Daily Coding Problem: Problem #2

problem2.py

"""
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. Solve it without using division and in O(n) time.

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].

https://dailycodingproblem.com/
"""

def problem2solver(numbers):
  numbers_product = []
  for i in range(0, len(numbers)):
    product = 1
    for j in range(0, len(numbers)):
      if not i is j:
        product *= numbers[j]
    numbers_product.append(product)
  return numbers_product

if __name__ == '__main__':
  # Read input, numbers separated by commas
  numbers = [int(n) for n in input().split(',')]
  print(problem2solver(numbers))

Simple solution
Step 1: First find the product of all the array elements and store in some variable let's sat prod.
Step 2: Divide prod variable by each array element.
Algo:
Given Arr = [1,2,3,4,5];
let prod = 1;
let outArr = [];
for(let count =0 ; count < Arr.length ; count++){
prod *= Arr[count];
}

for(let count =0 ; count < Arr.length ; count++){
outArr[count] = prod / Arr[count];
}
console.log(outArr)