Find the two non-repeating elements in an array of repeating elements

Asked by SG
Given an array in which all numbers except two are repeated once. (i.e. we have 2n+2 numbers and n numbers are occurring twice and remaining two have occurred once). Find those two numbers in the most efficient way.

Method 1(Use Sorting)
First, sort all the elements. In the sorted array, by comparing adjacent elements we can easily get the non-repeating elements. Time complexity of this method is O(nLogn)

Method 2(Use XOR)
Let x and y be the non-repeating elements we are looking for and arr[] be the input array. First, calculate the XOR of all the array elements.



     xor = arr[0]^arr[1]^arr[2].....arr[n-1]

All the bits that are set in xor will be set in one non-repeating element (x or y) and not in others. So if we take any set bit of xor and divide the elements of the array in two sets – one set of elements with same bit set and another set with same bit not set. By doing so, we will get x in one set and y in another set. Now if we do XOR of all the elements in the first set, we will get the first non-repeating element, and by doing same in other sets we will get the second non-repeating element.

Let us see an example.
   arr[] = {2, 4, 7, 9, 2, 4}
1) Get the XOR of all the elements.
     xor = 2^4^7^9^2^4 = 14 (1110)
2) Get a number which has only one set bit of the xor.   
   Since we can easily get the rightmost set bit, let us use it.
     set_bit_no = xor & ~(xor-1) = (1110) & ~(1101) = 0010
   Now set_bit_no will have only set as rightmost set bit of xor.
3) Now divide the elements in two sets and do xor of         
   elements in each set and we get the non-repeating 
   elements 7 and 9. Please see the implementation for this step.

Implementation:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

#include <bits/stdc++.h>
using namespace std;
  
/* This finction sets the values of 
*x and *y to nonr-epeating elements
in an array arr[] of size n*/
void get2NonRepeatingNos(int arr[], int n, int *x, int *y) 
{ 
    int Xor = arr[0]; /* Will hold Xor of all elements */
    int set_bit_no; /* Will have only single set bit of Xor */
    int i; 
    *x = 0; 
    *y = 0; 
      
    /* Get the Xor of all elements */
    for(i = 1; i < n; i++) 
    Xor ^= arr[i]; 
      
    /* Get the rightmost set bit in set_bit_no */
    set_bit_no = Xor & ~(Xor-1); 
      
    /* Now divide elements in two sets by comparing rightmost set 
    bit of Xor with bit at same position in each element. */
    for(i = 0; i < n; i++) 
    { 
        if(arr[i] & set_bit_no) 
        *x = *x ^ arr[i]; /*Xor of first set */
        else
        *y = *y ^ arr[i]; /*Xor of second set*/
    } 
} 
  
/* Driver code */
int main() 
{ 
    int arr[] = {2, 3, 7, 9, 11, 2, 3, 11}; 
    int *x = new int[(sizeof(int))]; 
    int *y = new int[(sizeof(int))]; 
    get2NonRepeatingNos(arr, 8, x, y); 
    cout<<"The non-repeating elements are "<<*x<<" and "<<*y; 
} 
  
// This code is contributed by rathbhupendra

chevron_right






                        filter_none









C














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















#include <stdio.h> 


#include <stdlib.h> 


  


/* This finction sets the values of *x and *y to nonr-epeating 


 elements in an array arr[] of size n*/


void get2NonRepeatingNos(int arr[], int n, int *x, int *y) 


{ 


  int xor = arr[0]; /* Will hold xor of all elements */


  int set_bit_no;  /* Will have only single set bit of xor */


  int i; 


  *x = 0; 


  *y = 0; 


  


  /* Get the xor of all elements */


  for(i = 1; i < n; i++) 


   xor ^= arr[i]; 


  


  /* Get the rightmost set bit in set_bit_no */


  set_bit_no = xor & ~(xor-1); 


  


  /* Now divide elements in two sets by comparing rightmost set 


   bit of xor with bit at same position in each element. */


  for(i = 0; i < n; i++) 


  { 


    if(arr[i] & set_bit_no) 


     *x = *x ^ arr[i]; /*XOR of first set */


    else


     *y = *y ^ arr[i]; /*XOR of second set*/


  } 


} 


  


/* Driver program to test above function */


int main() 


{ 


  int arr[] = {2, 3, 7, 9, 11, 2, 3, 11}; 


  int *x = (int *)malloc(sizeof(int)); 


  int *y = (int *)malloc(sizeof(int)); 


  get2NonRepeatingNos(arr, 8, x, y); 


  printf("The non-repeating elements are %d and %d", *x, *y); 


  getchar(); 


} 



















                        chevron_right







                        filter_none










Output:


The non-repeating elements are 7 and 9


Time Complexity: O(n)

Auxiliary Space: O(1)


Please refer below post for detailed explanation :

Find the two numbers with odd occurrences in an unsorted array



          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts: