Table Of Contents

• Multiples of 3 or 5
  • 📃 Description
  • 🤔 Thinking
  • 👨‍💻 Code
  • 🐞 Bugs
• Validate Credit Card Number
  • 📃 Description
  • 🤔 Thinking
  • 👨‍💻 Code
  • 🐞 Bugs
  • 🏁 Finally

Multiples of 3 or 5 : ✍ by jhoffner

📃 Description

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

🤔 Thinking

I'm gonna make a for loop will check if a number less than the limit divisible % by 3 or || 5
Then push it to an array
Then add it

👨‍💻 Code

const solution = num => {
  let divBy3Or5 = [];

  for(let i = 1; i < num; i++) {
    ((i % 3 === 0) || (i % 5 === 0)) ? divBy3Or5.push(i) : null;  
  }

  return divBy3Or5.reduce((acc, elm) => acc + elm, 0);
}

🐞 Bugs

I think it's a stupid code like it has to be a better way to solve this kata
Please if you have a better solution let me know in the comments

Validate Credit Card Number : ✍ by mcclaskc

📃 Description

In this Kata, you will implement the Luhn Algorithm, which is used to help validate credit card numbers.
Given a positive integer of up to 16 digits, return true if it is a valid credit card number, and false if it is not.

Here is the algorithm:

Double every other digit, scanning from right to left, starting from the second digit (from the right).
Another way to think about it is: if there are an even number of digits, double every other digit starting with the first; if there are an odd number of digits, double every other digit starting with the second:

1714 ==> [1*, 7, 1*, 4] ==> [2, 7, 2, 4]

12345 ==> [1, 2*, 3, 4*, 5] ==> [1, 4, 3, 8, 5]

891 ==> [8, 9*, 1] ==> [8, 18, 1]

If a resulting number is greater than 9, replace it with the sum of its own digits (which is the same as subtracting 9 from it):

[8, 18*, 1] ==> [8, (1+8), 1] ==> [8, 9, 1]

// OR

[8, 18*, 1] ==> [8, (18-9), 1] ==> [8, 9, 1]

Sum all of the final digits:

[8, 9, 1] ==> 8 + 9 + 1 = 18

Finally, take that sum and divide it by 10. If the remainder equals zero, the original credit card number is valid.

18 (modulus) 10 ==> 8 , which is not equal to 0, so this is not a valid credit card number

🤔 Thinking

I will create an array out of the credit card number
Then check its length if it's even I will loop over it starting in the first index 0 jumping one index at a time like 0, 2, 4, 6, n.length if its odd I will do the same but starting in the second element index number 1
Then double it and add them into another array and then sum it
Then divide it by 10 and check if its remainder is equal to 0

👨‍💻 Code

const validate = num => {
  let numArr = Array.from(String(num), Number);

  if (numArr.length % 2 === 0) {
    for(let i = 0; i< numArr.length; i+=2) {
      numArr[i] *= 2;
    }
  } else {
    for(let i = 1; i< numArr.length; i+=2) {
      numArr[i] *= 2;
    }
  }

  const lessThan18Arr = numArr.map(num => num > 9 ? num - 9 : num)

  const sum = lessThan18Arr.reduce((acc, elm) => acc + elm, 0)

  return sum % 10 === 0;
}

🐞 Bugs

I think it's the Time complexity (Both Solutions takes about 1000ms give or take 100ms)
And there is repetitive code
Not DRY (Don't Repeat Yourself)

🏁 Finally

const validate = num => {
  let numArr = Array.from(String(num), Number);  
  let i = numArr.length % 2 === 0 ? 0 : 1; 

  while(i < numArr.length) {
    numArr[i] *= 2;
    i+=2;
  }

  const lessThan18Arr = numArr.map(num => num > 9 ? num - 9 : num)

  const sum = lessThan18Arr.reduce((acc, elm) => acc + elm, 0)

  return sum % 10 === 0;
}

If you know a better way to solve any of the previous katas let me know in the comment