## 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:

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 `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:

1. Find out how many times I need the 25 cents coin
2. Find out how many times I need the 10 cents coin
3. Find out how many times I need the 5 cents coin
4. Find out how many times I need the 1 cents coin
5. 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 β½

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!