Find the number which occurs once in an array while all other elements occurs twice using XOR

 1. Find the number which occurs once in an array while all other elements occurs twice using XOR,

CODE: 

#include <iostream>

using namespace std;

int main()

{

    int a[11]={1,3,9,9,3,6,7,8,7,8,1};//declaring the array//

    int result=a[0];

    for(int i=1;i<11;i++)

    {

        result=result^a[i];

    }

    cout<<result<<endl;

     return 0;

}

OUTPUT:

NOTE: In XOR operation two same elements gives an output of 0 . Also, XOR operation is commutative and associative, so all the elements appearing twice will results in 0 giving the number which occurs once as output. 

In this problem, 

            

                        1^3^9^9^3^6^7^8^7^8^1

                        =(1^1)^(3^3)^(9^9)^6^(7^7)^(8^8)   (by commutative and associative property)

                        =0^0^0^6^0^0

                        =6^0

                        =6      (giving the output as 6 which occurs once)