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Find the XOR of Numbers Which Appear Twice

EasyUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

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.

Examples

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.

Solution

Method 1 – Hash Map Counting and XOR

Intuition

We need to find all numbers that appear exactly twice in the array and return their XOR. We can use a hash map (or array) to count occurrences, then XOR all numbers with count 2.

Approach

  1. Use a hash map (or array) to count the occurrences of each number in nums.
  2. Initialize ans = 0.
  3. For each number with count 2, XOR it into ans.
  4. Return ans (if no number appears twice, ans will be 0).

Code

C++
class Solution {
public:
    int duplicateNumbersXOR(vector<int>& nums) {
        unordered_map<int, int> cnt;
        for (int x : nums) cnt[x]++;
        int ans = 0;
        for (auto& [k, v] : cnt) if (v == 2) ans ^= k;
        return ans;
    }
};
Go
func duplicateNumbersXOR(nums []int) int {
    cnt := map[int]int{}
    for _, x := range nums { cnt[x]++ }
    ans := 0
    for k, v := range cnt {
        if v == 2 { ans ^= k }
    }
    return ans
}
Java
class Solution {
    public int duplicateNumbersXOR(int[] nums) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int x : nums) cnt.put(x, cnt.getOrDefault(x, 0) + 1);
        int ans = 0;
        for (var e : cnt.entrySet()) if (e.getValue() == 2) ans ^= e.getKey();
        return ans;
    }
}
Kotlin
class Solution {
    fun duplicateNumbersXOR(nums: IntArray): Int {
        val cnt = mutableMapOf<Int, Int>()
        for (x in nums) cnt[x] = cnt.getOrDefault(x, 0) + 1
        var ans = 0
        for ((k, v) in cnt) if (v == 2) ans = ans xor k
        return ans
    }
}
Python
class Solution:
    def duplicateNumbersXOR(self, nums: list[int]) -> int:
        from collections import Counter
        cnt = Counter(nums)
        ans = 0
        for k, v in cnt.items():
            if v == 2:
                ans ^= k
        return ans
Rust
impl Solution {
    pub fn duplicate_numbers_xor(nums: Vec<i32>) -> i32 {
        use std::collections::HashMap;
        let mut cnt = HashMap::new();
        for &x in &nums { *cnt.entry(x).or_insert(0) += 1; }
        let mut ans = 0;
        for (&k, &v) in &cnt {
            if v == 2 { ans ^= k; }
        }
        ans
    }
}
TypeScript
class Solution {
    duplicateNumbersXOR(nums: number[]): number {
        const cnt: Record<number, number> = {};
        for (const x of nums) cnt[x] = (cnt[x] || 0) + 1;
        let ans = 0;
        for (const k in cnt) if (cnt[k] === 2) ans ^= +k;
        return ans;
    }
}

Complexity

  • ⏰ Time complexity: O(n) — One pass to count, one pass to XOR.
  • 🧺 Space complexity: O(n) — For the count map.

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