<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 len(grid) == 1:
            return sum(grid[0])
        if len(grid[0]) == 1:
            total = 0
            for row in grid:
                total += row[0]
            return total
        pos = defaultdict(lambda: float("inf"))
        stack = [(0, 0, grid[0][0])]
        while stack:
            curr_x, curr_y, path_total = stack.pop()
            init_moves = [(curr_x+1, curr_y), (curr_x, curr_y+1)]
            cont = []
            for x, y in init_moves:#check if its a valid spot to move to
                if 0<= x < len(grid) and 0<=y<len(grid[0]):
                    cont.append((x, y))
                else:
                    continue
            for x, y in cont:#check if its a value less than an existing position in pos
                value = path_total + grid[x][y]
                if value < pos[(x, y)]:
                    pos[(x, y)] = value
                    stack.append((x, y, value))
                else:
                    continue
        #print(pos)
        #print(len(grid), len(grid[0]))
        return pos[(len(grid)-1,len(grid[0])-1)]

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