<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 {
    void Find_neigh(int i, int j, int n, vector<pair<int,int>> &ni)
    {
        if(i!=0 && j!=0)
            ni.push_back({i-1, j-1});
        if(i!=0)
            ni.push_back({i-1,j});
        if(i!=0 && j!=n-1)
            ni.push_back({i-1,j+1});
        if(i!=n-1 && j!=0)
            ni.push_back({i+1, j-1});
        if(i!=n-1)
            ni.push_back({i+1,j});
        if(i!=n-1 && j!=n-1)
            ni.push_back({i+1,j+1});
        if(j!=0)
            ni.push_back({i,j-1});
        if(j!=n-1)
            ni.push_back({i,j+1});
    }
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int n=grid.size();
        if(grid[0][0]==1)
            return -1;
        vector<vector<int>> answer(n, vector<int> (n,INT_MAX));
        answer[0][0]=1;
        queue<pair<int,pair<int,int>>> upcoming;
        upcoming.push({1,{0,0}});
        while(!upcoming.empty())
        {
            pair<int,int> node=upcoming.front().second;
            int dis=upcoming.front().first;
            upcoming.pop();
            vector<pair<int,int>> neighbours;
            Find_neigh(node.first, node.second, n, neighbours);
            for(auto nigh:neighbours)
            {
                if(answer[nigh.first][nigh.second]>dis+1 && grid[nigh.first][nigh.second]==0)
                {
                    answer[nigh.first][nigh.second]=dis+1;
                    upcoming.push({dis+1, nigh});
                }
            }
        }
        return answer[n-1][n-1]==INT_MAX?-1:answer[n-1][n-1];
    }
};
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