Algorithm challenge week 4 solution

arrayProduct.js

const arrayProduct = nums => {
  if(!Array.isArray(nums)) throw new Error('Invalid input')
  
  let product = 1
  let countOfZeros = 0;
  let productWithoutZeros = 1

  for (let i = 0, j = nums.length - 1; i <= j; i++, j--) {
    if (typeof nums[i] != 'number' || typeof nums[j] != 'number') throw new Error('Invalid input')
    
    if (nums[i] == 0) {
      countOfZeros++
    }

    product *= nums[i]
    productWithoutZeros = nums[i] != 0 ? productWithoutZeros * nums[i] : productWithoutZeros

    if (i != j) {
      if (nums[j] == 0) {
        countOfZeros++
      }

      product *= nums[j]
      productWithoutZeros = nums[j] != 0 ? productWithoutZeros * nums[j] : productWithoutZeros
    }
  }

  return nums.map(num => {
    return countOfZeros > 1 ? 0 : num != 0 ? product / num : productWithoutZeros
  })
}

Hello @chygoz2, thank you for participating in Week 4 of Algorithm Fridays.

This is a robust and decent solution - use of two pointers, the check on line 9, the check for each entry of the array, the use of map function.

However, the one test your solution fails is when one of the entries in the array is 0. For example:

arrayProduct ([4, 3, 0]); ❌ // should return [0, 0, 12] but yours returns [0, 0, NaN]

What do you think?

@meekg33k I missed that edge case. I've now updated the solution.