<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:
        

        #top down recursion
        #what do you want to store in the dp?
        #write out the recursion and find out
        rows, cols = len(grid), len(grid[0])

        #idea -> store r, c -> minm seen path sum 
        dp = defaultdict()

        def recurse(r, c):
            # print(r,c)
            #reached bottom right
            if r == rows - 1 and c == cols -1:
                return grid[r][c]

            #exhausted rows or cols
            #need not consider this path so return inf, coz minm
            if r == rows or c == cols:
                return float("inf")
            
            if (r, c) in dp:
                return dp[(r, c)]
            
            #directions only down or right
            directions = [[0, 1], [1, 0]]
            #without loop -> way faster?
            result = grid[r][c] + min(recurse(r+1, c), recurse(r, c+1))
            # print(r, c, grid[r][c])
            dp[(r, c)] = result
            return result
        # print(dp)
        return recurse(0, 0)


23400