<p>Given a <code>m x n</code> <code>grid</code> filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path.</p>

<p><strong>Note:</strong> You can only move either down or right at any point in time.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/05/minpath.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[1,3,1],[1,5,1],[4,2,1]]
<strong>Output:</strong> 7
<strong>Explanation:</strong> Because the path 1 &rarr; 3 &rarr; 1 &rarr; 1 &rarr; 1 minimizes the sum.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,2,3],[4,5,6]]
<strong>Output:</strong> 12
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 200</code></li>
</ul>

from heapq import heappush, heappop
class Solution:
    def minPathSum(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        visited = set([(0, 0)])
        queue = [(grid[0][0], (0, 0))]
        directions = (
            (0, 1),
            (1, 0),
        )
        while queue:
            path_sum, cell = heappop(queue)
            if cell == (m-1, n-1):
                return path_sum
            for direction in directions:
                next_cell = (cell[0]+direction[0], cell[1]+direction[1])
                if (-1 < next_cell[0] < m) and (-1 < next_cell[1] < n) and next_cell not in visited:
                    curr_path_sum = path_sum + grid[next_cell[0]][next_cell[1]]
                    visited.add(next_cell)
                    heappush(queue, (curr_path_sum, next_cell))

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