<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 bfs(int i,int j,vector<vector<int>>&grid,vector<vector<int>>&vis){
        queue<pair<int,pair<int,int>>>pq;
        int m=grid.size();
        int n=grid[0].size();
        if(i==0 && j==0)return 1;
        pq.push({1,{i,j}});
        while(!pq.empty()){
            auto mp=pq.front();
            pq.pop();
            int cnt=mp.first;
            int cr=mp.second.first;
            int cc=mp.second.second;
            
            for(int dr=-1;dr<=1;dr++){
                for(int dc=-1;dc<=1;dc++){
                    int nr=cr+dr;
                    int nc=cc+dc;
                    if(nr>=0 && nr<m && nc>=0 && nc<n && grid[nr][nc]==0 && !vis[nr][nc]){
                        if(nr==0 && nc==0)return cnt+1;
                        vis[nr][nc]=1;
                        pq.push({cnt+1,{nr,nc}});
                    }
                }
            }
        }
        return -1;
    }
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int m=grid.size(),n=grid[0].size();
        
        vector<vector<int>>vis(m,vector<int>(n,0));
        if(grid[m-1][n-1]==0) {
            vis[m-1][n-1]=1;
            return bfs(m-1,n-1,grid,vis);
        }


        return -1;

    }
};
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