20. Modified Game of Nim

The problem can be found at the following link: Question Link

My Approach

This problem is a variation of the classic Nim game. The key observation is that if the XOR of all the elements in the array is 0, the first player will always win; otherwise, the winner depends on whether the number of elements is even or odd.

Here are the steps for impolementing above logic

• Calculate the XOR of all elements in the array.

• If XOR is 0, return 1 (indicating the first player wins).

• If XOR is not 0, return 1 if the number of elements is even; otherwise, return 2.

Time and Auxiliary Space Complexity

• Time Complexity: O(n) - We iterate through the array once.

• Auxiliary Space Complexity: O(1) - We use only a constant amount of extra space.

Code (C++)

class Solution {
public:
    int findWinner(int n, int A[]) {
        int XOR = 0;
        for(int i = 0; i < n; i++)
            XOR ^= A[i];

        if(XOR == 0)
            return 1;

        return n % 2 == 0 ? 1 : 2;
    }
};

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