<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>

from collections import deque
import heapq
class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:   
        # using dijkastra
        # bcz distance is constant and in minheap distance will always pushed in sorted order.
        n = len(grid)-1
        if grid[0][0] == 1 or grid[n][n] ==1:
            return -1
            
        minheap = []
        heapq.heapify(minheap)
        heapq.heappush(minheap,[1,0,0]) # distance,i,j
        directions = [(0,-1),(-1,-1),(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1)]
        # which ever reaches first return their distance.
        # result array.
        result = [[float('inf') for _ in range(len(grid))]for _ in range(len(grid))]
        result[0][0] = 1

        while minheap: 
            d,i,j = heapq.heappop(minheap)
            for x,y in directions:
                dx = i+x
                dy = j+y    
                if 0 <= dx <= n and 0<=dy <= n and grid[dx][dy] == 0:
                    if result[dx][dy] > d+1: # checking if distance is less than previous distance.
                        result[dx][dy] = d+1
                        heapq.heappush(minheap,[d+1,dx,dy])

        if result[n][n] == float('inf'):   
            return -1 # srounding of source or destination are 1.
        return result[n][n]

17300