{"componentChunkName":"component---src-templates-blog-post-js","path":"/algorithm/codility/lesson6/16-triangle/","result":{"data":{"site":{"siteMetadata":{"title":"Zayden","author":"[Your Name]","siteUrl":"https://gatsby-starter-bee.netlify.com","comment":{"disqusShortName":"","utterances":"JaeYeopHan/gatsby-starter-bee"},"sponsor":{"buyMeACoffeeId":"jbee"}}},"markdownRemark":{"id":"d4755943-3581-59c1-8ee2-29888e570fe1","excerpt":"문제 설명 N개의 정수로 구성된 배열 A가 주어집니다. 삼항식(P, Q, R)은 0 ≤ P < Q < R < N이면 삼각형입니다: 예를 들어 다음과 같은 배열 A를 생각해 봅시다: 삼항(0, 2, 4)은 삼각형입니다. 함수를 작성합니다: 이 함수는 N개의 정수로 구성된 배열 A가 주어졌을 때, 이 배열에 대한 삼각형 삼중항이 존재하면 1을 반환하고 그렇지 않으면 0을 반환합니다. 예를 들어, 배열 A가 주어졌을 때 이면 위에서 설명한 대로 함수는 1을 반환해야 합니다. 다음과 같은 배열 A…","html":"<h2 id=\"문제-설명\" style=\"position:relative;\"><a href=\"#%EB%AC%B8%EC%A0%9C-%EC%84%A4%EB%AA%85\" aria-label=\"문제 설명 permalink\" class=\"anchor before\"><svg aria-hidden=\"true\" focusable=\"false\" height=\"16\" version=\"1.1\" viewBox=\"0 0 16 16\" width=\"16\"><path fill-rule=\"evenodd\" d=\"M4 9h1v1H4c-1.5 0-3-1.69-3-3.5S2.55 3 4 3h4c1.45 0 3 1.69 3 3.5 0 1.41-.91 2.72-2 3.25V8.59c.58-.45 1-1.27 1-2.09C10 5.22 8.98 4 8 4H4c-.98 0-2 1.22-2 2.5S3 9 4 9zm9-3h-1v1h1c1 0 2 1.22 2 2.5S13.98 12 13 12H9c-.98 0-2-1.22-2-2.5 0-.83.42-1.64 1-2.09V6.25c-1.09.53-2 1.84-2 3.25C6 11.31 7.55 13 9 13h4c1.45 0 3-1.69 3-3.5S14.5 6 13 6z\"></path></svg></a>문제 설명</h2>\n<p>N개의 정수로 구성된 배열 A가 주어집니다. 삼항식(P, Q, R)은 0 ≤ P &#x3C; Q &#x3C; R &#x3C; N이면 삼각형입니다:</p>\n<div class=\"gatsby-highlight\" data-language=\"text\"><pre class=\"language-text\"><code class=\"language-text\">A[P] + A[Q] > A[R],\nA[Q] + A[R] > A[P],\nA[R] + A[P] > A[Q].</code></pre></div>\n<p>예를 들어 다음과 같은 배열 A를 생각해 봅시다:</p>\n<div class=\"gatsby-highlight\" data-language=\"text\"><pre class=\"language-text\"><code class=\"language-text\">  A[0] = 10 A[1] = 2 A[2] = 5\n  A[3] = 1 A[4] = 8 A[5] = 20</code></pre></div>\n<p>삼항(0, 2, 4)은 삼각형입니다.</p>\n<p>함수를 작성합니다:</p>\n<div class=\"gatsby-highlight\" data-language=\"javascript\"><pre class=\"language-javascript\"><code class=\"language-javascript\"><span class=\"token keyword\">function</span> <span class=\"token function\">solution</span><span class=\"token punctuation\">(</span><span class=\"token parameter\"><span class=\"token constant\">A</span></span><span class=\"token punctuation\">)</span><span class=\"token punctuation\">;</span></code></pre></div>\n<p>이 함수는 N개의 정수로 구성된 배열 A가 주어졌을 때, 이 배열에 대한 삼각형 삼중항이 존재하면 1을 반환하고 그렇지 않으면 0을 반환합니다.</p>\n<p>예를 들어, 배열 A가 주어졌을 때</p>\n<div class=\"gatsby-highlight\" data-language=\"text\"><pre class=\"language-text\"><code class=\"language-text\">  A[0] = 10 A[1] = 2 A[2] = 5\n  A[3] = 1 A[4] = 8 A[5] = 20</code></pre></div>\n<p>이면 위에서 설명한 대로 함수는 1을 반환해야 합니다. 다음과 같은 배열 A가 주어졌습니다:</p>\n<div class=\"gatsby-highlight\" data-language=\"text\"><pre class=\"language-text\"><code class=\"language-text\">  A[0] = 10 A[1] = 50 A[2] = 5\n  A[3] = 1</code></pre></div>\n<p>이면 함수는 0을 반환해야 합니다.</p>\n<p>다음 가정에 대한 효율적인 알고리즘을 작성합니다:</p>\n<ul>\n<li>N은 [0..100,000] 범위 내의 정수입니다;</li>\n<li>배열 A의 각 요소는 [-2,147,483,648..2,147,483,647] 범위 내의 정수입니다.