Nim Game

EASY

Description

You are playing the following Nim Game with your friend:

Initially, there is a heap of stones on the table.
You and your friend will alternate taking turns, and you go first.
On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
The one who removes the last stone is the winner.

Given n, the number of stones in the heap, return true if you can win the game assuming both you and your friend play optimally, otherwise return false.

Example 1:

Input: n = 4
Output: false
Explanation: These are the possible outcomes:
1. You remove 1 stone. Your friend removes 3 stones, including the last stone. Your friend wins.
2. You remove 2 stones. Your friend removes 2 stones, including the last stone. Your friend wins.
3. You remove 3 stones. Your friend removes the last stone. Your friend wins.
In all outcomes, your friend wins.

Example 2:

Input: n = 1
Output: true

Example 3:

Input: n = 2
Output: true

Constraints:

1 <= n <= 2³¹ - 1

Approaches

Checkout 3 different approaches to solve Nim Game. Click on different approaches to view the approach and algorithm in detail.

Recursive Solution

We can solve this problem using recursion by trying all possible moves and checking if any of them leads to a winning position.

Algorithm

Check base cases:
- If n ≤ 0, return false
- If n ≤ 3, return true
Recursively check if we can win by trying all possible moves
Return true if any move leads to a winning position

In this approach, we use recursion to simulate all possible moves and determine if we can win. For each turn:

Base cases:
- If n = 0, return false (no stones left)
- If n = 1, 2, or 3, return true (can take all stones)
For each possible move (1, 2, or 3 stones), we:
- Make the move by subtracting stones
- Recursively check if opponent can win with remaining stones
- If opponent can't win, we win

Here's the implementation:

public boolean canWinNim(int n) {
    if (n <= 0) return false;
    if (n <= 3) return true;
    
    // Try removing 1, 2, or 3 stones
    return !(canWinNim(n-1) && canWinNim(n-2) && canWinNim(n-3));
}

Complexity Analysis

Time Complexity: O(3^n) - For each position, we make 3 recursive callsSpace Complexity: O(n) - Maximum depth of recursion tree

Pros and Cons

Pros:

Easy to understand and implement
Directly simulates the game process
Works for small inputs

Cons:

Extremely inefficient for large inputs
Will cause stack overflow for large n
Many redundant calculations

Code Solutions

Checking out 3 solutions in different languages for Nim Game. Click on different languages to view the code.

class Solution {
public
  boolean canWinNim(int n) { return n % 4 != 0; }
}

Video Solution

Watch the video walkthrough for Nim Game

Patterns:

MathGame Theory