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11.3 &nbsp; 버블 정렬
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11.4 &nbsp; 삽입 정렬
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<h1 id="107">10.7 &nbsp; 연습 문제<a class="headerlink" href="#107" title="고유 링크">&para;</a></h1>
<h2 id="1071">10.7.1 &nbsp; 개념 확인<a class="headerlink" href="#1071" title="고유 링크">&para;</a></h2>
<h3 id="1">1. &nbsp; 이진 탐색은 탐색 구간을 어떻게 줄이는가<a class="headerlink" href="#1" title="고유 링크">&para;</a></h3>
<p>정렬된 배열 <code>[2, 5, 8, 12, 16, 23, 38]</code>에서 16을 찾는다.
양 끝을 포함하는 닫힌 구간 <code>[i, j]</code>를 사용하며, 중간 인덱스는
<span class="arithmatex">\(m=i+(j-i)/2\)</span>다. 나눗셈 결과의 소수 부분은 버린다.</p>
<p>목표 값을 찾을 때까지 매 단계의 <code>(i, j, m)</code>과 중간 원소,
다음 단계에서 구간을 어떻게 줄이는지 적어 보자.</p>
<details class="success">
<summary>참고 답안</summary>
<p>각 단계의 탐색 과정은 다음과 같다.</p>
<table>
<thead>
<tr>
<th>단계</th>
<th><code>(i, j, m)</code></th>
<th>중간 원소</th>
<th>다음 동작</th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td><code>(0, 6, 3)</code></td>
<td>12</td>
<td><code>12 &lt; 16</code>이므로 <code>i = 4</code>로 설정</td>
</tr>
<tr>
<td>2</td>
<td><code>(4, 6, 5)</code></td>
<td>23</td>
<td><code>23 &gt; 16</code>이므로 <code>j = 4</code>로 설정</td>
</tr>
<tr>
<td>3</td>
<td><code>(4, 4, 4)</code></td>
<td>16</td>
<td>목표 값을 찾았으므로 인덱스 4를 반환</td>
</tr>
</tbody>
</table>
<p>배열이 정렬되어 있으므로 중간 값이 목표 값보다 작으면 중간 원소와 그 왼쪽을 제외할 수 있다.
중간 값이 목표 값보다 크면 중간 원소와 그 오른쪽을 제외할 수 있다.</p>
</details>
<h3 id="2">2. &nbsp; 중복 원소의 왼쪽 경계와 오른쪽 경계<a class="headerlink" href="#2" title="고유 링크">&para;</a></h3>
<p>배열 <code>[1, 2, 2, 2, 4, 6]</code>에서 숫자 2를 찾는다.
한 학생이 이진 탐색으로 인덱스 2에서 목표 값을 찾자마자 반환한 뒤,
“인덱스 2가 숫자 2의 왼쪽 경계다.”라고 말했다.</p>
<!-- numbered-subquestions -->
<ol>
<li>이 학생의 말이 맞는가? 숫자 2의 왼쪽 경계와 오른쪽 경계는 각각 어디인지 이유와 함께 설명하라.</li>
<li>왼쪽 경계를 찾을 때 중간 원소가 목표 값과 같다면, 다음에는 어느 쪽을 계속 탐색해야 하는가?</li>
<li>오른쪽 경계를 찾을 때는 어느 쪽을 계속 탐색해야 하는가? 탐색 과정 전체를 적지 말고 방향만 설명하라.</li>
</ol>
<details class="success">
<summary>참고 답안</summary>
<ol>
<li>
<p>이 학생의 말은 틀렸다. 2를 하나 찾자마자 반환하면 어떤 2를 찾았다는 것만 보장할 수 있고, 가장 왼쪽이나 가장 오른쪽의 2를 찾았다고 보장할 수는 없다.
이 문제에서 왼쪽 경계는 인덱스 1, 오른쪽 경계는 인덱스 3이다.</p>
</li>
<li>
<p>왼쪽 경계를 찾을 때는 중간 원소가 2와 같더라도 왼쪽을 계속 탐색해야 한다.
예를 들어 양 끝을 포함하는 닫힌 구간을 사용하는 경우 <code>j = m - 1</code>로 설정한다.</p>
</li>
<li>
<p>오른쪽 경계를 찾을 때는 중간 원소가 2와 같으면 오른쪽을 계속 탐색해야 한다.
