How to meet properly if condition - kotlin

Question

I'm having a problem with this task. My goal is to calculate car price providing its mileage.
The formula is:
price lowers by 200$ for every 10000 kilometers that the car passed. For example, for 19999 km the price decreases by 200 dollars, but for 20000 km the price lowers by 400 dollars. And additionally the prices lowers due to car's age multiplied by 200.
What I did is:
when (input) {
in 0..9999 -> {
moneyToRemove = 2000
price -= moneyToRemove
}
else -> {
var counter = 0
moneyToRemove = old * 2000
for (x in 10000..input) {
if (x % 10000 == 0 ) {
counter++
if (x == input) {
println(counter)
price = price - moneyToRemove - counter*200
}
}
}
}
}
input - mileage provided from scanner
In first when condition moneyToRemove variable is set to 2000 ( 5years *200)
In else condition I tried to count how many extra dollars will lower the car price using counter. But then I have a problem with e.g. value 299 999. The counter in debugger mode shows 29 which is correct, but I can't meet the second condition and properly count car's price. Can you help?
This is from JetBrains academy Kotlin dev track.

Answer 1

You don't need any if conditions for that. You can put it in a simple formula
import kotlin.math.*
fun main() {
var originalprice = 30000; // the original price
var miles = 12345.6; // the mileage
var age = 7; //the age in years
var reducedprice = originalprice -
(floor(miles / 10000.0)) * 200 - //substract 200$ for each full 10000 miles
age * 200 //substract 200$ foreach year
print(reducedprice);
}
floor(miles/ 10000.0) will divide the miles by 10000 and round it downwards to the next integer. So it will be 0 for 0-9999, 1 for 10000 - 19999 and so on ... floor
If miles is an integer value, you don't even need floor, because the integer division will only return whole numbers. Ie 9999/10000 = 0, 19999/10000 = 1 and so on ...
var reducedprice = originalprice -
(miles / 10000) * 200 - //substract 200$ for each full 10000 miles
age * 200 //substract 200$ foreach year
EDIT
Eventhough I consider your approach as quite complicated and not easily maintainable
You have code duplication for the age part,
You do a lot of unnecessary % calculations for each value from 10000 up to your input
The when isn't necessary, because all the code exectued in the first branch is also executed in the second branch. Just could just do the loop, starting at 10000, so if the input is < 10000 it won't be executed ...
But all in all, your approach will in principle work too. But there are two issues with it
You don't consider the age of the car, when the mileage is below 10000, but you just withdraw a fixed amount.
When the mileage is above 10000, you calculate the reduced price only if the input is an exact multiple of 10000. You must put that calculation after the loop.
when (input) {
in 0..9999 -> {
moneyToRemove = old * 2000
price -= moneyToRemove
}
else -> {
moneyToRemove = old * 2000
for (x in 10000..input) {
if (x % 10000 == 0 ) {
moneyToRemove += 200
}
}
println(counter)
price -= moneyToRemove
}
}