<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>

import heapq

class Solution:
    def minPathSum(self, grid: List[List[int]]) -> int:
        n = len(grid)
        m = len(grid[0])
        q = [(0, (0, 0), [])]
        
        while q:
            last_score, (next_i, next_j), his = heapq.heappop(q)

            if not (0 <= next_i < n and 0 <= next_j < m):
                continue

            if grid[next_i][next_j] is False:  # visited
                continue

            score = last_score + grid[next_i][next_j]
            if next_i == n-1 and next_j == m-1: 
                # print(his)
                return score

            grid[next_i][next_j] = False

            heapq.heappush(q, (score, (next_i+1, next_j), his + [(next_i, next_j)]))
            # heapq.heappush(q, (score, (next_i-1, next_j), his + [(next_i, next_j)]))
            heapq.heappush(q, (score, (next_i, next_j+1), his + [(next_i, next_j)]))
            # heapq.heappush(q, (score, (next_i, next_j-1), his + [(next_i, next_j)]))

        return -1
21000