Problem Description

This problem is about simulating an elimination game known as the Josephus problem.

You have n friends sitting in a circle, numbered from 1 to n in clockwise order. They play an elimination game with the following rules:

1. The game starts with friend number 1
2. From the current position, count k friends in the clockwise direction (including the friend you start counting from)
3. The k-th friend counted is eliminated from the circle
4. If more than one friend remains, continue the game starting from the friend immediately clockwise to the eliminated friend
5. Repeat until only one friend remains - this friend is the winner

The counting wraps around the circle, so after friend n, you continue with friend 1.

Example visualization:

• If n = 5 and k = 2:
  • Start at friend 1, count 2 friends (1, 2) → friend 2 is eliminated
  • Start at friend 3, count 2 friends (3, 4) → friend 4 is eliminated
  • Start at friend 5, count 2 friends (5, 1) → friend 1 is eliminated
  • Start at friend 3, count 2 friends (3, 5) → friend 5 is eliminated
  • Friend 3 is the winner

The solution uses a recursive approach based on the mathematical property of the Josephus problem. The key insight is that if we know the winner's position for n-1 friends, we can calculate the winner's position for n friends using the formula: (k + winner(n-1, k)) % n. The base case is when n = 1, where the only friend (friend 1) is the winner.

The modulo operation handles the circular nature of the problem, and the special case when the result is 0 is adjusted to return n (since friends are numbered 1 to n, not 0 to n-1).

Intuition

The key insight comes from recognizing a pattern in how positions shift after each elimination.

Let's think about this problem step by step. When we have n friends and eliminate one, we're left with n-1 friends. The critical observation is that the winner among these n-1 friends will be the same as the original winner - just with a shifted position.

Consider what happens after the first elimination:

• We start at position 1 and count k friends
• Friend at position k is eliminated (using 1-based indexing)
• Now we have n-1 friends, and we start counting from position k+1

Here's the breakthrough: If we solve the smaller problem with n-1 friends first, we can map that solution back to our original problem. The winner's position in the n-1 game corresponds to a specific position in the original n game.

Think of it this way:

1. In the n-1 friends game, positions are numbered 1 to n-1
2. But in the original game, after eliminating the k-th friend, the new "position 1" in the reduced game is actually position k+1 in the original game
3. So if position p wins in the n-1 game, its actual position in the n game would be (k + p - 1) % n + 1

This leads to the recursive relationship: winner(n, k) = (k + winner(n-1, k)) % n

The modulo operation handles the circular wrapping. When the result is 0, we need to return n instead because we're using 1-based indexing (positions 1 to n) rather than 0-based.

The base case is straightforward: when there's only 1 friend (n = 1), that friend at position 1 is the winner.

This recursive approach elegantly reduces the problem size by 1 at each step, building up the solution from the simplest case to the full problem.

Learn more about Recursion, Queue and Math patterns.

Solution Approach

The solution implements a recursive approach to solve the Josephus problem efficiently.

Recursive Formula: The core of the solution relies on the mathematical relationship:

• winner(n, k) = (k + winner(n-1, k)) % n

Implementation Details:

1. Base Case: When n = 1, there's only one friend remaining, so friend 1 is the winner:

   if n == 1:
       return 1

2. Recursive Case: For n > 1, we calculate the winner's position using the recursive formula:

   ans = (k + self.findTheWinner(n - 1, k)) % n

   This formula works because:

   • self.findTheWinner(n - 1, k) gives us the winner's position in a game with n-1 friends
   • Adding k adjusts for the shift in positions after the first elimination
   • The modulo n handles the circular nature of the problem

3. Position Adjustment: Since we're using 1-based indexing (friends numbered 1 to n) but modulo gives us 0-based results:

   return n if ans == 0 else ans

   When ans == 0, it actually represents the n-th position in 1-based indexing.

Time and Space Complexity:

• Time Complexity: O(n) - We make exactly n recursive calls, one for each value from n down to 1
• Space Complexity: O(n) - The recursion stack depth is n

Example Walkthrough: For n = 5, k = 2:

• findTheWinner(5, 2) calls findTheWinner(4, 2)
• findTheWinner(4, 2) calls findTheWinner(3, 2)
• findTheWinner(3, 2) calls findTheWinner(2, 2)
• findTheWinner(2, 2) calls findTheWinner(1, 2)
• findTheWinner(1, 2) returns 1 (base case)
• Back-propagating:
  • findTheWinner(2, 2) = (2 + 1) % 2 = 1
  • findTheWinner(3, 2) = (2 + 1) % 3 = 0 → returns 3
  • findTheWinner(4, 2) = (2 + 3) % 4 = 1
  • findTheWinner(5, 2) = (2 + 1) % 5 = 3

The winner is friend 3.

