Problem Description

This is a two-player game called Nim Game where you and your friend take turns removing stones from a heap.

The game rules are:

• There is initially a heap containing n stones on the table
• You always go first
• Players alternate turns
• On each turn, a player must remove 1, 2, or 3 stones from the heap
• The player who removes the last stone wins the game

Given an integer n representing the initial number of stones, you need to determine if you can win the game when both players play optimally (meaning both players make the best possible moves).

The solution reveals an interesting pattern: you can win if and only if n % 4 != 0.

The key insight is that positions where the number of stones is divisible by 4 are losing positions for the current player. Here's why:

• If n < 4 (i.e., n = 1, 2, or 3), you can take all stones and win immediately
• If n = 4, no matter how many stones you take (1, 2, or 3), your opponent can take the remaining stones and win
• If n = 5, 6, or 7, you can take 1, 2, or 3 stones respectively to leave exactly 4 stones for your opponent, putting them in a losing position
• If n = 8, any move you make (taking 1, 2, or 3 stones) leaves 7, 6, or 5 stones for your opponent, all winning positions for them

This pattern continues: multiples of 4 are always losing positions because any legal move leaves the opponent with a non-multiple of 4 (a winning position), while non-multiples of 4 are winning positions because you can always force your opponent into a multiple of 4 (a losing position).

Intuition

The key to solving this problem is recognizing that it's a classic game theory problem where we need to identify winning and losing positions.

Let's think about this step by step by analyzing small values of n:

• When n = 1, 2, or 3: You can take all stones in one move and win immediately. These are winning positions.
• When n = 4: This is interesting - no matter what you do (take 1, 2, or 3 stones), you leave your opponent with 3, 2, or 1 stones respectively. Your opponent can then take all remaining stones and win. So n = 4 is a losing position.

Now let's continue:

• When n = 5: You can take 1 stone, leaving 4 for your opponent (forcing them into the losing position we just identified)
• When n = 6: You can take 2 stones, leaving 4 for your opponent
• When n = 7: You can take 3 stones, leaving 4 for your opponent

So positions 5, 6, and 7 are all winning positions because you can force your opponent into the losing position of 4.

What about n = 8? No matter what move you make:

• Take 1 stone → opponent gets 7 (winning position for them)
• Take 2 stones → opponent gets 6 (winning position for them)
• Take 3 stones → opponent gets 5 (winning position for them)

So n = 8 is another losing position!

The pattern emerges: positions that are multiples of 4 (4, 8, 12, ...) are losing positions, while all other positions are winning positions. This happens because:

1. From any multiple of 4, you can only move to a non-multiple of 4 (since you can only take 1, 2, or 3 stones)
2. From any non-multiple of 4, you can always move to a multiple of 4 (by taking the remainder when divided by 4)

Therefore, the solution is simply checking if n % 4 != 0. If the remainder is not zero, you're in a winning position and can force your opponent into a losing position (multiple of 4). If the remainder is zero, you're already in a losing position and any move you make gives your opponent a winning position.

Solution Approach

Based on the pattern we discovered, the implementation is remarkably simple and elegant.

The winning strategy revolves around the mathematical relationship with the number 4. From our analysis:

1. Losing positions: When n is divisible by 4 (i.e., n = 4, 8, 12, 16, ...)
2. Winning positions: When n is not divisible by 4 (i.e., n = 1, 2, 3, 5, 6, 7, 9, 10, 11, ...)

The implementation uses the modulo operator to check this condition:

class Solution:
    def canWinNim(self, n: int) -> bool:
        return n % 4 != 0

Why this works:

• If n % 4 = 0: You're at a losing position. Any move you make (removing 1, 2, or 3 stones) will leave n-1, n-2, or n-3 stones, all of which have non-zero remainders when divided by 4. This puts your opponent in a winning position.

• If n % 4 = 1, 2, or 3: You're at a winning position. You can remove exactly n % 4 stones, leaving a multiple of 4 for your opponent, forcing them into a losing position.

Mathematical proof by induction:

• Base case: For n < 4, the first player wins by taking all stones
• When n = 4k (multiple of 4): Any legal move leaves 4k-1, 4k-2, or 4k-3 stones, none of which are divisible by 4, giving the opponent a winning position
• When n = 4k + r (where r = 1, 2, or 3): The first player can remove r stones, leaving exactly 4k stones for the opponent, forcing them into a losing position

The time complexity is O(1) since we only perform a single modulo operation, and the space complexity is also O(1) as we don't use any additional data structures.

