<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:
    void path(vector<vector<int>>& grid, int i, int j, int len, int &ans, int &flag)
    {
        if(i == grid.size()-1 && j == grid.size()-1)
        {
            flag = 1;
            ans = min(ans, len);
        }
        if(i+1 < grid.size() && grid[i+1][j] == 0) path(grid, i+1, j, len+1, ans,flag);
        if(j+1 < grid.size() && grid[i][j+1] == 0) path(grid, i, j+1, len+1, ans, flag);
        if(i+1 < grid.size() && j+1 < grid.size() && grid[i+1][j+1] == 0) path(grid, i+1, j+1, len+1, ans, flag);
    }
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        vector<int> adj[grid.size()*grid.size()];
        int n = grid.size();
        if(grid[0][0] != 0) return -1;
        for(int i = 0; i < grid.size(); i++)
        {
            for(int j = 0; j < grid.size(); j++)
            {
                int node = i*n + j;
                if(i+1 < n && grid[i+1][j] == 0) adj[node].push_back((i+1)*n + j);
                if(j+1 < n && grid[i][j+1] == 0) adj[node].push_back(i*n + j+1);
                if(i+1 < n && j+1 < n && grid[i+1][j+1] == 0) adj[node].push_back((i+1)*n + j+1);
                if(i-1 >= 0 && grid[i-1][j] == 0) adj[node].push_back((i-1)*n + j);
                if(j-1 >= 0 && grid[i][j-1] == 0) adj[node].push_back(i*n + j - 1);
                if(i-1 >= 0 && j-1 >= 0 && grid[i-1][j-1] == 0) adj[node].push_back((i-1)*n + j -1);
                if(i-1 >= 0 && j+1 < n && grid[i-1][j+1] == 0) adj[node].push_back((i-1)*n + j+1);
                if(i+1 < n && j-1 >= 0 && grid[i+1][j-1] == 0) adj[node].push_back((i+1)*n + j - 1);
            }
        }
        vector<int> ans(n*n, -1);
        ans[0] = 1;
        queue<pair<int, int>> q;
        q.push({0, 1});
        vector<int> vst(n*n, 0);
        vst[0] = 1;
        while(!q.empty())
        {
            int node = q.front().first;
            int len = q.front().second;
            q.pop();
            for(auto it : adj[node])
            {
                if(vst[it] == 0)
                {
                    vst[it] = 1;
                    q.push({it, len+1});
                    ans[it] = 1+len;
                }
            }
        }
        return ans[n*n-1];
    }
};
71580