<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>

# define pii pair<int , int>

class Solution {
public:

    int res = INT_MAX;
    vector<vector<int>> offset = { {1 , 1} , {1 , -1} , {-1 , 1} , {-1 , -1} , {1, 0} , {0 , 1} , {-1 , 0} , {0 , -1}};

    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        queue<int> q;
        int n = grid.size();

        if (grid[0][0] == 1){
            return -1;
        }
        else if (grid[0][0] == 0 && n == 1){
            return 1;
        }

        q.push(0);
        grid[0][0] = 2;
        int counter = 1;


        while (!q.empty()){

            int size = q.size();

            for (int k = 0 ; k < size ; k++){
                int curr = q.front();
                q.pop();
                int i = curr / n , j = curr % n;


                for (auto value:offset){
                    int new_row = i + value[0] , new_col = j + value[1];
                    // do the check to ensure that its valid

                    if (new_row >= 0 && new_row < n && new_col >= 0 && new_col < n && (grid[new_row][new_col] == 0) && grid[new_row][new_col] == 0){

                        if (new_row == (n - 1) && new_col == (n - 1)){
                            res = counter + 1;
                            return res;
                        }

                        q.push(n*new_row+new_col);
                        grid[new_row][new_col] = 2;
                    }
                }
            }
            counter++;

        }

        return -1;
    }

    

};
80091