Find duplicates in O(n) time and O(1) extra space

Question: Given an array of n elements which contains elements from 0 to n-1, with any of these numbers appearing any number of times. Find these repeating numbers in O(n) and using only constant memory space.

For example, let n be 7 and array be {1, 2, 3, 1, 3, 6, 6}, the answer should be 1, 3 and 6.

Algorithm:

traverse the list for i= 0 to n-1 elements
{
check for sign of A[abs(A[i])] ;
if positive then
make it negative by A[abs(A[i])]=-A[abs(A[i])];
else // i.e., A[abs(A[i])] is negative
this element (ith element of list) is a repetition
}

Solution:

1 2 3 4 5 6 7 8 9 10 11 12

void printRepeating(int arr[], int size) {   int i;   printf("The repeating elements are: \n");   for (i = 0; i < size; i++)   {     if (arr[abs(arr[i])] >= 0)       arr[abs(arr[i])] = -arr[abs(arr[i])];     else       printf(" %d ", abs(arr[i]));   } }