<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>>& arr) {

    if(arr[0][0] == 1)
    return -1;

    int n = arr.size();

      map<pair<int,int>, bool> visited;
      map<pair<int,int>, int> distance;
      queue<pair<int,int>> q;
      q.push({0,0});
      visited[{0,0}] = true;
      distance[{0,0}] = 1;
      while(!q.empty())
      {
        auto temp = q.front();
        int i = temp.first;
        int j = temp.second;
        q.pop();

        if(j < n-1 && arr[i][j+1] == 0 && !visited[{i,j+1}])
        {
            distance[{i,j+1}] = distance[{i,j}]+1;
            visited[{i,j+1}] = true;
            q.push({i,j+1});
        }

        if(j < n-1 && i<n-1 && arr[i+1][j+1] == 0 && !visited[{i+1,j+1}])
        {
            distance[{i+1,j+1}] = distance[{i,j}]+1;
            visited[{i+1,j+1}] = true;
            q.push({i+1,j+1});
        }


        if(i < n-1 && arr[i+1][j] == 0 && !visited[{i+1,j}])
        {
            distance[{i+1,j}] = distance[{i,j}]+1;
            visited[{i+1,j}] = true;
            q.push({i+1,j});
        }

        if(i<n-1 && j>0 && arr[i+1][j-1] == 0 && !visited[{i+1,j-1}])
        {
            distance[{i+1,j-1}] = distance[{i,j}]+1;
            visited[{i+1,j-1}] = true;
            q.push({i+1,j-1});
        }

        if(j > 0 && arr[i][j-1] == 0 && !visited[{i,j-1}])
        {
            distance[{i,j-1}] = distance[{i,j}]+1;
            visited[{i,j-1}] = true;
            q.push({i,j-1});
        }


        if(j >0 && i>0 && arr[i-1][j-1] == 0 && !visited[{i-1,j-1}])
        {
            distance[{i-1,j-1}] = distance[{i,j}]+1;
            visited[{i-1,j-1}] = true;
            q.push({i-1,j-1});
        }


        if(i > 0 && arr[i-1][j] == 0 && !visited[{i-1,j}])
        {
            distance[{i-1,j}] = distance[{i,j}]+1;
            visited[{i-1,j}] = true;
            q.push({i-1,j});
        }


        if( i > 0 && j < n-1 && arr[i-1][j+1] == 0 && !visited[{i-1,j+1}])
        {
            distance[{i-1,j+1}] = distance[{i,j}]+1;
            visited[{i-1,j+1}] = true;
            q.push({i-1,j+1});
        }


      }

      if(visited[{n-1,n-1}] == false)
      return -1;

      return distance[{n-1,n-1}];
        
    }
};
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