<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:
        cache = {(y, x) :float("inf") for y in range(len(grid)) for x in range(len(grid[0]))}
        pos = (0, 0)
        cache[pos] = grid[0][0]
        queue = [pos]

        while queue:
            row, column = queue.pop(0)
            pos = (row, column)
            for dr, dc in [(0, 1), (1, 0)]:
              new_pos = (row + dr, column + dc)
              if 0<= row + dr < len(grid) and 0<= column + dc < len(grid[0]):
                if cache[new_pos] > grid[row+dr][column+dc] + cache[pos]:
                  cache[new_pos] = grid[row+dr][column+dc] + cache[pos]
                if new_pos not in queue:
                  queue.append(new_pos)
        return cache[(len(grid))-1, len(grid[0])-1]
        
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