<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:
        m=len(grid)
        n=len(grid[0])
        dp={}
        dp[(0,0)]=grid[0][0]
        for r in range(m):
            for c in range(n):
              if r==0 and c==0: dp[(0,0)]=grid[0][0]
              else:
                up=float('inf')
                left=float('inf')
                if r>0: up=dp[(r-1,c)]+grid[r][c]
                if c>0: left=dp[(r,c-1)]+grid[r][c]
                dp[(r,c)]=min(up,left)
        return dp[(m-1,n-1)]
23000