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livedash-node/.opencode/skills/vercel-react-best-practices/rules/js-min-max-loop.md
Kaj Kowalski 5bfd762e55 fix: comprehensive TypeScript/build fixes and modernization
- Update tsconfig to ES2024 target and bundler moduleResolution
- Add dynamic imports for chart.js and recharts (bundle optimization)
- Consolidate 17 useState into useReducer in sessions page
- Fix 18 .js extension imports across lib files
- Add type declarations for @rapideditor/country-coder
- Fix platform user types (PlatformUserRole enum)
- Fix Calendar component prop types
- Centralize next-auth type augmentation
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- Fix Prisma JSON null handling with Prisma.DbNull
- Fix various type mismatches (SessionMessage, ImportRecord, etc.)
- Export ButtonProps from button component
- Update next-themes import path
- Replace JSX.Element with React.ReactElement
- Remove obsolete debug scripts and pnpm lockfile
- Downgrade eslint to v8 for next compatibility
2026-01-20 07:28:10 +01:00

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title, impact, impactDescription, tags
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.