<p>Given an integer <code>rowIndex</code>, return the <code>rowIndex<sup>th</sup></code> (<strong>0-indexed</strong>) row of the <strong>Pascal&#39;s triangle</strong>.</p>

<p>In <strong>Pascal&#39;s triangle</strong>, each number is the sum of the two numbers directly above it as shown:</p>
<img alt="" src="https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif" style="height:240px; width:260px" />
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> rowIndex = 3
<strong>Output:</strong> [1,3,3,1]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> rowIndex = 0
<strong>Output:</strong> [1]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> rowIndex = 1
<strong>Output:</strong> [1,1]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= rowIndex &lt;= 33</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you optimize your algorithm to use only <code>O(rowIndex)</code> extra space?</p>

class Solution:
    def getRow(self, rowIndex: int) -> int:
        if rowIndex == 0:
            return [1]
        prev_row = [1]
        for _ in range(1, rowIndex + 1):
            curr_row = [1]
            for j in range(1, len(prev_row)):
                # Compute the current element using the two elements above
                curr_row.append(prev_row[j - 1] + prev_row[j])
            curr_row.append(1)
            prev_row = curr_row
        return prev_row

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