</li>\n</ul>\n<h2 id=\"문제-접근\" style=\"position:relative;\"><a href=\"#%EB%AC%B8%EC%A0%9C-%EC%A0%91%EA%B7%BC\" aria-label=\"문제 접근 permalink\" class=\"anchor before\"><svg aria-hidden=\"true\" focusable=\"false\" height=\"16\" version=\"1.1\" viewBox=\"0 0 16 16\" width=\"16\"><path fill-rule=\"evenodd\" d=\"M4 9h1v1H4c-1.5 0-3-1.69-3-3.5S2.55 3 4 3h4c1.45 0 3 1.69 3 3.5 0 1.41-.91 2.72-2 3.25V8.59c.58-.45 1-1.27 1-2.09C10 5.22 8.98 4 8 4H4c-.98 0-2 1.22-2 2.5S3 9 4 9zm9-3h-1v1h1c1 0 2 1.22 2 2.5S13.98 12 13 12H9c-.98 0-2-1.22-2-2.5 0-.83.42-1.64 1-2.09V6.25c-1.09.53-2 1.84-2 3.25C6 11.31 7.55 13 9 13h4c1.45 0 3-1.69 3-3.5S14.5 6 13 6z\"></path></svg></a>문제 접근</h2>\n<p><em>배열 A 를 오름차순을 정렬한다</em></p>\n<p><strong>왜?! 오름차순을 정렬한다면 P &#x3C; Q &#x3C; R 이 조건을 만족한다. 그래서 아래 두조건을 무조건 만족한다.</strong></p>\n<ul>\n<li>P + R > Q</li>\n<li>Q + R > P</li>\n</ul>\n<p><em>그러므로 <strong>P + Q > R</strong> 이 조건만 확인하면 된다.</em></p>\n<ol>\n<li>배열 A 를 오름 차순을 정렬한다.</li>\n<li>배열 A 를 순회하면서 음수 일 경우는 제외한다.</li>\n<li>\n<p><strong>P + Q > R</strong> 조건만 확인하여</p>\n<ul>\n<li>만족하면 1 반환</li>\n<li>만족하지 않으면 0 반환</li>\n</ul>\n</li>\n</ol>\n<div class=\"gatsby-highlight\" data-language=\"javascript\"><pre class=\"language-javascript\"><code class=\"language-javascript\"><span class=\"token keyword\">function</span> <span class=\"token function\">solution</span><span class=\"token punctuation\">(</span><span class=\"token parameter\"><span class=\"token constant\">A</span></span><span class=\"token punctuation\">)</span> <span class=\"token punctuation\">{</span>\n  <span class=\"token keyword\">const</span> arr <span class=\"token operator\">=</span> <span class=\"token constant\">A</span><span class=\"token punctuation\">.</span><span class=\"token function\">slice</span><span class=\"token punctuation\">(</span><span class=\"token punctuation\">)</span><span class=\"token punctuation\">.</span><span class=\"token function\">sort</span><span class=\"token punctuation\">(</span><span class=\"token punctuation\">(</span><span class=\"token parameter\">a<span class=\"token punctuation\">,</span> b</span><span class=\"token punctuation\">)</span> <span class=\"token operator\">=></span> a <span class=\"token operator\">-</span> b<span class=\"token punctuation\">)</span>\n\n  <span class=\"token keyword\">for</span> <span class=\"token punctuation\">(</span><span class=\"token keyword\">let</span> i <span class=\"token operator\">=</span> <span class=\"token number\">0</span><span class=\"token punctuation\">;</span> i <span class=\"token operator\">&lt;</span> <span class=\"token constant\">A</span><span class=\"token punctuation\">.