Description Link to heading

260. Single Number III (Medium)

Given an integer array nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. You can return the answer in any order.

You must write an algorithm that runs in linear runtime complexity and uses only constant extra space.

Example 1:

Input: nums = [1,2,1,3,2,5]
Output: [3,5]
Explanation:  [5, 3] is also a valid answer.

Example 2:

Input: nums = [-1,0]
Output: [-1,0]

Example 3:

Input: nums = [0,1]
Output: [1,0]

Constraints:

  • 2 <= nums.length <= 3 * 10⁴
  • -2³¹ <= nums[i] <= 2³¹ - 1
  • Each integer in nums will appear twice, only two integers will appear once.

Solution Link to heading

Initially, perform bitwise XOR on all numbers. Assuming the elements to be identified are denoted as $a$ and $b$, the result of the XOR operation is $c = a \oplus b$. Consider obtaining the lowbit of $c$, which is $c & (-c)$. This enables us to categorize all elements into two groups: one with $nums[i] & lowbit = 0$ and the other with non-zero values. Notably, $a$ and $b$ each belong to one of these groups. If the XOR result of all elements in the first group is $a$, then the XOR result of all elements in the second group is $b$.

Code Link to heading 

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
       	long res = 0;
       	int n = nums.size();
       	for (int i = 0; i < n; ++i) {
       		res = res ^ nums[i];
       	} 
       	long c = (res & (-res));
       	int res1 = 0, res2 = 0;
       	for (int i = 0; i < n; ++i) {
       		if (nums[i] & c) {
       			res1 = res1 ^ nums[i];
       		} else {
       			res2 = res2 ^ nums[i];
       		}
       	}
       	return {res1, res2};
    }
};