
            return -1;
        q.push({0,0,1});
        int currx, curry, dist;
        while(!q.empty())
        {
            currx = q.front()[0];
            curry = q.front()[1];
            dist = q.front()[2];
            if(currx == n-1 && curry == n-1)
                return dist;
            q.pop();
            if(grid[currx][curry])
                continue;
            grid[currx][curry] = 1;
            if(currx < n - 1 && !grid[currx + 1][curry])
            {
                q.push({currx + 1, curry, dist + 1});
            }  
            if(curry < n - 1 && currx < n - 1 && !grid[currx + 1][curry + 1])
            {
                q.push({currx + 1, curry + 1, dist + 1});
            }   
            if(curry > 0 && currx < n - 1 && !grid[currx + 1][curry - 1])
            {
                q.push({currx + 1, curry - 1, dist + 1});
            }   
            if(currx > 0 && !grid[currx - 1][curry ])
            {
                q.push({currx - 1, curry, dist + 1});
            }
            if(curry < n - 1 && currx > 0 && !grid[currx - 1][curry + 1])
            {
                q.push({currx - 1, curry + 1, dist + 1});
            }  
            if(curry > 0 && currx > 0 && !grid[currx - 1][curry - 1])
            {
                q.push({currx - 1, curry - 1, dist + 1});
            }
            if(curry < n - 1 && !grid[currx][curry + 1])
            {
                q.push({currx, curry + 1, dist + 1});
            }
            if(curry > 0 && !grid[currx][curry - 1])
            {
                q.push({currx, curry - 1, dist + 1});
            }
        }
        return -1;
    }
};
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