<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:
    vector<vector<int>> dir = {{0, -1}, {-1, 0}, {0, 1}, {1, 0}, {1, -1}, {-1, 1}, {1, 1}, {-1, -1}};
    
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int row = grid.size();
        int col = grid[0].size();
        
        if(grid[0][0] == 1)
            return -1;
        
        queue<pair<int, int>> q;
        map<pair<int, int>, int> visited;
        q.push({0, 0});
        visited[{0, 0}] = 1;
        
        bool reached = false;
        int dist = numeric_limits<int>::max();
        
        while(!q.empty()){
            auto curr = q.front();
            q.pop();
            
            if(curr.first == row - 1 && curr.second == col - 1){
                reached = true;
                dist = visited[{curr.first, curr.second}];
            }
            
            for(int i = 0; i < 8; i++){
                int next_r = curr.first + dir[i][0];
                int next_c = curr.second + dir[i][1];
                
                if(next_r >= 0 && next_r < row &&
                  next_c >= 0 && next_c < col &&
                  grid[next_r][next_c] == 0 &&
                  visited.count({next_r, next_c}) == 0){
                    int new_dist = visited[{curr.first, curr.second}] + 1;
                    // Cost comparison is NOT needed, coz BFS gives shortest path when no weights are there
                    //if(visited.count({next_r, next_c}) == 0){// || visited[{next_r, next_c}] > new_dist){
                        visited[{next_r, next_c}] = new_dist;
                        q.push({next_r, next_c});
                    //}
                }
            }
        }
        
        
        return (reached ? dist : -1);
    }
};
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