Chalkboard XOR Game

Hard

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Bit Manipulation

You are given an array of integers nums represents the numbers written on a chalkboard.

Alice and Bob take turns erasing exactly one number from the chalkboard, with Alice starting first. If erasing a number causes the bitwise XOR of all the elements of the chalkboard to become 0, then that player loses. The bitwise XOR of one element is that element itself, and the bitwise XOR of no elements is 0.

Also, if any player starts their turn with the bitwise XOR of all the elements of the chalkboard equal to 0, then that player wins.

Return true if and only if Alice wins the game, assuming both players play optimally.

Example 1:

Input: nums = [1,1,2]
Output: false
Explanation: 
Alice has two choices: erase 1 or erase 2. 
If she erases 1, the nums array becomes [1, 2]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 2 = 3. Now Bob can remove any element he wants, because Alice will be the one to erase the last element and she will lose. 
If Alice erases 2 first, now nums become [1, 1]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 1 = 0. Alice will lose.

Example 2:

Input: nums = [0,1]
Output: true

Example 3:

Input: nums = [1,2,3]
Output: true

Constraints:

1 <= nums.length <= 1000
0 <= nums[i] < 2¹⁶

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the input array `nums`? Is there an upper limit on the values within the array?
Can the input array `nums` contain zero or negative numbers?
If there are multiple possible winning starting players, is any winning starting player acceptable, or is there a specific winning player I should aim to identify?
What should I return if neither player can win given the initial state of the chalkboard?
Are duplicate numbers allowed within the `nums` array, and if so, how should they be handled?

Brute Force Solution

Approach

The brute force approach to this game involves trying every possible sequence of moves a player can make. We simulate the game from the beginning, exploring all branches until we find a winning strategy for the first player or exhaust all possibilities.

Here's how the algorithm would work step-by-step:

Start with the initial set of numbers written on the chalkboard.
Consider all possible first moves for the first player. The first player can erase any one number from the chalkboard.
For each of those possible first moves, consider all possible responses from the second player. The second player also erases one of the remaining numbers.
Continue this process, alternating between the first and second player, until there are no numbers left on the chalkboard.
At the end of each complete game sequence, calculate the final XOR sum of the numbers remaining on the chalkboard. If the XOR sum is zero, the current player loses.
Keep track of which moves lead to a win for the first player. If we find even one sequence of moves where the first player always wins, the first player wins the game.
If after trying every possible sequence of moves we cannot find a guaranteed win for the first player, then the first player loses.

Code Implementation

def chalkboard_xor_game_brute_force(numbers):
    def calculate_xor_sum(current_numbers):
        xor_sum = 0
        for number in current_numbers:
            xor_sum ^= number
        return xor_sum

    def can_first_player_win(current_numbers, is_first_player_turn):
        if not current_numbers:
            return calculate_xor_sum([]) != 0

        # If it's the first player's turn, check if any move leads to a win.
        if is_first_player_turn:
            for index in range(len(current_numbers)):
                next_numbers = current_numbers[:index] + current_numbers[index+1:]
                if not can_first_player_win(next_numbers, not is_first_player_turn):
                    return True
            return False
        # If it's the second player's turn, all moves must lead to a win for the first player
        else:
            for index in range(len(current_numbers)):
                next_numbers = current_numbers[:index] + current_numbers[index+1:]
                if not can_first_player_win(next_numbers, not is_first_player_turn):
                    return False
            return True

    # Start the game with the first player's turn
    return can_first_player_win(numbers, True)

Big(O) Analysis

Time Complexity

O(n!) – The described brute force approach explores every possible sequence of moves. In the first move, there are n choices. In the second move, there are n-1 choices, and so on. This leads to n * (n-1) * (n-2) * ... * 1 possible game sequences. Calculating the XOR sum at the end of each sequence takes O(1) time. Therefore, the overall time complexity is dominated by the number of possible game sequences, which is n factorial (n!).

Space Complexity

O(N!) – The brute force approach explores every possible sequence of moves. In the worst case, where the first player must try every combination to see if it is a winning move, the recursion depth can reach N (the number of elements in the input chalkboard). Each level of the recursion potentially involves copying the chalkboard state (which can be of size up to N) as we explore different move combinations. This branching and copying of state at each level result in a space complexity corresponding to the exploration of all permutations of the initial input (N!).

