Algo friday 3 solution

startAndEnd.js

const startAndEnd = (nums, val) => {
  if (!Array.isArray(nums) || typeof val !== 'number') throw new Error('Invalid input')

  const min = Math.min(...nums);

  let positions = [-1, -1]

  let countLessThanVal = 0
  let countOfVal = 0
  let countOfMin = 0

  for (let num of nums) {
    if (num === min) countOfMin++
    if (num < val) countLessThanVal++
    if (num === val) countOfVal++
  }

  if (!countOfVal) return positions

  positions[0] = val === min ? 0 : countLessThanVal + countOfMin - 1
  positions[1] = positions[0] + countOfVal - 1

  return positions
}

Well sorting the array results in a O(nlogn) time complexity, which basically supersedes the linear time complexity due to looping through the array. Unless there's a solution that doesn't involve sorting the array, this is the best I think is obtainable.

Update: Solution has been updated, so previous comments are no longer valid.

Hello @chygoz2, congratulations, you are one of the winners of the $20 award for Week 3 of #AlgorithmFridays. 🎉🎉

Your solution was selected because it is most optimal in terms of time complexity. You were able to find a way to solve the problem without sorting the array.

I have written an in-depth blog post here about the solution. The post also announces you as one of the winners for Week 3 of #AlgorithmFridays.

We will contact you in less than 24 hours for your award.

Congratulations once again and thanks for participating!

Thank you.