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

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.

This problem is an extended version of following problem.

Find the two repeating elements in a given array

Method 1 and Method 2 of the above link are not applicable as the question says O(n) time complexity and O(1) constant space. Also, Method 3 and Method 4 cannot be applied here because there can be more than 2 repeating elements in this problem. Method 5 can be extended to work for this problem. Below is the solution that is similar to the Method 5.

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
}

Implementation:

#include <stdio.h> #include <stdlib.h>   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]));   } }   int main() {   int arr[] = {1, 2, 3, 1, 3, 6, 6};   int arr_size = sizeof(arr)/sizeof(arr[0]);   printRepeating(arr, arr_size);   getchar();   return 0; }

Note: The above program doesn’t handle 0 case (If 0 is present in array). The program can be easily modified to handle that also. It is not handled to keep the code simple.

Output:
The repeating elements are:
1 3 6

Time Complexity: O(n)
Auxiliary Space: O(1)

Please write comments if you find the above codes/algorithms incorrect, or find better ways to solve the same problem.