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10.1 &nbsp; 이진 탐색
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10.2 &nbsp; 이진 탐색으로 삽입 위치 찾기
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10.3 &nbsp; 이진 탐색으로 경계 찾기
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10.4 &nbsp; 해시를 이용한 최적화 전략
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10.5 &nbsp; 탐색 알고리즘 다시 보기
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제 11 장 &nbsp; 정렬
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11.2 &nbsp; 선택 정렬
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11.3 &nbsp; 버블 정렬
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11.4 &nbsp; 삽입 정렬
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11.5 &nbsp; 퀵 정렬
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11.6 &nbsp; 병합 정렬
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11.8 &nbsp; 버킷 정렬
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11.9 &nbsp; 계수 정렬
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11.10 &nbsp; 기수 정렬
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11.11 &nbsp; 요약
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제 12 장 &nbsp; 분할 정복
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12.1 &nbsp; 분할 정복 알고리즘
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12.2 &nbsp; 분할 정복 탐색 전략
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12.3 &nbsp; 이진 트리 구성 문제
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12.4 &nbsp; 하노이의 탑 문제
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12.5 &nbsp; 요약
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제 13 장 &nbsp; 백트래킹
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13.4 &nbsp; n-퀸 문제
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14.1 &nbsp; 동적 계획법 첫걸음
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14.2 &nbsp; 동적 계획법 문제의 특성
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<h1 id="65">6.5 &nbsp; 연습 문제<a class="headerlink" href="#65" title="고유 링크">&para;</a></h1>
<h2 id="651">6.5.1 &nbsp; 개념 확인<a class="headerlink" href="#651" title="고유 링크">&para;</a></h2>
<h3 id="1">1. &nbsp; 해시 충돌이 일어난 뒤 탐색하는 방법<a class="headerlink" href="#1" title="고유 링크">&para;</a></h3>
<p>버킷이 5개인 해시 테이블의 해시 함수는 <span class="arithmatex">\(h(x)=x \bmod 5\)</span>다. 충돌이 나면 해당 버킷의 리스트에 원소를 순서대로 넣는다.
<code>[1, 6, 11, 7]</code>을 차례대로 삽입한다.</p>
<!-- numbered-subquestions -->
<ol>
<li>0~4번 버킷의 내용을 적어 보자.</li>
<li>6을 찾을 때 어느 버킷으로 들어가고, 어떤 원소를 차례대로 확인하는가?</li>
<li>1번에서 적은 버킷의 내용을 보면 나중에 삽입한 원소가 먼저 삽입한 원소를 덮어썼는가? 이 충돌 처리 방식과 연결하여 이유를 설명하라.</li>
</ol>
<details class="success">
<summary>참고 답안</summary>
<ol>
<li>
<p><span class="arithmatex">\(1\bmod5=6\bmod5=11\bmod5=1\)</span>이고 <span class="arithmatex">\(7\bmod5=2\)</span>이므로 각 버킷의 내용은 다음과 같다.</p>
<div class="highlight"><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a>0: []
<a id="__codelineno-0-2" name="__codelineno-0-2" href="#__codelineno-0-2"></a>1: [1, 6, 11]
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a>2: [7]
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a>3: []
<a id="__codelineno-0-5" name="__codelineno-0-5" href="#__codelineno-0-5"></a>4: []
</code></pre></div>
</li>
<li>
<p>6을 찾을 때는 먼저 1번 버킷으로 들어가 1, 6을 차례대로 비교한다. 두 번째 비교에서 목표를 찾는다.</p>
</li>
<li>
<p>해시값이 같다는 것은 같은 버킷에 들어간다는 뜻일 뿐, 원소들이 같다는 뜻은 아니다. 체이닝은 충돌한 원소를 모두 버킷에 저장하고,
탐색할 때 하나씩 비교한다. 따라서 1, 6, 11은 서로를 덮어쓰지 않는다.</p>
</li>
</ol>
</details>
<h3 id="2">2. &nbsp; 해시 테이블의 용량을 늘리면 원소는 어디로 가는가<a class="headerlink" href="#2" title="고유 링크">&para;</a></h3>
<p>체이닝을 사용하는 해시 테이블에 원래 5개의 버킷이 있고, 해시 함수는 <span class="arithmatex">\(h(x)=x\bmod5\)</span>다.
