<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>

// Dijkstra + PQ
class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int n = grid.size();
        if (n == 0 || grid[0][0] == 1 || grid[n-1][n-1] == 1) return -1;

        // 8 directions
        vector<vector<int>> dirs = {{1,0},{0,1},{-1,0},{0,-1},{-1,1},{1,-1},{1,1},{-1,-1}};

        // keep a min heap pq<{i*n+j, len}> that stores and pop the shortest path to node
        priority_queue<pair<int,int>, vector<pair<int,int>>, Compare> pq;
        pq.push({0, 1});  //start -> start = 1 cell (len is the visited cells in the path)

        // keep a visited vector to keep track of visited nodes
        vector<bool> visited(n*n, false);

        while (!pq.empty())
        {
            auto cur = pq.top();
            pq.pop();
            int x = cur.first / n;
            int y = cur.first % n;
            int len = cur.second;

            if (!visited[cur.first])
            {
                visited[cur.first] = true;
                if (x == n-1 && y == n-1) return len;  // found the path!

                for (auto dir : dirs)
                {
                    int nx = x + dir[0];
                    int ny = y + dir[1];

                    if (inBound(nx,ny,n) && !visited[nx*n+ny] && grid[nx][ny] == 0)
                    {
                        pq.push({nx*n+ny, len+1});
                    }
                }
            }
        }

        return -1;
    }
private:
    bool inBound(int i, int j, int n)
    {
        return i >= 0 && j >= 0 && i < n && j < n;
    }

    class Compare
    {
    public:
        bool operator()(pair<int,int> a, pair<int,int> b)
        {
            return a.second > b.second;
        }
    };
};
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