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:
- 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 💭
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, because2 * 25 = 50=>1left - Find out how many times I need the 10 cents coin:
0, because I got only1left =>1left - Find out how many times I need the 5 cents coin:
0, because I got only1left =>1left - Find out how many times I need the 1 cents coin:
1, because1 * 1 = 1=>0left - Return the array with the combination of the coins:
[1, 0, 0, 2] - Output:
[1, 0, 0, 2]✅
Implementation ⛑
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] ✅
console.log(coinCombo(26));
// [1, 0, 0, 1] ✅
Playground ⚽
You can play around with the code here
Next Part ➡️
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 📖
Questions ❔
- 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
reduceRightin 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)])
}