- Add 8 new skills (62 total, up from 58) - Official Anthropic skills: docx, pdf, pptx, xlsx, brand-guidelines, internal-comms - Vercel Labs skills: react-best-practices, web-design-guidelines - Implement dual-versioning: -official/-anthropic and -community suffixes - Update README with new skill registry and credits - Regenerate skills_index.json (62 skills validated) - Add comprehensive walkthrough.md BREAKING CHANGE: Document skills (docx/pdf/pptx/xlsx) renamed with version suffixes
2.1 KiB
2.1 KiB
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 for very large arrays due to spread operator limitations. Use the loop approach for reliability.