/**
 *
 * Ashish Patel
 * e: ashishsushilPatel@gmail.com
 * w: https://ashish.me
 *
 */

/* 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.

We have [1, 2, 3, 4, 5, 6, 7, 8] -> [3, 7, 11, 15] -> [21, 165] -> [186], so the answer is 186.
*/

function arrayConversion(values) {
  let isOdd = true
  while(values.length !== 1){
    values = calculateSumOrProduct(values, isOdd)
    isOdd = !isOdd
  }
  return values
}

function calculateSumOrProduct(array, isOdd){
  let calculatedArray = []
  if(isOdd){
    for (let index = 0; index < array.length; index+=2) {
      calculatedArray.push(array[index] + array[index+1])
    }
  } else {
    for (let index = 0; index < array.length; index+=2) {
      calculatedArray.push(array[index] * array[index+1])
    }
  }
  return calculatedArray
}

test('array Conversion', () => {
  expect(arrayConversion([1, 2, 3, 4, 5, 6, 7, 8])).toEqual([186])
});