Example Walkthrough

Let's walk through a small example with n = 4 friends and k = 3 to illustrate the solution approach.

Initial Setup:

• 4 friends sitting in a circle: [1, 2, 3, 4]
• We count 3 positions each time to eliminate someone

Step-by-step Simulation (to understand the problem):

1. Start at friend 1, count 3: [1, 2, 3] → Friend 3 is eliminated
   • Remaining: [1, 2, 4]
2. Start at friend 4, count 3: [4, 1, 2] → Friend 2 is eliminated
   • Remaining: [1, 4]
3. Start at friend 4, count 3: [4, 1, 4] → Friend 4 is eliminated
   • Remaining: [1]
4. Friend 1 is the winner!

Now let's trace through our recursive solution:

The recursion builds from the base case upward:

findTheWinner(4, 3)
  ├── calls findTheWinner(3, 3)
      ├── calls findTheWinner(2, 3)
          ├── calls findTheWinner(1, 3)
              └── returns 1 (base case: only 1 friend)
          └── returns (3 + 1) % 2 = 0 → returns 2
      └── returns (3 + 2) % 3 = 2 → returns 2
  └── returns (3 + 2) % 4 = 1 → returns 1

Breaking down each recursive return:

1. findTheWinner(1, 3) = 1

   • Base case: with 1 friend, friend 1 wins

2. findTheWinner(2, 3) = (3 + 1) % 2 = 0 → returns 2

   • With 2 friends, after adding the shift and taking modulo
   • Since result is 0, we return n=2 (adjusting for 1-based indexing)

3. findTheWinner(3, 3) = (3 + 2) % 3 = 2

   • With 3 friends, the winner is at position 2

4. findTheWinner(4, 3) = (3 + 2) % 4 = 1

   • With 4 friends, the winner is at position 1

Why the formula works:

• Each recursive call represents a game with one fewer friend
• The formula (k + winner(n-1, k)) % n accounts for:
  • The position shift that occurs after each elimination (the k term)
  • The winner's position in the smaller game (the winner(n-1, k) term)
  • The circular nature of the problem (the % n operation)

The final answer is friend 1, which matches our manual simulation!

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The recursive function makes exactly n calls, starting from n down to the base case when n = 1. Each recursive call performs a constant amount of work (modulo operation and addition), so the total time complexity is O(n).

Space Complexity: O(n)

The space complexity is determined by the recursion call stack. Since the function recursively calls itself n times before reaching the base case, the maximum depth of the recursion stack is n. Therefore, the space complexity is O(n) due to the recursive calls.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Index Confusion: 0-based vs 1-based Indexing

The most common pitfall in implementing the Josephus problem is mixing up 0-based and 1-based indexing. The mathematical formula works with 0-based indices internally, but the problem requires 1-based output (friends numbered 1 to n).

Incorrect Implementation:

def findTheWinner(self, n: int, k: int) -> int:
    if n == 1:
        return 1
    # Forgetting to handle the 0 case properly
    return (k + self.findTheWinner(n - 1, k)) % n

This would return 0 for certain cases, which is invalid since friends are numbered from 1 to n.

Correct Solution: Always check if the modulo result is 0 and convert it to n:

ans = (k + self.findTheWinner(n - 1, k)) % n
return n if ans == 0 else ans

2. Stack Overflow for Large n

The recursive solution has O(n) space complexity due to the call stack. For very large values of n (typically > 10^4), this could cause a stack overflow.

Iterative Solution to Avoid Stack Overflow:

def findTheWinner(self, n: int, k: int) -> int:
    # Start with 0-indexed position for 1 person
    winner = 0
  
    # Build up from 2 people to n people
    for i in range(2, n + 1):
        winner = (winner + k) % i
  
    # Convert from 0-indexed to 1-indexed
    return winner + 1

3. Incorrect Base Case Return Value

Some might incorrectly return 0 for the base case, thinking in 0-based terms:

Incorrect:

if n == 1:
    return 0  # Wrong! Friends are numbered 1 to n

Correct:

if n == 1:
    return 1  # The only friend is friend #1

4. Simulation Approach Performance Issues

While simulating the actual elimination process is intuitive, it's inefficient:

Inefficient Simulation Approach:

def findTheWinner(self, n: int, k: int) -> int:
    friends = list(range(1, n + 1))
    idx = 0
  
    while len(friends) > 1:
        idx = (idx + k - 1) % len(friends)
        friends.pop(idx)
  
    return friends[0]

This has O(n²) time complexity due to the list removal operations, compared to O(n) for the recursive/iterative mathematical solution.