<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 row[8] = {-1, 1, 0, 0, -1, 1, -1, 1};
    int col[8] = {0, 0, 1, -1, -1, 1, 1, -1};
    vector<pair<int,int>> adj(int i,int j,vector<vector<int>>& grid){
        vector<pair<int,int>> ans;
        for(int k=0;k<8;k++){
            int r = i + row[k];
            int c = j + col[k];
            if(r>=0 && r<grid.size() && c>=0 && c<grid[0].size())
                ans.push_back({r,c});
        }
        return ans;
    }
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        if(grid[0][0]==1)
            return -1;
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> vis(n,vector<int>(m,0));
        queue<pair<int,int>> q;
        q.push({0,0});
        int dist = 1;
        while(!q.empty()){
            int size = q.size();
            for(int k=0;k<size;k++){
                int i = q.front().first;
                int j = q.front().second;
                q.pop();
                if(i==n-1 && j==m-1)
                    return dist;
                vector<pair<int,int>> neighbour = adj(i,j,grid);
                for(auto it: neighbour){
                    if(grid[it.first][it.second]==0 & vis[it.first][it.second]==0){
                        vis[it.first][it.second] = 1;
                        q.push({it.first,it.second});
                    }
                }
            }
            dist++;
        }
        return -1;
    }
};
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