Josephus problem

def josephus(n,k):
    arr=list(range(1,n+1))
    p=0
    while len(arr)>1:
        p=(p+k-1)%len(arr)
        arr.pop(p)
    return arr[0]

• (p+k-1) % len(arr) — pointer reset in O(1), always one step
• arr.pop(p) — mid-array deletion in O(n), shifts all elements after index p
• Outer while loop runs n−1 times → O(n)

Combined: O(n) × O(n) = $O(n^2)$ time, O(n) space.