Monday, March 31, 2014

Find duplicates in the ranged array

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.
Input : array  =  {1, 2, 3, 1, 3, 6, 6}, n = 7
Output = 1,3,6

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.

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
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])];
      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.




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