Add detailed landing page development plan in docs/frontend_landing_plan.md: - Complete landing page structure (Hero, Problem/Solution, Features, Demo, CTA) - Design guidelines from downloaded skills (typography, color, motion, composition) - Security considerations (XSS prevention, input sanitization, CSP) - Performance targets (LCP <2.5s, bundle <150KB, Lighthouse >90) - Responsiveness and accessibility requirements (WCAG 2.1 AA) - Success KPIs and monitoring setup - 3-week development timeline with daily tasks - Definition of Done checklist Download 10+ frontend/UI/UX skills via universal-skills-manager: - frontend-ui-ux: UI/UX design without mockups - frontend-design-guidelines: Production-grade interface guidelines - frontend-developer: React best practices (40+ rules) - frontend-engineer: Next.js 14 App Router patterns - ui-ux-master: Comprehensive design systems and accessibility - ui-ux-systems-designer: Information architecture and interaction - ui-ux-design-user-experience: Platform-specific guidelines - Plus additional reference materials and validation scripts Configure universal-skills MCP with SkillsMP API key for curated skill access. Safety first: All skills validated before installation, no project code modified. Refs: Universal Skills Manager (github:jacob-bd/universal-skills-manager) Next: Begin Sprint 3 landing page development
83 lines
2.1 KiB
Markdown
83 lines
2.1 KiB
Markdown
---
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title: Use Loop for Min/Max Instead of Sort
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impact: LOW
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impactDescription: O(n) instead of O(n log n)
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tags: javascript, arrays, performance, sorting, algorithms
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---
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## Use Loop for Min/Max Instead of Sort
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Finding the smallest or largest element only requires a single pass through the array. Sorting is wasteful and slower.
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**Incorrect (O(n log n) - sort to find latest):**
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```typescript
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interface Project {
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id: string
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name: string
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updatedAt: number
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}
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function getLatestProject(projects: Project[]) {
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const sorted = [...projects].sort((a, b) => b.updatedAt - a.updatedAt)
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return sorted[0]
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}
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```
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Sorts the entire array just to find the maximum value.
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**Incorrect (O(n log n) - sort for oldest and newest):**
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```typescript
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function getOldestAndNewest(projects: Project[]) {
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const sorted = [...projects].sort((a, b) => a.updatedAt - b.updatedAt)
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return { oldest: sorted[0], newest: sorted[sorted.length - 1] }
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}
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```
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Still sorts unnecessarily when only min/max are needed.
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**Correct (O(n) - single loop):**
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```typescript
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function getLatestProject(projects: Project[]) {
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if (projects.length === 0) return null
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let latest = projects[0]
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for (let i = 1; i < projects.length; i++) {
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if (projects[i].updatedAt > latest.updatedAt) {
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latest = projects[i]
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}
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}
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return latest
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}
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function getOldestAndNewest(projects: Project[]) {
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if (projects.length === 0) return { oldest: null, newest: null }
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let oldest = projects[0]
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let newest = projects[0]
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for (let i = 1; i < projects.length; i++) {
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if (projects[i].updatedAt < oldest.updatedAt) oldest = projects[i]
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if (projects[i].updatedAt > newest.updatedAt) newest = projects[i]
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}
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return { oldest, newest }
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}
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```
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Single pass through the array, no copying, no sorting.
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**Alternative (Math.min/Math.max for small arrays):**
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```typescript
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const numbers = [5, 2, 8, 1, 9]
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const min = Math.min(...numbers)
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const max = Math.max(...numbers)
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```
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This works for small arrays but can be slower for very large arrays due to spread operator limitations. Use the loop approach for reliability.
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