<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>

//URDL8
vector<int> dx = {-1, -1, -1, 0, +1, +1, +1, 0};
vector<int> dy = {-1, 0, +1, +1, +1, 0, -1, -1};
typedef pair<int, int> PII;

class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        //My approach : Model the matrix to a graph and apply bfs to find shortest path 
        //considering unit distances. (small dry run is enough)
        //T.C. = O(8*n*n) + O(8*n*n) = O(n*n)

        //edge case
        if(grid[0][0] == 1)
            return -1;

        int n = grid.size();
        vector<vector<int>> adj(n*n);
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                int node = i*n + j;
                for(int k = 0; k < 8; k++){
                    int x = i + dx[k];
                    int y = j + dy[k];
                    if(x >= 0 && x < n && y >= 0 && y < n){
                        if(grid[x][y] == 0){
                            int adjNode = x*n + y;
                            adj[node].push_back(adjNode); //the reverse edge will be added later automatically
                        }
                    }
                }
            }
        }

        //perform bfs (and keep track of levels => levels give the ans)
        queue<PII> q; //{node, level}
        vector<bool> visited(n*n, false);
        q.push({0, 1}); //starting with level 1 as question considers it as a move
        visited[0] = true;
        while(!q.empty()){
            PII front = q.front();
            q.pop();

            int node = front.first;
            int level = front.second;
            
            if(node == n*n - 1)
                return level;

            for(int neighbour : adj[node]){
                if(!visited[neighbour]){
                    q.push({neighbour, level+1});
                    visited[neighbour] = true;
                }
            }
        }

        return -1;
    }
};
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