Single Number III LeetCode Solution

Problem – Single Number III LeetCode Solution

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.

Single Number III LeetCode Solution in C++

class Solution
{
public:
    vector<int> singleNumber(vector<int>& nums) 
    {
        // Pass 1 : 
        // Get the XOR of the two numbers we need to find
        int diff = accumulate(nums.begin(), nums.end(), 0, bit_xor<int>());
        // Get its last set bit
        diff &= -diff;

        // Pass 2 :
        vector<int> rets = {0, 0}; // this vector stores the two numbers we will return
        for (int num : nums)
        {
            if ((num & diff) == 0) // the bit is not set
            {
                rets[0] ^= num;
            }
            else // the bit is set
            {
                rets[1] ^= num;
            }
        }
        return rets;
    }
};

Single Number III LeetCode Solution in Java

public class Solution {
    public int[] singleNumber(int[] nums) {
        // Pass 1 : 
        // Get the XOR of the two numbers we need to find
        int diff = 0;
        for (int num : nums) {
            diff ^= num;
        }
        // Get its last set bit
        diff &= -diff;
        
        // Pass 2 :
        int[] rets = {0, 0}; // this array stores the two numbers we will return
        for (int num : nums)
        {
            if ((num & diff) == 0) // the bit is not set
            {
                rets[0] ^= num;
            }
            else // the bit is set
            {
                rets[1] ^= num;
            }
        }
        return rets;
    }
}

Single Number III LeetCode Solution in Python

class Solution:
    def singleNumber(self, nums):
        s = reduce(xor, nums)
        nz = s & (s-1) ^ s
        num1 = reduce(xor, filter(lambda n: n & nz, nums))
        return(num1, s ^ num1)

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