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
class Solution:
def xorGame(self, nums: List[int]) -> bool:
return len(nums) % 2 == 0 or reduce(xor, nums) == 0
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;
}
};
func xorGame(nums []int) bool {
if len(nums)%2 == 0 {
return true
}
x := 0
for _, v := range nums {
x ^= v
}
return x == 0
}