Problem
Given an array of n elements which contains elements from 0 to n-1, with any of these numbers appearing any number of times.
Follow UP - Find these repeating numbers in O(n) and using only constant memory space.
Example
Input : array = {1, 2, 3, 1, 3, 6, 6}, n = 7
Output = 1,3,6
Solution
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:**
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]));
}
}
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.
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.
References
[http://www.geeksforgeeks.org/find-duplicates-in-on-time-and-constant-extra-space/](http://www.geeksforgeeks.org/find-duplicates-in-on-time-and-constant-extra-space/)