<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Bitmask on j2234's log</title><link>https://log.j2234.cc/tags/bitmask/</link><description>Recent content in Bitmask on j2234's log</description><generator>Hugo -- gohugo.io</generator><language>en</language><copyright>© 2026</copyright><lastBuildDate>Sat, 04 Apr 2026 21:02:54 +0000</lastBuildDate><atom:link href="https://log.j2234.cc/tags/bitmask/index.xml" rel="self" type="application/rss+xml"/><item><title>1915. Number of Wonderful Substrings</title><link>https://log.j2234.cc/posts/velog/016-1915.-number-of-wonderful-substrings/</link><pubDate>Tue, 30 Apr 2024 21:40:10 +0000</pubDate><guid>https://log.j2234.cc/posts/velog/016-1915.-number-of-wonderful-substrings/</guid><description>&lt;figure&gt;&lt;img
 class="my-0 rounded-md"
 loading="lazy"
 decoding="async"
 fetchpriority="auto"
 alt=""
 width="165"
 height="172"
 src="https://log.j2234.cc/posts/velog/016-1915.-number-of-wonderful-substrings/img/image-01.webp"
 srcset="https://log.j2234.cc/posts/velog/016-1915.-number-of-wonderful-substrings/img/image-01.webp 800w, https://log.j2234.cc/posts/velog/016-1915.-number-of-wonderful-substrings/img/image-01.webp 1280w"
 sizes="(min-width: 768px) 50vw, 65vw"
 data-zoom-src="https://log.j2234.cc/posts/velog/016-1915.-number-of-wonderful-substrings/img/image-01.webp"&gt;&lt;/figure&gt;
&lt;p&gt;&lt;a href="https://leetcode.com/problems/number-of-wonderful-substrings/description/" target="_blank" rel="noreferrer"&gt;문제링크&lt;/a&gt;&lt;/p&gt;

&lt;h2 class="relative group"&gt;설명
 &lt;div id="설명" class="anchor"&gt;&lt;/div&gt;
 
 &lt;span
 class="absolute top-0 w-6 transition-opacity opacity-0 -start-6 not-prose group-hover:opacity-100 select-none"&gt;
 &lt;a class="text-primary-300 dark:text-neutral-700 !no-underline" href="#%ec%84%a4%eb%aa%85" aria-label="Anchor"&gt;#&lt;/a&gt;
 &lt;/span&gt;
 
&lt;/h2&gt;
&lt;blockquote&gt;&lt;p&gt;문자열 중에 최대 한 글자만 홀수 번 나타나면 &lt;code&gt;완벽한 문자열&lt;/code&gt;임
단어 &lt;code&gt;word&lt;/code&gt;가 주어졌을때, 여기에서 나타날 수 있는 모든 완벽한 문자열의 수를 구하셈 (world는 a~j만으로 구성)&lt;/p&gt;
&lt;/blockquote&gt;
&lt;h2 class="relative group"&gt;풀이
 &lt;div id="풀이" class="anchor"&gt;&lt;/div&gt;
 
 &lt;span
 class="absolute top-0 w-6 transition-opacity opacity-0 -start-6 not-prose group-hover:opacity-100 select-none"&gt;
 &lt;a class="text-primary-300 dark:text-neutral-700 !no-underline" href="#%ed%92%80%ec%9d%b4" aria-label="Anchor"&gt;#&lt;/a&gt;
 &lt;/span&gt;
 
&lt;/h2&gt;
&lt;ol&gt;
&lt;li&gt;단어를 순회하면서, 해당 단어를 비트로 바꿔 bitmask 집합에 넣는다.
이 비트마스를 key로 하는 dict에서, dict[key] 가 존재한다면, 그만큼 더한다
dict[key]를 1증가시킨다&lt;/li&gt;
&lt;/ol&gt;
&lt;blockquote&gt;&lt;p&gt;왜?
&lt;code&gt;XOR&lt;/code&gt;연산은, 같은 숫자가 짝수번 연산되면 모두 상쇄되어 사라진다.
비트마스크는 &lt;strong&gt;현재의 상태&lt;/strong&gt; 를 의미하는데, 이전에 똑같은 마스크가 존재했다는 뜻은, 그 사이에 있는 모든 단어들은 짝수번 등장한다는 뜻이다 &lt;em&gt;근데 왜 dict[key]만큼 더함??&lt;/em&gt;&lt;/p&gt;</description><media:content xmlns:media="http://search.yahoo.com/mrss/" url="https://log.j2234.cc/posts/velog/016-1915.-number-of-wonderful-substrings/featured.webp"/></item></channel></rss>