josephos-survivor.js

/*
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.
*/

function josephusSurvivor(n,k){
  let array = []
  let elIndex = k - 1
  for (let i = 1; i <= n; i++){
    array[i - 1] = i
  }
  
    if (elIndex > array.length){
      elIndex = elIndex - (array.length * Math.floor(elIndex/array.length))
    }
    else if (elIndex == array.length){
      elIndex = 0
    }

  while (array.length > 1){
    if (elIndex == 0){
      array.shift()
    }
    else{
      array.splice(elIndex, 1)
    }
      
    elIndex += k - 1
    
    if (elIndex > array.length){
      elIndex = elIndex - (array.length * Math.floor(elIndex/array.length)) 
    }
    else if (elIndex == array.length){
      elIndex = 0
    }
  }
  return array[0]
}