Question

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?

Answer 1

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);