mirror of
https://github.com/kennethnym/aris.git
synced 2026-02-02 21:21:21 +00:00
- react-best-practices: Performance optimization patterns (client-side only) - web-design-guidelines: UI review against Web Interface Guidelines Co-authored-by: Ona <no-reply@ona.com>
83 lines
2.1 KiB
Markdown
83 lines
2.1 KiB
Markdown
---
|
|
title: Use Loop for Min/Max Instead of Sort
|
|
impact: LOW
|
|
impactDescription: O(n) instead of O(n log n)
|
|
tags: 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):**
|
|
|
|
```typescript
|
|
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):**
|
|
|
|
```typescript
|
|
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):**
|
|
|
|
```typescript
|
|
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):**
|
|
|
|
```typescript
|
|
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.
|