Question:
Given an array of integers, return a new array such that each element at index i of the new array is the product of all the numbers in the original array except the one at i.

For example, if our input was [1, 2, 3, 4, 5], the expected output would be [120, 60, 40, 30, 24]. If our input was [3, 2, 1], the expected output would be [2, 3, 6]. What if you can't use division?

Back again with some coding practice. With this problem, I opted in to not using division and my thought process was that the easiest way to solve the problem was with a nested for loop. Although, I believe there is a faster way that I am not seeing yet.

const a = [1, 2, 3, 4, 5]

const product = array => {
  const b = []
  for (i = 0; i < array.length; i++) {
    let n = 1
    for (j = 0; j < array.length; j++) {
      if (j !== i) {
        n *= array[j]
      }
    }
    b.push(n)
  }
  console.log(b)
  return b
}

product(a)