<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>

#from collections import defaultdict
from functools import cache

class Solution:
    def minPathSum(self, grid: List[List[int]]) -> int:
        # (row, col) grid[row][col]
        m = len(grid)
        n = len(grid[0])
        start = (0,0)
        end = ( m-1, n-1 )

        @cache
        def best_path(posn):
            (r, c) = posn
            cost = grid[r][c]
            print(posn, cost)

            if posn == start:
                return cost # only start node

            # boundary
            if r < 0 or c < 0:
                return inf
            
            up = (r-1, c)
            left = (r, c-1)
            return cost + min(best_path(up), best_path(left))

        return best_path(end)



"""
values 0-200

[[1,3,1],
 [1,5,1],
 [4,2,1]]

backtracking
dp
abandon too long
can't do exhaustive
A* can't manhattan
q

up and left moves?
    Note: You can only move either down or right at any point in time.
"""
24500