<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();
        queue<pair<int, int>> q;
        q.push(make_pair(0,0));
        int steps = 1;

        while(!q.empty()) {
            vector<pair<int, int>> curr;
            
            while(!q.empty()) {
                pair<int, int> back = q.front();
                q.pop();

                if(back.first<0 || back.first>=n || back.second<0 || back.second>=n || grid[back.first][back.second] != 0) {
                    continue;
                }
                if(back.first == n-1 && back.second == n-1) {
                    return steps;
                }
                grid[back.first][back.second] = -1;
                curr.push_back(make_pair(back.first-1,back.second-1));
                curr.push_back(make_pair(back.first,back.second-1));
                curr.push_back(make_pair(back.first+1,back.second-1));
                curr.push_back(make_pair(back.first+1,back.second));
                curr.push_back(make_pair(back.first+1,back.second+1));
                curr.push_back(make_pair(back.first,back.second+1));
                curr.push_back(make_pair(back.first-1,back.second+1));
                curr.push_back(make_pair(back.first-1,back.second));
            }
            cout<<endl;
            int x = curr.size();
            for(int i = 0 ; i<x ; i++) {
                q.push(curr[i]);
            }
            steps++;
        }

        return -1;
    }
};
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