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+<h2><a href="https://leetcode.com/problems/triangle">120. Triangle</a></h2><h3>Medium</h3><hr><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?