<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 = [["a" for _ in range(n)] for _ in range(m)]
        
        
        def fun(i,j):
            if  i==m-1 and j==n-1:
                return grid[i][j]
            if i>=m or j>=n:
                return float('inf')  
               
            
            if dp[i][j]!="a":
                
                return dp[i][j]
            opt=opt2=float('inf')    
            if j<n-1:

              opt=fun(i,j+1)
            if i<m-1:  
              opt2=fun(i+1,j)  
            
            dp[i][j]=min(opt,opt2)+grid[i][j]

            return dp[i][j]
           
        return fun(0,0) 
        
        
18700