Using XOR operator for finding duplicate elements in a array fails in many cases

Question

I came across a post How to find a duplicate element in an array of shuffled consecutive integers? but later realized that this fails for many input.

For ex:
arr[] = {601,602,603,604,605,605,606,607}

#include <stdio.h>
int main()
{
int arr[] = {2,3,4,5,5,7};
int i, dupe = 0;
for (i = 0; i < 6; i++) {
    dupe = dupe ^ a[i] ^ i;
}
printf ("%d\n", dupe);
return 0;
}

How can I modify this code so that the duplicate element can be found for all the cases ?

Answer 1

Remember these two properties of XOR operator :

(1) If you take xor of a number with 0 ( zero ) , it would return the same number again.

Means , n ^ 0 = n

(2) If you take xor of a number with itself , it would return 0 ( zero ).

Means , n ^ n = 0

Now , Coming to the problem :

   Let    Input_arr = { 23 , 21 , 24 , 27 , 22 , 27 , 26 , 25 }    

   Output should be 27 ( because 27 is the duplicate element in the Input_arr ).

Solution :

Step 1 : Find “min” and “max” value in the given array. It will take O(n).

Step 2 : Find XOR of all integers from range “min” to “max” ( inclusive ).

Step 3 : Find XOR of all elements of the given array.

Step 4 : XOR of Step 2 and Step 3 will give the required duplicate number.

Description :

Step1 : min = 21 , max = 27

Step 2 : Step2_result = 21 ^ 22 ^ 23 ^ 24 ^ 25 ^ 26 ^ 27 = 20

Step 3 : Step3_result = 23 ^ 21 ^ 24 ^ 27 ^ 22 ^ 27 ^ 26 ^ 25 = 15

Step 4 : Final_Result = Step2_result ^ Step3_result = 20 ^ 15 = 27

But , How Final_Result calculated the duplicate number ?

Final_Result = ( 21 ^ 22 ^ 23 ^ 24 ^ 25 ^ 26 ^ 27 ) ^ ( 23 ^ 21 ^ 24 ^ 27 ^ 22 ^ 27 ^ 26 ^ 25 )

Now , Remember above two properties : n ^ n = 0 AND n ^ 0 = n

So , here ,

Final_Result = ( 21 ^ 21 ) ^ ( 22 ^ 22 ) ^ ( 23 ^ 23 ) ^ ( 24 ^ 24 ) ^ ( 25 ^ 25 ) ^ ( 26 ^ 26 ) ^ ( 27 ^ 27 ^ 27 )

             = 0 ^ 0 ^ 0 ^ 0 ^ 0 ^ 0 ^ ( 27 ^ 0 ) ( property applied )

             = 0 ^ 27 ( because we know 0 ^ 0 = 0 )

             = 27 ( Required Result )

Answer 2

A XOR statement has the property that 'a' XOR 'a' will always be 0, that is they cancel out, thus, if you know that your list has only one duplicate and that the range is say x to y, 601 to 607 in your case, it is feasible to keep the xor of all elements from x to y in a variable, and then xor this variable with all the elements you have in your array. Since there will be only one element which will be duplicated it will not be cancelled out due to xor operation and that will be your answer.

void main()
{
    int a[8]={601,602,603,604,605,605,606,607};
    int k,i,j=601;

    for(i=602;i<=607;i++)
    {
        j=j^i;
    }

    for(k=0;k<8;k++)
    {
        j=j^a[k];
    }

    printf("%d",j);
}

This code will give the output 605, as desired!