<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] == 1)
            return -1;
        
        list <tuple<int,int,int>> q;
        q.emplace_back (0, 0, 1);
        
        vector <pair<int,int>> drc = {{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{-1,1},{1,-1},{1,1}};
        
        // mistake: there is no need to worry about diagonal paths separately.
        // ie there is no need to prioritize diagonal transitions other vertical/horizontal,
        // using a priority queue or deque.
        // because diagonal transitions are visited first, visited[][] will already be -1,
        // when we get there thru a horizontal/vertical transition.
        // there is no need to worry about duplicates in the queue either, we are just not going to
        // enter diagonally visited cells again in the queue.
        
        while (!q.empty())
        {
            auto [r, c, d] = q.front();
            q.pop_front();

            if (r == grid.size()-1 && c == grid[0].size()-1)
                return d;
            
            // if (grid[r][c] == -1)
            //     continue;
            // grid[r][c] = -1;
            
            for (int i = 0; i < drc.size(); i++)
            {
                auto [dr,dc] = drc[i];
                auto nr = r + dr, nc = c + dc;
                
                if (nr < 0 || nr >= grid.size() || nc < 0 || nc >= grid[0].size() || grid[nr][nc] != 0)
                    continue;
                grid[nr][nc] = -1;
                
                q.emplace_back(nr, nc, d+1);
            }
        }
        
        return -1;
    }
};
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