<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 {
private:
    void traverse(int row, int col, queue<pair<int, int>>& que, int n,vector<vector<int>>& grid){
        const vector<pair<int, int>> directions = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}, {1, 1}, {-1, -1}, {-1, 1}, {1, -1}};
        for(const auto& dir : directions){
            int newRow = row + dir.first;
            int newCol = col + dir.second;
            if(newRow >= 0 && newCol >= 0 && newRow < n and newCol < n && grid[newRow][newCol] != 1){
            que.push({newRow, newCol});
            }
        }
    }
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        
        int n = grid.size();
        queue<pair<int, int>> que;

        set<pair<int, int>> visitedCells;

        // set "1" cell as visited to avoid it
        for(int row = 0; row < n; row++){
            for(int col = 0; col < n; col++){
                if(grid[row][col] == 1)
                    visitedCells.insert({row, col});
            }
        }

        que.push({0, 0}); // push the source 
        int level = 1;
        // bfs
        while(!que.empty()){
            int currentLevelSize = que.size();
            while(currentLevelSize--){
                pair<int, int> currentCell = que.front();
                que.pop();
                int currentCellRow = currentCell.first;
                int currentCellCol = currentCell.second;
                if (visitedCells.find({currentCellRow, currentCellCol})!= visitedCells.end())
                    continue; // if visited skip 
                else if(currentCell.first == n-1 && currentCellCol == n - 1) // target found
                    return level;
                
                // else mark visited
                visitedCells.insert({currentCellRow, currentCellCol});

                // traverse all direction and push to queue
                traverse(currentCellRow, currentCellCol, que, n, grid);
            }
            level++;
        }

        return -1;
    }
};
71580