Solution Explanation for LeetCode 810: Chalkboard XOR Game

This problem explores game theory concepts within the context of bitwise XOR operations. The core idea revolves around the parity of the number of elements and the final XOR sum.

Understanding the Game

Two players, Alice and Bob, take turns removing a single number from an array nums. The game ends when the XOR sum of the remaining numbers becomes 0. The player who makes the XOR sum 0 loses. If a player starts their turn with the XOR sum already being 0, they win immediately.

Winning Strategy Analysis

The solution leverages a crucial observation about the XOR operation and the parity of the array's length.

• Initial XOR Sum is 0: If the XOR sum of all numbers in nums is initially 0, Alice wins immediately.

• Even Length: If the length of nums is even, Alice will always win. This is because no matter which number Alice chooses, Bob will always be left with an odd number of elements. Alice can then strategically choose numbers to ensure the XOR sum remains non-zero until Bob is forced to make the XOR sum 0, resulting in Bob's loss.

• Odd Length and Non-Zero Initial XOR Sum: If the length of nums is odd and the initial XOR sum is not 0, Alice will always lose. This is because no matter what Alice chooses, Bob will be left with an even number of elements, putting Bob in a winning position as described above.

Code Implementation

The code efficiently implements this winning strategy. It first checks if the initial XOR sum is 0. If not, it checks the parity (even or odd) of the array's length.

Time and Space Complexity

• Time Complexity: O(n), where n is the length of nums. This is because calculating the XOR sum takes linear time.

• Space Complexity: O(1). The solution uses only a constant amount of extra space.

Code Examples (with explanations)

Python:

from functools import reduce
import operator
 
class Solution:
    def xorGame(self, nums: List[int]) -> bool:
        """
        Determines if Alice wins the Chalkboard XOR Game.
 
        Args:
            nums: A list of integers representing the numbers on the chalkboard.
 
        Returns:
            True if Alice wins, False otherwise.
        """
        xor_sum = reduce(operator.xor, nums, 0)  #Efficient XOR sum calculation
        return len(nums) % 2 == 0 or xor_sum == 0
 

Java:

import java.util.Arrays;
 
class Solution {
    public boolean xorGame(int[] nums) {
        int xorSum = Arrays.stream(nums).reduce(0, (a, b) -> a ^ b); //Java streams for efficient XOR
        return nums.length % 2 == 0 || xorSum == 0;
    }
}

C++:

#include <numeric> //for accumulate function
 
class Solution {
public:
    bool xorGame(vector<int>& nums) {
        int xorSum = accumulate(nums.begin(), nums.end(), 0, bit_xor<int>()); //Efficient XOR using accumulate and bit_xor
        return nums.size() % 2 == 0 || xorSum == 0;
    }
};

Go:

func xorGame(nums []int) bool {
    xorSum := 0
    for _, num := range nums {
        xorSum ^= num
    }
    return len(nums)%2 == 0 || xorSum == 0
}

These codes all follow the same logic: calculate the XOR sum and check the array's length parity to determine the winner. The use of reduce (Python), streams (Java), accumulate (C++), and a simple loop (Go) showcase different approaches to efficient XOR sum calculation in various languages.