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Js Min Max Loop

ASecurity

Use Loop for Min/Max Instead of Sort. Use when you need help with js min max loop.

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  • Added September 8, 2026
developmenttypescriptgo

Security analysis

A100/100

Scanned September 8, 2026

npx -y skills add anubhavg-icpl/vibe --skill js-min-max-loop --agent claude-code

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SKILL.md
---
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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