The function returns products of element in an array

productOfArray.js

//Given an integer array nums, 
//write a function that returns an array products,
//such that each entry at position i, in products is a product of all the other elements in nums except num[i]
// #algorithm friday set4


const productOfArray = (nums) => {
//  testing for array or empty array
   if (!Array.isArray(nums)  || nums.length == 0 ) return ' Invalid Input '

  let count = nums.length  
  let start = new Array(count);
  let end = new Array(count);
  let product = new Array(count);
  
//   setting the intials of the array
     start[0] = 1;
     end[count-1] = 1   

// the start array
    for (let i = 1; i < count; i++){
        start[i] = nums[i - 1] * start[i - 1];
     }
// the end array        
    for (let j = count - 2; j >= 0; j--){
         end[j] = nums[j + 1] * end[j + 1];
     }
//  the product array = start * end
    for (let i = 0; i < count; i++){
         product[i] = start[i] * end[i];
    }
        return product
}

console.log(productOfArray([4,5,10,2])) //return [ 100, 80, 40, 200 ]

Hello @Abiola-Farounbi, thank you for participating in Week 4 of Algorithm Fridays.

This is a really decent solution that passes all the test cases. Very sweet!

I like your use of edge case checks on line 9. One thing to say about the checks however is that, if an empty array is passed, it's okay to return an empty array as your result instead of throwing 'Invalid Input' errors.

Also, your solution is optimal in terms of time complexity O(N). However, your solution uses three arrays start, end and the final product. Do you think you can come up with a better solution in terms of memory efficiency?

That said, this was really good. Kudos!

Thanks a lot for the feedback,
I have been pondering over it for some time, so far this is what I have been able to come up with, hopefully in due time, I will be able to come with a more memory efficient approach..