<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) {        
        // NeetCodeIO
        // Simple BFS
        
        // Dijkstra's with unit weight on implicit graph
        // Time: O(n.n.logn)
        // Space: O(m.n[dist] + m.n[queue])
        
        int path = INT_MAX;
        int n = grid.size();
 
        if (grid[0][0] == 1 or grid[n-1][n-1] == 1)
            return -1;
        
        set<pair<int,int>> visited;
        queue<vector<int>> pq;
        
        pq.push({1,0,0}); // {d,i,j}
        visited.insert({0,0});
        while (!pq.empty()) {
            auto u = pq.front();
            int &d = u[0], &i = u[1], &j = u[2];
            pq.pop();
            
            // Process the node
            if (i == n-1 and j == n-1)
                return d;
            
            // Explore neigbours
            vector<pair<int,int>> neighbors = {
                {i-1,j-1},
                {i-1,j},
                {i-1,j+1},
                {i,j-1},
                {i,j+1},
                {i+1,j-1},
                {i+1,j},
                {i+1,j+1},
            };
            for (auto& [n_i, n_j]: neighbors) {
                if (n_i >= 0 and n_j >= 0 and n_i < n and n_j < n and 
                    grid[n_i][n_j] != 1 and visited.find({n_i,n_j}) == visited.end()) {
                    pq.push({d+1,n_i,n_j});
                    visited.insert({n_i, n_j});
                }
            }
        }
        
        return -1;        
    }
};
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