class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int n = nums.size();
        set<int> set;
        int a=0, b=0;
		
		//inserting all the nums in the set
		//and computing the sum of all the nums        
        for(auto x: nums){
            set.insert(x);
            b+=x;
        }
        //now re-computing the sum of all the unique elements
        //present in the set and since every duplicate number has
        //frequncy equals to 2 so by multiplying the sum we will get
        //the sum if all the nums are duplicate
        
        for(auto x: set)
            a+=x;
        a = a*2;
        //now if we divide the sum a and b we will get the num 
        //that appears only once
        return a-b;
    }
};

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int res = 0;
        for(auto x: nums)
	        //by using xor operator we can find 
	        //the num that appear only once
            res^=x;
        return res;
    }
};