sangwook yoo 3ac66feba2 chore: .agents/.claude 설정·CLAUDE.md·AGENTS.md git 추적 추가
gitignore 의 *.md / .claude/ 차단을 풀어 다음 파일을 git 으로 추적.
- /CLAUDE.md, /AGENTS.md (루트 에이전트 가이드)
- .agents/**/*.md (skills 문서 95개)
- .claude/settings.json (Claude Code 프로젝트 설정)
.claude/worktrees/ 는 git worktree 메타이므로 신규 ignore 라인으로 제외.

Co-Authored-By: Claude Opus 4.7 (1M context) <noreply@anthropic.com>
2026-05-12 18:05:03 +09:00

2.2 KiB

title impact impactDescription tags
Use Loop for Min/Max Instead of Sort LOW O(n) instead of O(n log n) javascript, arrays, performance, sorting, algorithms

Use Loop for Min/Max Instead of Sort

Finding the smallest or largest element only requires a single pass through the array. Sorting is wasteful and slower.

Incorrect (O(n log n) - sort to find latest):

interface Project {
  id: string
  name: string
  updatedAt: number
}

function getLatestProject(projects: Project[]) {
  const sorted = [...projects].sort((a, b) => b.updatedAt - a.updatedAt)
  return sorted[0]
}

Sorts the entire array just to find the maximum value.

Incorrect (O(n log n) - sort for oldest and newest):

function getOldestAndNewest(projects: Project[]) {
  const sorted = [...projects].sort((a, b) => a.updatedAt - b.updatedAt)
  return { oldest: sorted[0], newest: sorted[sorted.length - 1] }
}

Still sorts unnecessarily when only min/max are needed.

Correct (O(n) - single loop):

function getLatestProject(projects: Project[]) {
  if (projects.length === 0) return null
  
  let latest = projects[0]
  
  for (let i = 1; i < projects.length; i++) {
    if (projects[i].updatedAt > latest.updatedAt) {
      latest = projects[i]
    }
  }
  
  return latest
}

function getOldestAndNewest(projects: Project[]) {
  if (projects.length === 0) return { oldest: null, newest: null }
  
  let oldest = projects[0]
  let newest = projects[0]
  
  for (let i = 1; i < projects.length; i++) {
    if (projects[i].updatedAt < oldest.updatedAt) oldest = projects[i]
    if (projects[i].updatedAt > newest.updatedAt) newest = projects[i]
  }
  
  return { oldest, newest }
}

Single pass through the array, no copying, no sorting.

Alternative (Math.min/Math.max for small arrays):

const numbers = [5, 2, 8, 1, 9]
const min = Math.min(...numbers)
const max = Math.max(...numbers)

This works for small arrays, but can be slower or just throw an error for very large arrays due to spread operator limitations. Maximal array length is approximately 124000 in Chrome 143 and 638000 in Safari 18; exact numbers may vary - see the fiddle. Use the loop approach for reliability.