<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>>& g, int steps = 0) {
      queue<pair<int, int>> q;
      q.push({ 0, 0 });
      while (!q.empty()) {
        ++steps;
        queue<pair<int, int>> q1;
        while (!q.empty()) {
          auto c = q.front();
          q.pop();
          if (exchange(g[c.first][c.second], 1) == 1) continue;
          // g[c.first][c.second] = 1;
          if (c.first == g.size() - 1 && c.second == g.size() - 1) 
            return steps;

          for (auto i = c.first - 1; i <= c.first + 1; ++i)
            for (auto j = c.second - 1; j <= c.second + 1; ++j)
              if (i != c.first || j != c.second) {
                if (i >= 0 && j >= 0 && i < g.size() && j < g.size() 
                    && !g[i][j]) {
                  q1.push({ i, j });
                }
              }
        }
        swap(q, q1);
      }
      return -1;
    }
};
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