CODEWITHSUNDEEP
c++algorithmxor
Ajai
Ajai

Reputation: 3500

XOR to find Duplicates in an array

I have seen the solution for this problem in this thread -> How to find a duplicate element in an array of shuffled consecutive integers?

But the problem I am now having is little varied from it. 

int arr[10] = {1,2,3,4,5,6,7,8,4,9};
    int a= 0;
    for(int i=0;i<10;i++) {
     a= a^ arr[i] ^i;
    }
    cout<<a;

Consider the above mentioned code snippet. Things work fine as it is. But when I add a 0 to the above mentioned array like, int arr[11] = {0,1,2,3,4,5,6,7,8,4,9}; I am not getting the proper duplicate element. Can somebody correct me the mistake I am making here?

Upvotes: 0

Views: 9570

Answers (3)

Vaughn Cato
Vaughn Cato

Reputation: 64308

The trick relies on the values being between 1 and n. If the numbers are in some other range you'll have to offset them.

static const int n = 11;
int arr[n] = {0,1,2,3,4,5,6,7,8,4,9};
int offset = 1;
int a= 0;
for(int i=0;i<n;i++) {
 a= a^ (arr[i]+offset) ^i;
}
cout<< (a-offset);

Upvotes: 4

aleph_null
aleph_null

Reputation: 5786

I think i see what happened. you probably changed the loop to be

for(int i=0; i<11; i++)
  ...

since you added an extra element to the loop. the problem is that it's not XORing with 10, which is not a number in your array. So, the algorithm stops working.

int arr[11] = {0,1,2,3,4,5,6,7,8,4,9};
int a= 0;
for(int i=0;i<10;i++) {
  a= a^ arr[i] ^i;
}
a = a^arr[10];

cout<<a;

now, its XORing nums from 0 to 9 twice (which equals zero) and 4 an additional time, so the result should be 4.

Upvotes: 0

DanZimm
DanZimm

Reputation: 2578

I'm guessing it could have to do with the fact that the i value would then be the same as arr[i]

thats like doing:

00000001 ^ 00000001

which equals 00000000

and I may be thinking incorrectly but wouldn't that screw up the process?

Upvotes: 1

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