<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) {
        //BFS from (0,0) to search (m-1, n-1)
        int m = grid.size();
        int n = grid[0].size();
        queue<pair<vector<int>, int>> q;
        vector<vector<int>> visited(m, vector<int>(n,0));
        if(grid[0][0] == 0){
            q.push({{0,0}, 0});
            visited[0][0] = 1;
        }

        while(q.size()){
            auto curr = q.front();
            q.pop();
            int x = curr.first[0];
            int y = curr.first[1];

            int depth = curr.second;
            //if reached destination return answer
            if(x == m-1 && y == n-1)
                return depth+1;

            //check connected
            for(int a = -1; a <= 1; ++a){
                for(int b = -1; b <= 1; ++b){
                    //not the center cell
                    if(a == b && a == 0) continue;

                    int ptX = x + a;
                    int ptY = y + b;
                    //out of range
                    if(ptX >= m || ptY >= n || ptX < 0 || ptY < 0) continue;

                    //only if 0
                    if(grid[ptX][ptY] == 1) continue;

                    //check if visited
                    if(visited[ptX][ptY]) continue;
                    
                    //if reached destination return answer
                    if(ptX == m-1 && ptY == n-1)
                        return depth+2;

                    //now push
                    q.push({{ptX, ptY}, depth+1});

                    //mark visited
                    visited[ptX][ptY] = 1;
                }
            }
        }

        return -1;
    }
};
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