β Blog## JavaScript Katas: Find Odd Digits

## Intro π

## Understanding the Exerciseβ

### Today's exercise

## Thinking about the Solution π

## Implementation (Explicit) β

### Result

## Implementation (Implicit) β

### Result

## Playground β½

## Next Part β‘οΈ

## Further Reading π

## Questions β

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.

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

My personal method:

- Input: What do I put in?
- Output: What do I want to get out?

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.

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:

- Find all odd digits
- Find the first
`k`

digits - Handle all edge cases
- 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`

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);
}
```

```
console.log(findOddDigits(123456789111, 5));
// 13579 β
console.log(findOddDigits(0, 100));
// 0 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));
}
```

```
console.log(findOddDigits(123456789111, 5));
// 13579 β
console.log(findOddDigits(0, 100));
// 0 β
```

You can play around with the code here

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!

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

Hi! I'm Michael π I'm a Mentor & Senior Web Developer - I help you to reach your (career) goals.