Use Loop for Min/Max Instead of Sort. Use when you need help with js min max loop.
Scanned 9/8/2026
Install to Claude Code
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---
name: js-min-max-loop
description: Use Loop for Min/Max Instead of Sort. Use when you need help with js min max loop.
license: CC-BY-NC-SA-4.0
metadata:
risk: unknown
source: community
kind: mode
category: rules
---
## 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.
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