# 3158. Find the XOR of Numbers Which Appear Twice

## Description

You are given an array nums, where each number in the array appears either once or twice.

Return the bitwise XOR of all the numbers that appear twice in the array, or 0 if no number appears twice.

Example 1:

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

Output: 1

Explanation:

The only number that appears twice in nums is 1.

Example 2:

Input: nums = [1,2,3]

Output: 0

Explanation:

No number appears twice in nums.

Example 3:

Input: nums = [1,2,2,1]

Output: 3

Explanation:

Numbers 1 and 2 appeared twice. 1 XOR 2 == 3.

Constraints:

• 1 <= nums.length <= 50
• 1 <= nums[i] <= 50
• Each number in nums appears either once or twice.

## Solutions

### Solution 1: Counting

We define an array or hash table cnt to record the occurrence of each number.

Next, we traverse the array nums. When a number appears twice, we perform an XOR operation with the answer.

The time complexity is $O(n)$, and the space complexity is $O(M)$. Where $n$ is the length of the array nums, and $M$ is the maximum value in the array nums or the number of distinct numbers in the array nums.

• class Solution {
public int duplicateNumbersXOR(int[] nums) {
int[] cnt = new int[51];
int ans = 0;
for (int x : nums) {
if (++cnt[x] == 2) {
ans ^= x;
}
}
return ans;
}
}

• class Solution {
public:
int duplicateNumbersXOR(vector<int>& nums) {
int cnt[51]{};
int ans = 0;
for (int x : nums) {
if (++cnt[x] == 2) {
ans ^= x;
}
}
return ans;
}
};

• class Solution:
def duplicateNumbersXOR(self, nums: List[int]) -> int:
cnt = Counter(nums)
return reduce(xor, [x for x, v in cnt.items() if v == 2], 0)


• func duplicateNumbersXOR(nums []int) (ans int) {
cnt := [51]int{}
for _, x := range nums {
cnt[x]++
if cnt[x] == 2 {
ans ^= x
}
}
return
}

• function duplicateNumbersXOR(nums: number[]): number {
const cnt: number[] = Array(51).fill(0);
let ans: number = 0;
for (const x of nums) {
if (++cnt[x] === 2) {
ans ^= x;
}
}
return ans;
}