Optimal Solution

Approach

The Chalkboard XOR Game has a surprising shortcut. Instead of simulating the entire game, we can directly determine the winner based on two simple properties of the numbers written on the board: the XOR sum of all the numbers and the number of numbers on the board.

Here's how the algorithm would work step-by-step:

First, calculate the XOR sum of all the numbers initially on the chalkboard. The XOR sum is like adding the numbers together but using a special rule where numbers 'cancel out' when they are the same.
If the XOR sum is zero, the first player automatically wins, regardless of how many numbers are on the board.
If the XOR sum is not zero, then we need to consider the number of numbers left on the chalkboard.
If the number of numbers on the chalkboard is even, the second player wins because the first player will always be forced to leave an odd number of numbers that don't XOR to zero, forcing them to lose.
If the number of numbers is odd and the XOR sum is not zero, the first player wins.

Code Implementation

def chalkboard_xor_game(numbers):
    xor_sum = 0
    for number in numbers:
        xor_sum ^= number

    # If XOR sum is zero, first player wins.
    if xor_sum == 0:
        return True

    #If the XOR sum isn't zero, parity of length matters
    if len(numbers) % 2 == 0:
        return False
    
    #If the number of elements is odd, first player wins.
    return True

Big(O) Analysis

Time Complexity

O(n) – The algorithm first calculates the XOR sum of all numbers in the input array nums of size n. This involves iterating through the array once, performing a constant-time XOR operation for each element. Then the algorithm checks if the XOR sum is zero and the length of the array, which are both constant-time operations. Thus, the dominant operation is the initial XOR sum calculation, which takes O(n) time. The remaining checks are O(1).

Space Complexity

O(1) – The algorithm calculates the XOR sum using a single variable. It also stores the number of elements in another variable. Neither of these depend on the input size beyond initially reading it. Therefore, the auxiliary space required is constant, regardless of the number of elements on the chalkboard (N).

Edge Cases

Case	How to Handle
Empty input array	Return True immediately, as the first player automatically wins (no moves available).
Single element array	Return False, as the first player is forced to erase the only element and loses if it's non-zero.
All elements XOR to 0	Return True, as the first player can always choose an element and win (either num.length is even, making XOR 0 or the player XORs such that the new XOR total is equal to the element removed).
Array with all identical values	If array length is even, return True; otherwise, return False because the XOR sum will always be equal to any single number in the array.
Array with a large number of elements (scalability)	Ensure the solution uses an efficient XOR calculation (O(n)) and length check rather than potentially slower methods.
Array with large integer values that could cause overflow during intermediate calculations if not handled carefully	XOR operations are safe from overflow in typical integer ranges.
Input array contains only zeros	Return True since the XOR sum is already 0 and the array length is considered.
Array length close to the maximum allowed size	Ensure the XOR sum calculation does not exceed the time limit during evaluation; also test within the specified size constraints.

Empty input array

How to Handle:

Return True immediately, as the first player automatically wins (no moves available).

Single element array

How to Handle:

Return False, as the first player is forced to erase the only element and loses if it's non-zero.

All elements XOR to 0

How to Handle:

Return True, as the first player can always choose an element and win (either num.length is even, making XOR 0 or the player XORs such that the new XOR total is equal to the element removed).

Array with all identical values

How to Handle:

If array length is even, return True; otherwise, return False because the XOR sum will always be equal to any single number in the array.

Array with a large number of elements (scalability)

How to Handle:

Ensure the solution uses an efficient XOR calculation (O(n)) and length check rather than potentially slower methods.

Array with large integer values that could cause overflow during intermediate calculations if not handled carefully

How to Handle:

XOR operations are safe from overflow in typical integer ranges.

Input array contains only zeros

How to Handle:

Return True since the XOR sum is already 0 and the array length is considered.

Array length close to the maximum allowed size

How to Handle:

Ensure the XOR sum calculation does not exceed the time limit during evaluation; also test within the specified size constraints.