키 <code>[1, 6, 11]</code>은 모두 1번 버킷에 있다.</p>
<p>이제 버킷 수를 7개로 늘리고 해시 함수도 <span class="arithmatex">\(h(x)=x\bmod7\)</span>로 바꾼다.</p>
<!-- numbered-subquestions -->
<ol>
<li>1, 6, 11의 새 버킷 번호를 각각 계산하라.</li>
<li>용량을 늘린 뒤 어느 버킷에 원소가 들어 있는가?</li>
<li>기존 1번 버킷의 리스트를 새 1번 버킷에 그대로 복사해도 되는가? 1번과 2번의 결과를 바탕으로 이유를 설명하라.</li>
</ol>
<details class="success">
<summary>참고 답안</summary>
<ol>
<li>
<p>새 버킷 번호는 각각 다음과 같다.</p>
<ul>
<li><span class="arithmatex">\(1\bmod7=1\)</span></li>
<li><span class="arithmatex">\(6\bmod7=6\)</span></li>
<li><span class="arithmatex">\(11\bmod7=4\)</span></li>
</ul>
</li>
<li>
<p>1번 버킷에 1, 4번 버킷에 11, 6번 버킷에 6이 저장된다. 세 키가 더 이상 하나의 버킷에 모여 있지 않다.</p>
</li>
<li>
<p>그대로 복사해서는 안 된다. 버킷 번호는 키를 버킷 수로 나눈 나머지로 구한다. 버킷 수가 5에서 7로 바뀌면 같은 키의 버킷 번호도 달라질 수 있으므로,
모든 키의 위치를 다시 계산해야 한다. 이전 1번 버킷을 그대로 복사하면 새 식으로 6과 11을 찾을 때
각각 6번 버킷과 4번 버킷으로 가게 되어 찾지 못한다.</p>
</li>
</ol>
</details>
<h3 id="3-6-11">3. &nbsp; 6을 삭제한 뒤에도 11을 찾을 수 있는가<a class="headerlink" href="#3-6-11" title="고유 링크">&para;</a></h3>
<p>해시 테이블에 5개의 자리가 있고 인덱스는 <code>0~4</code>이며, 해시 함수는 <span class="arithmatex">\(h(x)=x\bmod5\)</span>다.
충돌이 나면 해시 함수로 계산한 인덱스부터 오른쪽으로 이동하며 처음 만나는 빈자리를 찾는다.</p>
<p><code>[1, 6, 11]</code>을 차례대로 삽입한다.</p>
<!-- numbered-subquestions -->
<ol>
<li>세 수는 최종적으로 각각 어느 인덱스에 저장되는가?</li>
<li>11을 찾을 때 어떤 인덱스를 차례대로 확인하는가?</li>
<li>6을 삭제하면서 그 자리를 ‘한 번도 사용하지 않은 빈자리’로 바꾸고, 탐색은 빈자리를 만나면 멈춘다고 하자.
다시 11을 찾으면 어떤 일이 일어나는가? 그 결과는 올바른가? 문제가 있다면 어떻게 피할 수 있는가?</li>
</ol>
<details class="success">
<summary>참고 답안</summary>
<ol>
<li>
<p>1은 인덱스 1에 저장된다. 6도 인덱스 1로 매핑되지만 충돌하므로 인덱스 2에 저장된다.
11 역시 인덱스 1부터 시작해 이미 사용 중인 인덱스 1과 2를 지나 인덱스 3에 저장된다.</p>
</li>
<li>
<p>11을 찾을 때는 인덱스 <code>1, 2, 3</code>을 차례대로 확인하고 인덱스 3에서 찾는다.</p>
</li>
<li>
<p>인덱스 2를 ‘한 번도 사용하지 않음’을 뜻하는 빈자리로 바꾸면, 11을 찾다가 인덱스 1을 확인한 뒤 인덱스 2에서 멈춘다.
따라서 11이 없다고 잘못 판단한다. 삭제할 때는 ‘삭제됨’ 표시를 남겨야 한다.
탐색 중 이 표시를 만나면 다음 인덱스를 계속 확인하고, 인덱스 4를 지나면 0으로 돌아간다. 이후 삽입할 때는 이 자리를 다시 사용할 수 있다.</p>
</li>
</ol>
</details>
<h2 id="652">6.5.2 &nbsp; 프로그래밍 연습<a class="headerlink" href="#652" title="고유 링크">&para;</a></h2>
<h3 id="1_1">1. &nbsp; 두 문자열의 문자 구성 비교하기<a class="headerlink" href="#1_1" title="고유 링크">&para;</a></h3>
<p>영어 소문자만 포함하는 두 문자열 <code>s</code>와 <code>t</code>가 주어진다.
<code>s</code>의 문자 위치는 자유롭게 바꿀 수 있지만 문자를 추가하거나 삭제하거나 다른 문자로 바꿀 수는 없다.</p>
<p>순서를 바꿔 <code>t</code>를 만들 수 있는지 판단하고, 가능하면 <code>true</code>, 그렇지 않으면 <code>false</code>를 반환하라.
문자열의 문자를 정렬하지 말고 해시 테이블로 각 문자의 등장 횟수를 기록하라.</p>
<details class="tip">
<summary>풀이 힌트</summary>
<ol>
<li>두 문자열의 길이가 다르면 문자 구성도 반드시 다르다.</li>
<li>해시 테이블에 각 문자의 개수를 기록한다. s를 순회할 때 해당 개수를 1씩 늘린다.</li>
<li>t를 순회할 때 해당 개수를 1씩 줄인다. 마지막에 모든 개수가 0이어야 문자 구성이 같다.</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.com/problems/valid-anagram/" rel="noopener noreferrer" target="_blank">LeetCode</a></p>
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