## Intro π

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 Exerciseβ

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's exercise

Today, another `7 kyu`

kata,

meaning we slightly increase the difficulty.

Source: Codewars

Write a function `coinCombo`

, that accepts one parameter: `cents`

.

Given a number of cents, e.g. `51`

,

return the minimum number of coins combination of the same value, e.g. `[1, 0, 0, 2]`

.

The function should return an array where:

```
coins[0] = 1 cent
coins[1] = 5 cents
coins[2] = 10 cents
coins[3] = 25 cents
```

Example: `coinCombo(51)`

needs two `25 cents`

and one `1 cent`

=> `[1, 0, 0, 2]`

Input: a number.

Output: an array of numbers.

## Thinking about the Solution π

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 out how many times I need the 25 cents coin
- Find out how many times I need the 10 cents coin
- Find out how many times I need the 5 cents coin
- Find out how many times I need the 1 cents coin
- Return the array with the combination of the coins

Example:

- Input:
`51`

- Find out how many times I need the 25 cents coin:
`2`

, because`2 * 25 = 50`

=>`1`

left - Find out how many times I need the 10 cents coin:
`0`

, because I got only`1`

left =>`1`

left - Find out how many times I need the 5 cents coin:
`0`

, because I got only`1`

left =>`1`

left - Find out how many times I need the 1 cents coin:
`1`

, because`1 * 1 = 1`

=>`0`

left - Return the array with the combination of the coins:
`[1, 0, 0, 2]`

- Output:
`[1, 0, 0, 2]`

β

## Implementation β

```
function coinCombo() {
// all coin values
const coinValues = [25, 10, 5, 1];
// array for the output, filled with zeros
const coins = Array(coinValues.length).fill(0);
let currentCents = cents;
// iterate over the coins
for (const coin of coinValues) {
// only do this if there are some coins left
while (currentCents >= coin) {
// find out how many cents are left
// and how many times the current coins fit into the current cents
const remainder = currentCents % coin;
const increaseBy = (currentCents - remainder) / coin;
currentCents = currentCents % coin;
const index = coinValues.length - 1 - coinValues.indexOf(coin);
coins[index] += increaseBy;
}
}
return coins;
}
```

### Result

```
console.log(coinCombo(51));
// [1, 0, 0, 2] β
console.log(coinCombo(26));
// [1, 0, 0, 1] β
```

## Playground β½

You can play around with the code here

## Next Part β‘οΈ

Great work!

We learned how to use `Array`

, `for of`

, `while`

, `indexOf`

, `%`

.

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 π

## Questions β

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

## Top comments (3)

This took me a while:

I believe that understanding the solution is part of the fun, so I usually don't add explanation here not to ruin that deduction exercise. Though, I'd be happy to explain/discuss it if someone asks. Is that ok? Michael, what do you think?Hey Kostia,

nice solution. Think I never used

`reduceRight`

in real life.Sure, you can do that!

look at this solution

let coinCombo = (number) => {

let count25 = number / 25

let reminderOf25 = number % 25

let count10 = reminderOf25 / 10

let reminderOf10 = reminderOf25 % 10

let count5 = reminderOf10 / 5

let reminderOf5 = reminderOf10 % 5

let count1 = reminderOf5 / 1

console.log([parseInt(count1), parseInt(count5), parseInt(count10), parseInt(count25)])

}