AngeloooooandDraculabo
8ed4593f73
* Refactor Gemini model display: filter old versions, group low/high pro, and improve naming * Fix potential negative minutes and refine Gemini 3 Pro grouping logic based on PR feedback * The account change process has been sped up * Address PR review: enhance error message and parameterize polling interval in process handler * Address PR review: fix potential race condition in database backup logic * fix: prevent renderer white screen and harden sentry gating * feat: localize model labels * chore: enable hoisted node-linker * chore: ignore .agent/.claude/.codex in prettier --------- Co-authored-by: Draculabo <[email protected]>
83 lines
2.2 KiB
Markdown
83 lines
2.2 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 or just throw an error for very large arrays due to spread operator limitations. Maximal array length is approximately 124000 in Chrome 143 and 638000 in Safari 18; exact numbers may vary - see [the fiddle](https://jsfiddle.net/qw1jabsx/4/). Use the loop approach for reliability.
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