rules--js-min-max-loop

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):

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):

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):

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):

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.