260. Single Number III LeetCode Solution

Last updated: February 15, 2025

In this guide, you will get 260. Single Number III LeetCode Solution with the best time and space complexity. The solution to Single Number III problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Single Number III solution in C++
Single Number III solution in Java
Single Number III solution in Python
Additional Resources

260. Single Number III LeetCode Solution image

Problem Statement of Single Number III

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.

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(1)

260. Single Number III LeetCode Solution in C++

class Solution {
 public:
  vector<int> singleNumber(vector<int>& nums) {
    const int xors = accumulate(nums.begin(), nums.end(), 0, bit_xor<>());
    const int lowbit = xors & -xors;
    vector<int> ans(2);

    // Seperate `nums` into two groups by `lowbit`.
    for (const int num : nums)
      if (num & lowbit)
        ans[0] ^= num;
      else
        ans[1] ^= num;

    return ans;
  }
};
/* code provided by PROGIEZ */

260. Single Number III LeetCode Solution in Java

class Solution {
  public int[] singleNumber(int[] nums) {
    final int xors = Arrays.stream(nums).reduce((a, b) -> a ^ b).getAsInt();
    final int lowbit = xors & -xors;
    int[] ans = new int[2];

    // Seperate `nums` into two groups by `lowbit`.
    for (final int num : nums)
      if ((num & lowbit) > 0)
        ans[0] ^= num;
      else
        ans[1] ^= num;

    return ans;
  }
}
// code provided by PROGIEZ

260. Single Number III LeetCode Solution in Python

class Solution:
  def singleNumber(self, nums: list[int]) -> list[int]:
    xors = functools.reduce(operator.xor, nums)
    lowbit = xors & -xors
    ans = [0, 0]

    # Seperate `nums` into two groups by `lowbit`.
    for num in nums:
      if num & lowbit:
        ans[0] ^= num
      else:
        ans[1] ^= num

    return ans
 # code by PROGIEZ

Additional Resources

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