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810. Chalkboard XOR Game
Description
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] < 2^{16}
Solutions

class Solution { public boolean xorGame(int[] nums) { return nums.length % 2 == 0  Arrays.stream(nums).reduce(0, (a, b) > a ^ b) == 0; } }

class Solution { public: bool xorGame(vector<int>& nums) { if (nums.size() % 2 == 0) return true; int x = 0; for (int& v : nums) x ^= v; return x == 0; } };

class Solution: def xorGame(self, nums: List[int]) > bool: return len(nums) % 2 == 0 or reduce(xor, nums) == 0

func xorGame(nums []int) bool { if len(nums)%2 == 0 { return true } x := 0 for _, v := range nums { x ^= v } return x == 0 }