Codewars: Multiples of 3 or 5

This problem on codewars asks the following:

A function should return the sum of all the multiples of 3 or 5 below a given number. If the number is negative, return 0. For numbers that are multiples of both 3 and 5, count them only once.

Example: (Input --> Output)

solution(10) = 23
solution(-1) = 0
solution(15) = 45 (Not including 15)

How to approach it with the PEDAC method:

P: Understand the problem -
The task is to compute the sum of all numbers below a given number that are multiples of 3 or 5, treating negatives as a special case where the result is always 0.

E: Give Example -
For a number like 10, you’d get 3, 5, 6, and 9 as the multiples, summing up to 23.
For -1, since it's negative, the result should be 0.

D: What data structure(s) would be needed -
You can simply use a variable to accumulate the total sum; no complex data structure is required.

A: Steps to solve it without language specific information -

• Initialize a sum variable at 0.
• If the input number is negative, return 0 immediately.
• Loop through numbers starting from 1 up to but not including the input number.
• For each number, check if it's a multiple of 3 or 5.
• If it is, add it to the sum.
• Return the sum after the loop.

C: The final code using the pseudocode (A) to guide me -

- Initialize sum and handle negative input

function solution(number){
  let total = 0;  
  if(number < 0){ return 0}
}

- Loop through numbers and check for multiples of 3 or 5

function solution(number){
  let total = 0;  
  if(number < 0){ return 0}

  for(let i = 1; i < number; i++){
    if(i % 3 === 0 || i % 5 === 0){
      total += i;
    }
  }

  return total; 
}

- Final step: returning the total sum

function solution(number){
  let total = 0;  
  if(number < 0){ return 0}

  for(let i = 1; i < number; i++){
    if(i % 3 === 0 || i % 5 === 0){
      total += i;
    }
  }

  return total; // This is where the sum gets returned
}