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Kotlin-based LeetCode algorithm problem solutions, regularly updated
120\. Triangle
Medium
Given a `triangle` array, return _the minimum path sum from top to bottom_.
For each step, you may move to an adjacent number of the row below. More formally, if you are on index `i` on the current row, you may move to either index `i` or index `i + 1` on the next row.
**Example 1:**
**Input:** triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
**Output:** 11
**Explanation:** The triangle looks like:
2
3 4
6 5 7
4 1 8 3
The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).
**Example 2:**
**Input:** triangle = [[-10]]
**Output:** -10
**Constraints:**
* `1 <= triangle.length <= 200`
* `triangle[0].length == 1`
* `triangle[i].length == triangle[i - 1].length + 1`
* -104 <= triangle[i][j] <= 104
**Follow up:** Could you do this using only `O(n)` extra space, where `n` is the total number of rows in the triangle?