DEV Community

Cover image for JavaScript Katas: Find Odd Digits
miku86
miku86

Posted on

JavaScript Katas: Find Odd Digits

Intro 🌐

Problem solving is an important skill, for your career and your life in general.

That's why I take interesting katas of all levels, customize them and explain how to solve them.


Understanding the Exercise❗

First, we need to understand the exercise!
If you don't understand it, you can't solve it!.

My personal method:

  1. Input: What do I put in?
  2. Output: What do I want to get out?

Today's exercise

Today, another 7 kyu kata,
meaning we slightly increase the difficulty.

Source: Codewars

Write a function findOddDigits, that accepts two parameter: n a number and k a number.

Given n, e.g. 123456789111, and k, e.g. 5, return the first k odd digits in n, e.g. 13579.

In the following cases the function should return 0:

  • there are no odd digits in n;
  • k is bigger than n;
  • k is 0;
  • k is bigger than the number of odd digits in n.

Input: two numbers.

Output: one number.


Thinking about the Solution πŸ’­

I think I understand the exercise (= what I put into the function and what I want to get out of it).

Now, I need the specific steps to get from input to output.

I try to do this in small baby steps:

  1. Find all odd digits
  2. Find the first k digits
  3. Handle all edge cases
  4. Return the number

Example:

  • Input: 123456789111, 5
  • Find all odd digits: 13579111
  • Find the first k (= 5) digits: 13579
  • Return the number: 13579
  • Output: 13579 βœ…

Implementation (Explicit) β›‘

function findOddDigits(n, k) {
  // k = 0;
  // k is bigger than a number of digits in n;
  if (k === 0 || k > n) return 0;

  // find all odd digits
  const str = String(n);
  const split = str.split("");
  const odds = split.filter((num) => num % 2);

  // there are no odd digits in a number n;
  // k is bigger than a number of odd digits in n.
  if (!odds.length || k > odds.length) return 0;

  // find the first `k` digits
  const joined = odds.join("");
  const sliced = joined.slice(0, k);

  // return the number
  return Number(sliced);
}
Enter fullscreen mode Exit fullscreen mode

Result

console.log(findOddDigits(123456789111, 5));
// 13579 βœ…

console.log(findOddDigits(0, 100));
// 0 βœ…
Enter fullscreen mode Exit fullscreen mode

Implementation (Implicit) β›‘

function findOddDigits(n, k) {
  // find all odd digits
  const odds = String(n)
    .split("")
    .filter((num) => num % 2);

  // handle all edge cases
  if (k === 0 || k > n || !odds.length || k > odds.length) return 0;

  // find the first `k` digits and return them as a number
  return Number(odds.join("").slice(0, k));
}
Enter fullscreen mode Exit fullscreen mode

Result

console.log(findOddDigits(123456789111, 5));
// 13579 βœ…

console.log(findOddDigits(0, 100));
// 0 βœ…
Enter fullscreen mode Exit fullscreen mode

Playground ⚽

You can play around with the code here


Next Part ➑️

Great work!

We learned how to use String, Number, split, join, filter, slice.

I hope you can use your new learnings to solve problems more easily!

Next time, we'll solve another interesting kata. Stay tuned!


If I should solve a specific kata, shoot me a message here.

If you want to read my latest stuff, get in touch with me!


Further Reading πŸ“–


Questions ❔

  • How often do you do katas?
  • Which implementation do you like more? Why?
  • Any alternative solution?

Top comments (2)

Collapse
 
gerges27 profile image
Gerges Nady

let stringN = numbers.toString()
let countODD = []
let result = ""

if (k > numbers || k === 0) {
    result = 0
} else {
    for (const number of stringN) {
        if (number % 2 !== 0) {
            countODD.push(number)
        }
    }
    for (let i = 0; i < k; i++) {
        result += countODD[i]
        if (countODD[i] === undefined){
            result = 0
        }
    }
}
console.log(Number(result))
Enter fullscreen mode Exit fullscreen mode
Collapse
 
jpantunes profile image
JP Antunes

And now in one single (almost readable) line!

const odds = (n, k) => n > Math.max(0, k) ? Number([...`${n}`].filter(num => num % 2).join('').slice(0, k)) : 0