Unique numbers using bitwise operators - c++

Question

In an array that has n numbers, all numbers are present twice except for two numbers. find them using bitwise operators.
i have tried it by taking a xor of the nos. in the array. then finding out the right most set bit in the xor.
then for all the nos. which have the same position of set bit in the array i have taken a xor of them together,
my code:-
#include<iostream>
using namespace std;
int main()
{
cout<<"enter the size of array."<<endl;
int n;
cin>>n;
int arr[n];
cout<<"enter the array elements."<<endl;
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
int myxor=0;
for(int i=0;i<n;i++)
{
myxor^=arr[i];
}
int set=0;
int k=0;
while(myxor)
{ k++;
set=myxor&1;
if(set)
{
break;
}
//k++;
myxor>>=1;
}
int t,p;
int xor2=0; int c=0;
int xor1=0;
for(int j=0;j<n;j++)
{
p=arr[j];
while(p)
{ c++;
t=p&1;
if(t)
{
break;
}
//c++;
p>>=1;
}
if(c==k)
{
xor1^=arr[j];
}
else{
xor2^=arr[j];
}
}
cout<<xor1<<endl;
cout<<xor2<<endl;
return 0;
}

Answer 1

#include <iostream>
using namespace std;
int main()
{
int n;
cin >> n;
int *arr = new int[n];
for (int i = 0; i < n; ++i)
cin >> arr[i];
int res = 0;
for (int i = 0; i < n; ++i)
res ^= arr[i];
/*
So far, we XOR-ed all elements in the array; the trick behind, though,
is that any two duplicates will eliminate themselves, so effectively
res now contains the two unique numbers XOR-ed together.
*/
/*
Now we need the index of a bit where these two unique numbers differ
i. e. the index of a one-bit in res; ANY one will do the trick,
so let's just pick the right most one...
*/
int set = 1;
while (!(set & res))
set <<= 1;
/*
Now the tricky part.
It is guaranteed that there are only 2 numbers which occur only once in the array.
Note that for any set bit in the xor of these two numbers,
that bit has to occur either in the 1st or 2nd.
Now traverse the array and check for which numbers this bit is set.
Note that for any number for which it is set, that number occurs twice unless it's one of the 2 unique numbers.
Hence if we keep "XOR-ing" `res` with numbers for which `set` bit is set, we acquire one of the 2 unique numbers.
Since the xor of both is stored in `res`, "XOR-ing" `ans` with `res` gives the 2nd unique number.
*/
int ans = res;
for (int i = 0; i < n; ++i)
if (arr[i] & set)
ans ^= arr[i];
cout << ans << ' ' << (ans ^ res) << endl;
}