<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])
        start=grid[0][0]
        least_path=0
        dp=[[-1 for i in range(n+1)] for j in range(m+1)]
        def recur(indexes):
            if dp[indexes[0]][indexes[1]]!=-1:
                return dp[indexes[0]][indexes[1]]
            if indexes[0]==m-1 and indexes [1]==n-1:
                return grid[indexes[0]][indexes[1]]
            elif indexes[0]==m-1:
                return grid[indexes[0]][indexes[1]]+recur([indexes[0],indexes[1]+1])
            elif indexes[1]==n-1:
                return grid[indexes[0]][indexes[1]]+recur([indexes[0]+1,indexes[1]])
            right_path=recur([indexes[0]+1,indexes[1]])
            bottom_path=recur([indexes[0],indexes[1]+1])
            least_path=min(right_path,bottom_path)
            dp[indexes[0]][indexes[1]]=grid[indexes[0]][indexes[1]]+least_path
            return grid[indexes[0]][indexes[1]]+least_path
        var=recur([0,0])
        return var

20400