Array Conversion

Given an array of 2k integers (for some integer k), perform the following operations until the array contains only one element:

• On the 1st, 3rd, 5th, etc. iterations (1-based) replace each pair of consecutive elements with their sum;
• On the 2nd, 4th, 6th, etc. iterations replace each pair of consecutive elements with their product. After the algorithm has finished, there will be a single element left in the array. Return that element.

Example
For inputArray = [1, 2, 3, 4, 5, 6, 7, 8], the output should be arrayConversion(inputArray) = 186.

function arrayConversion(inputArray) {
  let isOdd = true

  while (inputArray.length !== 1) {
    inputArray = someProduct(inputArray, isOdd)
    isOdd = !isOdd
  }

  return inputArray[0]
}

function someProduct(nums, isOdd) {
  const sumProducs = []
  if (isOdd) {
    for (let i = 0; i < nums.length; i += 2) {
      sumProducs.push(nums[i] + nums[i + 1])
    }
  } else {
    for (let i = 0; i < nums.length; i += 2) {
      sumProducs.push(nums[i] * nums[i + 1])
    }
  }
  return sumProducs
}
console.log(arrayConversion([1, 2, 3, 4, 5, 6, 7, 8])) => 186

We create function called arrayConversion with parameter inputArray and local variable isOdd = true

We loop the array given into the parameter inputArray with while loop then checks if inputArray.length is diferent than 1. If it is we give inputArray = someProduct(inputArray, isOdd) and toggle isOdd = !isOdd and in the end we return inputArray[0]

We create a helper function someProduct that takes an array parameter nums and boolean isOdd also we create a local variable sumProducs = []


We check if isOdd and if it is we loop with for loop where we start at 0 goes input i is less than nums.lenght and we increase i by 2 in every iteration.
So as the description says if it is odd we use sumProducs.push(nums[i] + nums[i + 1])

Else if is even number we do the same we loop with for loop where we start at 0 goes input i is less than nums.lenght and we increase i by 2 in every iteration.
So as in the description says if it is even we use sumProducs.push(nums[i] * nums[i + 1])

In the end we **return sumProducs