</span>length <span class=\"token operator\">-</span> <span class=\"token number\">2</span><span class=\"token punctuation\">;</span> i<span class=\"token operator\">++</span><span class=\"token punctuation\">)</span> <span class=\"token punctuation\">{</span>\n    <span class=\"token keyword\">if</span> <span class=\"token punctuation\">(</span>arr<span class=\"token punctuation\">[</span>i<span class=\"token punctuation\">]</span> <span class=\"token operator\">&lt;</span> <span class=\"token number\">0</span><span class=\"token punctuation\">)</span> <span class=\"token punctuation\">{</span>\n      <span class=\"token keyword\">continue</span>\n    <span class=\"token punctuation\">}</span>\n\n    <span class=\"token keyword\">for</span> <span class=\"token punctuation\">(</span><span class=\"token keyword\">let</span> j <span class=\"token operator\">=</span> i <span class=\"token operator\">+</span> <span class=\"token number\">1</span><span class=\"token punctuation\">;</span> j <span class=\"token operator\">&lt;</span> <span class=\"token constant\">A</span><span class=\"token punctuation\">.</span>length <span class=\"token operator\">-</span> <span class=\"token number\">1</span><span class=\"token punctuation\">;</span> j<span class=\"token operator\">++</span><span class=\"token punctuation\">)</span> <span class=\"token punctuation\">{</span>\n      <span class=\"token keyword\">if</span> <span class=\"token punctuation\">(</span>arr<span class=\"token punctuation\">[</span>j<span class=\"token punctuation\">]</span> <span class=\"token operator\">&lt;</span> <span class=\"token number\">0</span> <span class=\"token operator\">||</span> arr<span class=\"token punctuation\">[</span>j <span class=\"token operator\">+</span> <span class=\"token number\">1</span><span class=\"token punctuation\">]</span> <span class=\"token operator\">&lt;</span> <span class=\"token number\">0</span><span class=\"token punctuation\">)</span> <span class=\"token punctuation\">{</span>\n        <span class=\"token keyword\">continue</span>\n      <span class=\"token punctuation\">}</span>\n\n      <span class=\"token keyword\">if</span> <span class=\"token punctuation\">(</span>arr<span class=\"token punctuation\">[</span>i<span class=\"token punctuation\">]</span> <span class=\"token operator\">+</span> arr<span class=\"token punctuation\">[</span>j<span class=\"token punctuation\">]</span> <span class=\"token operator\">></span> arr<span class=\"token punctuation\">[</span>j <span class=\"token operator\">+</span> <span class=\"token number\">1</span><span class=\"token punctuation\">]</span><span class=\"token punctuation\">)</span> <span class=\"token punctuation\">{</span>\n        <span class=\"token keyword\">return</span> <span class=\"token number\">1</span>\n      <span class=\"token punctuation\">}</span>\n    <span class=\"token punctuation\">}</span>\n  <span class=\"token punctuation\">}</span>\n\n  <span class=\"token keyword\">return</span> <span class=\"token number\">0</span>\n<span class=\"token punctuation\">}</span></code></pre></div>","frontmatter":{"title":"Codility Lesson 6 - Triangle","date":"July 17, 2023"}}},"pageContext":{"slug":"/algorithm/codility/lesson6/16-triangle/","previous":{"fields":{"slug":"/algorithm/codility/lesson6/15-maxProductOfThree/"},"frontmatter":{"title":"Codility Lesson 6 - MaxProductOfThree"}},"next":{"fields":{"slug":"/algorithm/codility/lesson6/17-numberOfDiscIntersections/"},"frontmatter":{"title":"Codility Lesson 6 - NumberOfDiscIntersections"}}}}}