<p>Given a <code>triangle</code> array, return <em>the minimum path sum from top to bottom</em>.</p>

<p>For each step, you may move to an adjacent number of the row below. More formally, if you are on index <code>i</code> on the current row, you may move to either index <code>i</code> or index <code>i + 1</code> on the next row.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
<strong>Output:</strong> 11
<strong>Explanation:</strong> The triangle looks like:
   <u>2</u>
  <u>3</u> 4
 6 <u>5</u> 7
4 <u>1</u> 8 3
The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> triangle = [[-10]]
<strong>Output:</strong> -10
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= triangle.length &lt;= 200</code></li>
	<li><code>triangle[0].length == 1</code></li>
	<li><code>triangle[i].length == triangle[i - 1].length + 1</code></li>
	<li><code>-10<sup>4</sup> &lt;= triangle[i][j] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you&nbsp;do this using only <code>O(n)</code> extra space, where <code>n</code> is the total number of rows in the triangle?
"""
[
    [2],
    [3,4],
    [6,5,7],
    [4,1,8,3]
]
{ 
    3,0: 4
    3,1: 1
    3,2: 8
    3,3: 3
    2,0: 6 + min(4, 1)
    2,1: 5 + min(1, 8)
}
"""

class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        n = len(triangle)

        if n == 0:
            return 0
        if n == 1:
            return triangle[0][0]

        dp = {}
        for i, val in enumerate(triangle[-1]):
            dp[(n-1, i)] = val
        
        for y in range(n-2, -1, -1):
            for x, val in enumerate(triangle[y]):
                dp[(y, x)] = val + min(
                    dp[(y+1, x)],
                    dp[(y+1, x+1)],
                )
        
        return dp[(0,0)]
19500