ArrayConvertion

Given an array of 2k integers we perfrom the following operations until the array contains only one element.

1. On the 1st,3rd,5th etc. iterations (1-based) replace each pair of consecutive elements with their sum.
   
2. 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. We return that element.

Example:
console.log(ArrayConvertion([1,2,3,4,5,6,7,8])) 186

function ArrayConvertion(inputArray) {
  let isOdd = true;

  while(inputArray.length !== 1) {
    inputArray = sumProduct(inputArray,isOdd)
    isOdd = !isOdd;
  }
  return inputArray[0];
  
}

function sumProduct(nums,isOdd) {
  const sumProducts = [];
  if (isOdd) {
    for (let i = 0; i<nums.length;i+=2) {
      sumProducts.push(nums[i] + nums[i+1])
    }
  } else {
    for (let i = 0; i<nums.length;i+=2) {
      sumProducts.push(nums[i] * nums[i+1])
    }
  }
  return sumProducts;
}

console.log(ArrayConvertion([1,2,3,4,5,6,7,8])) 186
}

We create a function ArrayConversion with parameter of inputArray and we create a variable isOdd to be equal to true so we can keep track if the element is odd or even
We loop with while loop throuh the inputArray until the length is different than 1 (because as the algorithm suggest only one element remains as result)
We initialize inputArray to a function sumProduct that takes for arguments inputArray and isOdd then we switch isOdd to is not !isOdd
Function SumProduct takes as we said above two arguments nums array and boolean isOdd and also we create a variable to hold the sum and product of the array numbers.
We check if isOdd and if it is Odd we loop from 0 to the array length skipping one by doing for(let i = 0; i < nums.legth; i+= 2) Then we use sumProduct and push the first iteration (num[i]) + second iteration (num[i+1])
else if it is Even we loop from 0 to the array length skipping one by doing for(let i = 0; i < nums.legth; i+= 2) Then we use sumProduct and push the first iteration (num[i]) * second iteration (num[i+1])
We return the sumProduct
And in the ArrayConversion function we return inputArray[0]