<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) {
        int n = grid.size(), m = grid[0].size();
        if(grid[0][0]==1 || grid[n-1][m-1]==1) return -1;
        //i,j : m*i + j
        vector<vector<int>> adj(n*m);
        
        for(int i=0;i<n;++i){
            for(int j=0;j<m;++j){
                if(grid[i][j]) continue;
                for(int ni = i; ni<=i+1 ;++ni){
                    for(int nj=j-1; nj <= j+1 ;++nj){
                        if(ni==i && nj==j) continue;
                        if(ni<0 || ni>=n || nj<0 || nj>=m) continue;
                        if(grid[ni][nj]==0){
                            adj[i*m + j].push_back(ni*m + nj);
                            adj[ni*m + nj].push_back(i*m + j);
                        }
                    }
                }
            }
        }//adj list is complete

        vector<int> dist(n*m, 1e9);
        queue<int> q;

        q.push(0);
        dist[0] = 0;

        while(!q.empty()){
            int node = q.front();
            q.pop();

            for(int x : adj[node]){
                if(dist[node] + 1 < dist[x]){
                    dist[x] = 1 + dist[node];
                    q.push(x);
                }
            }
        }

        if(dist[n*m-1]==1e9) return -1;
        else return dist[n*m-1]+1;
    }
};
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