singleNumberIII.cpp

/*
Given an array of numbers 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.
For example:
Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].
Note:
The order of the result is not important. So in the above example, [5, 3] is also correct.
Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?*/


//my Solution 
#include<set>
using namespace std;
class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
       set<int> si;
       for(auto x:nums)
       {
            if(si.find(x)==si.end())
                si.insert(x);
            else
                si.erase(x);
       }
       return vector<int>(si.begin(),si.end());
    }
};






//better Solution 
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;
    }
};