Example Walkthrough

Let's walk through a concrete example with n = 10 stones to illustrate the solution approach.

Initial Setup: 10 stones on the table, you go first.

Step 1: Check if you can win

• Calculate 10 % 4 = 2
• Since the remainder is not 0, you're in a winning position

Step 2: Find the optimal move

• Since 10 % 4 = 2, you should remove 2 stones
• This leaves exactly 8 stones for your opponent

Step 3: Opponent's turn (8 stones)

• Your opponent now faces 8 stones
• 8 % 4 = 0, so they're in a losing position
• No matter what they do:
  • Remove 1 stone → 7 stones remain
  • Remove 2 stones → 6 stones remain
  • Remove 3 stones → 5 stones remain
• All these outcomes (7, 6, or 5) have non-zero remainders when divided by 4

Step 4: Your turn again (let's say opponent removed 3, leaving 5 stones)

• You face 5 stones
• 5 % 4 = 1, so you're in a winning position
• Remove 1 stone, leaving 4 for your opponent

Step 5: Opponent's turn (4 stones)

• Your opponent faces 4 stones (a losing position again)
• Whatever they remove (1, 2, or 3), you can take the remaining stones and win

Game conclusion:

• If opponent removes 1 → 3 stones left → you take all 3 and win
• If opponent removes 2 → 2 stones left → you take both and win
• If opponent removes 3 → 1 stone left → you take it and win

This example demonstrates how maintaining the pattern of leaving multiples of 4 for your opponent guarantees your victory when both players play optimally.

Solution Implementation

Time and Space Complexity

The time complexity is O(1) because the solution only performs a single modulo operation (n % 4) and a comparison operation (!= 0), both of which take constant time regardless of the input size n. The modulo operation on integers is a constant-time arithmetic operation.

The space complexity is O(1) because the solution uses no additional data structures or variables that scale with the input. It only stores the result of the modulo operation temporarily and returns a boolean value, which requires a fixed amount of memory regardless of the value of n.

Common Pitfalls

1. Overthinking the Solution

Many developers initially try to implement complex recursive or dynamic programming solutions with memoization, attempting to simulate all possible game states. This leads to unnecessary complexity and potential stack overflow or time limit exceeded errors for large values of n.

Incorrect approach:

class Solution:
    def canWinNim(self, n: int) -> bool:
        # Overly complex recursive solution
        memo = {}
        def canWin(stones):
            if stones <= 3:
                return True
            if stones in memo:
                return memo[stones]
          
            # Check if any move leads to opponent losing
            for move in [1, 2, 3]:
                if not canWin(stones - move):
                    memo[stones] = True
                    return True
          
            memo[stones] = False
            return False
      
        return canWin(n)

Why it's problematic: This recursive solution has O(n) time complexity and O(n) space complexity, and will fail for large inputs due to recursion depth limits or memory constraints.

2. Misunderstanding the Modulo Operation

Some might incorrectly implement the condition as n % 4 == 0 (returning True when divisible by 4) instead of n % 4 != 0, reversing the win/lose logic.

Incorrect implementation:

class Solution:
    def canWinNim(self, n: int) -> bool:
        return n % 4 == 0  # Wrong! This returns True for losing positions

3. Not Recognizing the Mathematical Pattern

Developers might try to handle special cases separately without recognizing the underlying pattern, leading to verbose and error-prone code:

Inefficient approach:

class Solution:
    def canWinNim(self, n: int) -> bool:
        if n <= 3:
            return True
        if n == 4:
            return False
        if n == 5 or n == 6 or n == 7:
            return True
        if n == 8:
            return False
        # Attempting to list all cases...

4. Assuming Linear Game Strategy

Some might think that taking the maximum number of stones (3) each turn is optimal, or that there's a greedy strategy based on the current stone count, missing the deeper mathematical insight.

Solution to Avoid Pitfalls:

1. Recognize that this is a classic game theory problem with a mathematical solution
2. Test the pattern with small values (n = 1 to 12) to verify the modulo 4 relationship
3. Understand that positions divisible by 4 are "cold" positions (losing for the current player)
4. Remember that optimal play means forcing your opponent into a losing position whenever possible