Problem Description
You are given an array of integers nums representing 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.
Key Insights
- The game is based on the properties of XOR: if the XOR of a set of numbers is
0, then the player who is about to play will win. - The XOR of an even number of identical elements is
0, which can influence the outcome of the game. - The game's outcome can be determined by the XOR of all elements in the array at the start of the game.
- If the XOR of all elements is
0, Alice wins immediately; otherwise, the outcome depends on the number of elements.
Space and Time Complexity
Time Complexity: O(n) - where n is the number of elements in the nums array, as we need to compute the XOR of all elements once.
Space Complexity: O(1) - we only use a constant amount of space for storing variables.
Solution
To determine if Alice wins the game, we can calculate the XOR of all numbers in the nums array. If the result is 0, Alice wins immediately. If it's not 0, we evaluate the number of elements in the array. If the count is even, Bob can mirror Alice's moves, leading Alice to lose; if the count is odd, Alice can ensure she wins.
The algorithm uses a single pass to compute the XOR and checks the length of the array, making it efficient.