<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])
        if n==1 or m==1:
            return sum(sum(x) for x in grid)
        r = [[201]*(n+1)]*(m+1)
        
        for i in range(m+1):
            r[i][0] = 0
        r[0] = [0]*(n+1)
        # r[0][0] = 0
        for i in range(1,m+1):
            for j in range(1,n+1):
                if i==1:
                    r[i][j] = r[i][j-1] + grid[i-1][j-1] 
                elif j==1:
                    r[i][j] = r[i-1][j] + grid[i-1][j-1] 
                elif i>1 and j>1:
                    r[i][j] = min(r[i][j-1],r[i-1][j]) + grid[i-1][j-1] 
        print(r)
        return int(r[m][n])

17700