In this article, we will deep dive into a popular problem on LeetCode, called the ‘Nim Game’. This problem, though simple, can be a litmus test for a candidate’s analytical skills and understanding of game theory and mathematical proofs. We will explore the problem statement, discuss the mathematics behind the problem, and analyze the solution’s time complexity.

## Introduction

First, let’s have a look at the problem statement:

You are playing the following Nim Game with your friend:

- Initially, there is a heap of
`n`

stones. - You and your friend will alternate taking turns, and you go first.
- On each turn, the player can remove 1 to 3 stones from the heap.
- The one who removes the last stone is the winner.

Given the number of stones `n`

, return `True`

if you can win the game, and `False`

if you cannot.

### Example:

Input: n = 4

Output: False

## Understanding the Problem

In this problem, we are playing a game where we can remove 1, 2, or 3 stones from a heap in each turn. We are the first player to move, and we want to know whether there is a strategy that guarantees our victory, regardless of how the other player plays.

## Mathematical Analysis

The key to solving this problem efficiently is to recognize the pattern that emerges when analyzing different values of `n`

.

Let’s take a look at different values of `n`

and whether the first player can win:

- n = 1: (1 stone) – Can win by taking 1.
- n = 2: (2 stones) – Can win by taking 2.
- n = 3: (3 stones) – Can win by taking 3.
- n = 4: (4 stones) – Cannot win as no matter how many stones we take, the opponent can always take the remaining stones.
- n = 5: (5 stones) – Can win by taking 1, as the opponent will then be in the position where n=4.
- n = 6: (6 stones) – Can win by taking 2.
- n = 7: (7 stones) – Can win by taking 3.
- n = 8: (8 stones) – Cannot win, as the opponent can force us into the position where n=4.

From this analysis, we can observe that we can always win if the number of stones is not divisible by 4. If the number of stones is divisible by 4, no matter how many stones we remove, the opponent can always remove a number of stones such that the number of stones remaining is again divisible by 4. This will eventually lead us to a situation where there are 4 stones remaining, and we have no winning move.

## Optimal Approach: Modulo Operation

With the above mathematical analysis, we can conclude that we can win if and only if the number of stones is not divisible by 4. We can easily check this by performing a modulo operation.

- Return
`True`

if`n`

modulo 4 is not equal to 0. - Otherwise, return
`False`

.

```
def canWinNim(n):
return n % 4 != 0
```

### Time Complexity:

O(1) – This approach only involves a single modulo operation, which is constant time.

### Space Complexity:

O(1) – No additional space is used.

## Conclusion:

The Nim Game problem on LeetCode is an elegant example of how mathematical analysis can lead to highly efficient algorithms. Through the pattern recognition and understanding of the underlying mathematics, we derived that the problem could be solved in constant time using a simple modulo operation. This problem emphasizes the importance of mathematics in algorithm design and showcases how sometimes a seemingly complex problem can have a very simple and efficient solution. Being familiar with such patterns and mathematical proofs is invaluable, especially in coding interviews and competitive programming.