### Daily Coding Problem 7MAY2020

I am subscribed to dailycodingproblem.com. From time to time I will share problems I attempt and have a discussion around it. My solution might have bugs or might not be the most efficient but it would be my attempt. Feel free to criticise.

Today's problem I got

This problem was asked by Uber.

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].

Follow-up: what if you can't use division?

This was my solution in javascript. I am using the Javascript for Android app.

var numbers = [1, 2, 3, 4, 5];
var total = numbers.reduce( (a,b) => a * b );

for (i = 0; i < numbers.length; i++)
console.log(divide(total, numbers[i]));

function divide (a, b) {
if (a == 0) return 0;
if (a < b) return 0;
if (b <= a) return 1 + divide(a-b, b);
}

I remembered from my uni days that division could be done by recursion and subtraction.