/*
In this kata you have to correctly return who is the "survivor", ie: the last element of a Josephus permutation.

Basically you have to assume that n people are put into a circle and that they are eliminated in steps of k elements, like this:

josephus_survivor(7,3) => means 7 people in a circle;
one every 3 is eliminated until one remains
[1,2,3,4,5,6,7] - initial sequence
[1,2,4,5,6,7] => 3 is counted out
[1,2,4,5,7] => 6 is counted out
[1,4,5,7] => 2 is counted out
[1,4,5] => 7 is counted out
[1,4] => 5 is counted out
[4] => 1 counted out, 4 is the last element - the survivor!
The above link about the "base" kata description will give you a more thorough insight about the origin of this kind of permutation, but basically that's all that there is to know to solve this kata.

Notes and tips: using the solution to the other kata to check your function may be helpful, but as much larger numbers will be used, using an array/list to compute the number of the survivor may be too slow; you may assume that both n and k will always be >=1.
*/

//My Solution
//===========

fun josephusSurvivor(n: Int, k: Int): Int {
    return (1..n).reduce { x, ni -> (x + k - 1) % ni + 1 }
}

//Best Solution
//=============
 
fun josephusSurvivor(n: Int, k: Int): Int = (1..n).fold(1){ i, j -> (i + k) % j } + 1
