<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:
        if not grid or not grid[0]:
            return 0
        
        m, n = len(grid), len(grid[0])
        
        # Initialize the dp table
        dp = [[0] * n for _ in range(m)]
        
        # Base case: starting point
        dp[0][0] = grid[0][0]
        
        # Fill the first row
        for j in range(1, n):
            dp[0][j] = dp[0][j-1] + grid[0][j]
        
        # Fill the first column
        for i in range(1, m):
            dp[i][0] = dp[i-1][0] + grid[i][0]
        
        # Fill the rest of the dp table
        for i in range(1, m):
            for j in range(1, n):
                dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1])
        
        # The bottom-right corner has the result
        return dp[m-1][n-1]
               

        
18500