예를 들어 <code>i = m + 1</code>로 설정한다.</p>
</li>
</ol>
</details>
<h3 id="3">3. &nbsp; 데이터에 따라 어떤 탐색 방법을 선택해야 하는가<a class="headerlink" href="#3" title="고유 링크">&para;</a></h3>
<p>다음 세 상황에 알맞은 방법을 ‘선형 탐색, 이진 탐색, 해시 테이블’ 중에서 고르고 이유를 설명하라.</p>
<!-- numbered-subquestions -->
<ol>
<li>이미 정렬되어 있고 더 이상 변하지 않는 정수 <span class="arithmatex">\(10^7\)</span>개에서 반복해서 값을 찾으며, 다른 자료 구조를 추가로 만들지 않는다.</li>
<li>삽입과 삭제가 빈번한 데이터 집합에서 특정 키가 있는지 반복해서 확인한다. 정렬 상태를 유지할 필요가 없고 범위 탐색도 하지 않는다.</li>
<li>정렬되지 않은 배열에서 특정 값을 한 번만 찾는다.</li>
</ol>
<details class="success">
<summary>참고 답안</summary>
<ol>
<li>이진 탐색: 데이터가 정렬되어 있고 변하지 않으므로 <span class="arithmatex">\(O(\log n)\)</span> 시간에 탐색할 수 있으며, 추가 공간도 필요하지 않다.</li>
<li>해시 테이블: 해시 함수가 키를 각 버킷에 비교적 고르게 분산한다면 삽입, 삭제, 키를 이용한 탐색의 평균 시간 복잡도가 모두 <span class="arithmatex">\(O(1)\)</span>이다.</li>
<li>
<p>처음부터 끝까지 직접 순회: 한 번만 찾는 경우에는 정렬하거나 해시 테이블을 만드는 데에도 먼저 배열 전체를 처리해야 하므로,
이번 작업의 전체 작업량이 줄어들지 않는다.</p>
<p>어떤 방법을 선택할지는 데이터의 정렬 여부, 추가 자료 구조의 허용 여부, 조회 횟수, 수행해야 하는 연산에 따라 달라진다.</p>
</li>
</ol>
</details>
<h2 id="1072">10.7.2 &nbsp; 프로그래밍 연습<a class="headerlink" href="#1072" title="고유 링크">&para;</a></h2>
<h3 id="1_1">1. &nbsp; 정렬된 배열의 이진 탐색<a class="headerlink" href="#1_1" title="고유 링크">&para;</a></h3>
<p>엄격한 오름차순으로 정렬된 정수 배열 <code>nums</code>와 목표 값 <code>target</code>이 주어진다. 이진 탐색으로 <code>target</code>을 찾아라. 존재하면 배열 인덱스를 반환하고, 존재하지 않으면 -1을 반환하라.</p>
<details class="tip">
<summary>풀이 힌트</summary>
<ol>
<li>초기 구간은 left = 0, right = n - 1이며, 구간이 비어 있지 않을 조건은 left &lt;= right다.</li>
<li>mid = left + (right - left) // 2로 중간 인덱스를 계산한다.</li>
<li><code>nums[mid]</code>가 <code>target</code>보다 작으면 왼쪽 경계를 <code>mid + 1</code>로 옮긴다. <code>nums[mid]</code>가 <code>target</code>보다 크면 오른쪽 경계를 <code>mid - 1</code>로 옮긴다. 같으면 즉시 반환한다.</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.com/problems/binary-search/" rel="noopener noreferrer" target="_blank">LeetCode</a></p>
<h3 id="2_1">2. &nbsp; 정렬된 배열의 삽입 위치<a class="headerlink" href="#2_1" title="고유 링크">&para;</a></h3>
<p>엄격한 오름차순으로 정렬된 정수 배열 <code>nums</code>와 목표 값 <code>target</code>이 주어진다.</p>
<p><code>target</code>이 이미 배열에 있으면 그 인덱스를 반환하라. 없으면 <code>target</code>을 삽입해도 엄격한 오름차순을 유지할 수 있는 위치를 반환하라. 답은 0일 수도 있고 배열 길이와 같을 수도 있다. 이진 탐색을 사용하라.</p>
<details class="tip">
<summary>풀이 힌트</summary>
<ol>
<li>답은 0일 수도 있고 배열 길이 n일 수도 있다.</li>
<li>양 끝을 포함하는 닫힌 구간을 사용할 때 <code>nums[mid]</code>가 <code>target</code>보다 크거나 같으면 <code>right = mid - 1</code>로 설정하여 더 왼쪽의 위치를 계속 확인한다. 그렇지 않으면 <code>left = mid + 1</code>로 설정한다.</li>
<li>반복문이 끝나면 left가 삽입 위치다.</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.com/problems/search-insert-position/" rel="noopener noreferrer" target="_blank">LeetCode</a></p>
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