Suppose we have n children standing in a circle, and they are waiting to get a balloon. The distribution is carried out starting with the kth child (first at index 0), and giving them a balloon they left the circle. Now every kth child gets a balloon going clockwise until there is only one child left that gets a balloon. So if we have n and k, we have to find the starting index of the child that receives the last balloon.
So, if the input is like n = 3 k = 2, then the output will be 1, in the first round, child 2 gets a balloon, and leave so the circle will be [0, 1]. In second round, child 0 gets a balloon, circle will be .
To solve this, we will follow these steps:
arr := a new list from range 0 to n
init := 0
while size of arr > 1, do
remove := (init + k) mod size of arr
init := remove
Let us see the following implementation to get better understanding:
class Solution: def solve(self, n, k): arr = list(range(0, n)) init = 0 while len(arr) > 1: remove = (init + k) % len(arr) del arr[remove] init = remove return arr ob = Solution() n = 3 k = 2 print(ob.solve(n, k))