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---
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.