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.
Comments