<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom" xmlns:content="http://purl.org/rss/1.0/modules/content/"><channel><title>LeetCode 338 on Jeanphilo Blog</title><link>https://shio-chan-dev.github.io/jeanblog/tags/leetcode-338/</link><description>Recent content in LeetCode 338 on Jeanphilo Blog</description><generator>Hugo -- 0.165.0</generator><language>en-us</language><lastBuildDate>Wed, 15 Jul 2026 00:00:00 +0800</lastBuildDate><atom:link href="https://shio-chan-dev.github.io/jeanblog/tags/leetcode-338/index.xml" rel="self" type="application/rss+xml"/><item><title>LeetCode 338: Counting Bits by Reusing Smaller Results</title><link>https://shio-chan-dev.github.io/jeanblog/alg/leetcode/hot100/338-counting-bits/</link><pubDate>Wed, 15 Jul 2026 00:00:00 +0800</pubDate><guid>https://shio-chan-dev.github.io/jeanblog/alg/leetcode/hot100/338-counting-bits/</guid><description>Start by counting every value in 0..n independently, then derive answer[i] = answer[i &amp;amp; (i - 1)] + 1 for an O(n) solution.</description></item></channel></rss>