<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>

class Solution:
    def minPathSum(self, grid: List[List[int]]) -> int:
        ROWS, COLS = len(grid), len(grid[0])

        def inBound(r, c):
            return min(r, c) >= 0 and r < ROWS and c < COLS

        cost = grid[0][0]
        minHeap = [[cost, 0, 0]]
        visit = set()
        visit.add((0, 0))

        while minHeap:
            cost, row, col = heapq.heappop(minHeap)
            if row == ROWS - 1 and col == COLS - 1:
                return cost
            
            nei = [[row + 1, col], [row, col + 1]]
            for r2, c2 in nei:
                if inBound(r2, c2) and (r2, c2) not in visit:
                    newCost = cost + grid[r2][c2]
                    visit.add((r2, c2))
                    heapq.heappush(minHeap, (newCost, r2, c2))
22600