<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?
class Solution:
    def minimumTotal(self, triangle: list[list[int]]) -> int:
        n = len(triangle)
        # ans1 = self.simpleRecursion(triangle, 0, 0)

        dp = [[-1] * (i + 1) for i in range(n)]
        ans2 = self.bottonUp(triangle, 0, 0, dp)
        return ans2

    def simpleRecursion(self, triangle, row, col):
        if row == len(triangle) - 1:
            return triangle[row][col]

        down = triangle[row][col] + self.simpleRecursion(triangle, row + 1, col)
        diagonal = triangle[row][col] + self.simpleRecursion(triangle, row + 1, col + 1)

        return min(down, diagonal)


    def bottonUp(self, triangle, row, col, dp):
        if row == len(triangle) - 1:
            return triangle[row][col]
        if dp[row][col] != -1:
            return dp[row][col]
        down = triangle[row][col] + self.bottonUp(triangle, row + 1, col, dp)
        diagonal = triangle[row][col] + self.bottonUp(triangle, row + 1, col + 1, dp)
        dp[row][col] = min(down, diagonal)
        return dp[row][col]
        
18000