<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 m = grid.size();
        int n = grid[0].size();
        
        if(grid[m-1][n-1] == 1 || grid[0][0] == 1) return -1;
        
        //BFS that prioritizes lower left
        queue<tuple<int, int, int>> q;  //stores coordinates + length
        q.push({0,0,1});
        
        while(!q.empty()){
            auto t = q.front();
            q.pop();
            int r = get<0>(t);
            int c = get<1>(t);
            int len = get<2>(t);
            
            cout << "[" << r << "," << c << "]:" << len << endl;
            
            if(r == m-1 && c == n-1){
                return len;
            }
            if(grid[r][c] == 1){
                continue;
            }
            grid[r][c] = 1;
            
            if(r > 0 && c > 0 && grid[r-1][c-1] == 0) q.push({r-1,c-1,len+1});
            if(r > 0 && grid[r-1][c] == 0) q.push({r-1,c,len+1});
            if(r > 0 && c < n-1 && grid[r-1][c+1] == 0) q.push({r-1,c+1,len+1});
            if(c > 0 && grid[r][c-1] == 0)  q.push({r,c-1,len+1});
            if(c < n-1 && grid[r][c+1] == 0)  q.push({r,c+1,len+1});
            if(r < m-1 && grid[r+1][c] == 0) q.push({r+1,c,len+1});
            if(r < m-1 && c > 0 && grid[r+1][c-1] == 0) q.push({r+1,c-1,len+1});
            if(r < m-1 && c < n-1 && grid[r+1][c+1] == 0) q.push({r+1,c+1,len+1});
        }
        return -1;
    }
};
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