Find the two non-repeating elements in an array of repeating elements/ Unique Numbers 2

Difficulty Level : Hard
Last Updated : 16 Sep, 2021

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)

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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.

Approach :
Step 1: Xor all the elements of the array into a variable sum thus all the elements present twice in an array will get removed as for example, 4 = “100” and if 4 xor 4 => “100” xor “100” thus answer will be “000”.
Step 2: Thus in the sum the final answer will be 3 xor 5 as both 2 and 4 are xor with itself giving 0, therefore sum = “011” xor “101” i.e sum = “110” = 6.
Step 3: Now we will take 2’s Complement of sum i.e (-sum) = “010”.
Step 4: Now bitwise And the 2’s of sum with the sum i.e “110” & “010” gives the answer “010” (Aim for bitwise & is that we want to get a number that contains only the rightmost set bit of the sum).
Step 5: bitwise & all the elements of the array with this obtained sum, 2 = “010” & “010” = 2, 3 = “011” & “010” = “010” , 4 = “100” & “010” = “000”, 5 = “101” & “010” = “000”.
Step 6: As we can see that the bitwise & of 2,3 > 0 thus they will be xor with sum1 and bitwise & of 4,5 is resulting into 0 thus they will be xor with sum2.
Step 7: As 2 is present two times so getting xor with sum1 two times only the result 3 is being stored in it and As 4 is also present two times thus getting xor with sum2 will cancel it’s value and thus only 5 will remain there.

Implementation:

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
/* This function sets the values of
*x and *y to non-repeating elements
in an array arr[] of size n*/
void get2NonRepeatingNos(int arr[], int n, int* x, int* y)
{
    /* Will hold Xor of all elements */
    int Xor = arr[0];
 
    /* Will have only single set bit of Xor */
    int set_bit_no;
    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++) {
 
        /*Xor of first set */
        if (arr[i] & set_bit_no)
            *x = *x ^ arr[i];
        /*Xor of second set*/
        else {
            *y = *y ^ arr[i];
        }
    }
}
 
/* Driver code */
int main()
{
    int arr[] = { 2, 3, 7, 9, 11, 2, 3, 11 };
    int n = sizeof(arr) / sizeof(*arr);
    int* x = new int[(sizeof(int))];
    int* y = new int[(sizeof(int))];
    get2NonRepeatingNos(arr, n, x, y);
    cout << "The non-repeating elements are " << *x
         << " and " << *y;
}
 
// This code is contributed by rathbhupendra

C

// C program for above approach
#include <stdio.h>
#include <stdlib.h>
 
/* This function sets the values of
*x and *y to non-repeating elements
in an array arr[] of size n*/
void get2NonRepeatingNos(int arr[], int n, int* x, int* y)
{
    /* Will hold Xor of all elements */
    int Xor = arr[0];
 
    /* Will have only single set bit of Xor */
    int set_bit_no;
    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++) {
 
        /*Xor of first set */
        if (arr[i] & set_bit_no)
            *x = *x ^ arr[i];
        /*Xor of second set*/
        else {
            *y = *y ^ arr[i];
        }
    }
}
 
/* 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();
}

Java

// Java Program for above approach
 
public class UniqueNumbers {
 
    // This function sets the values of
    // *x and *y to non-repeating elements
    // in an array arr[] of size n
    public static void UniqueNumbers2(int[] arr, int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++) {
 
            // Xor  all the elements of the array
            // all the elements occurring twice will
            // cancel out each other remaining
            // two unique numbers will be xored
            sum = (sum ^ arr[i]);
        }
 
        // Bitwise & the sum with it's 2's Complement
        // Bitwise & will give us the sum containing
        // only the rightmost set bit
        sum = (sum & -sum);
 
        // sum1 and sum2 will contains 2 unique
        // elements elements initialized with 0 box
        // number xored with 0 is number itself
        int sum1 = 0;
        int sum2 = 0;
 
        // traversing the array again
        for (int i = 0; i < arr.length; i++) {
 
            // Bitwise & the arr[i] with the sum
            // Two possibilities either result == 0
            // or result > 0
            if ((arr[i] & sum) > 0) {
 
                // if result > 0 then arr[i] xored
                // with the sum1
                sum1 = (sum1 ^ arr[i]);
            }
            else {
                // if result == 0 then arr[i]
                // xored with sum2
                sum2 = (sum2 ^ arr[i]);
            }
        }
 
        // print the the two unique numbers
        System.out.println("The non-repeating elements are "
                           + sum1 + " and " + sum2);
    }
 
    public static void main(String[] args)
    {
        int[] arr = new int[] { 2, 3, 7, 9, 11, 2, 3, 11 };
        int n = arr.length;
        UniqueNumbers2(arr, n);
    }
}
// This code is contributed by Parshav Nahta

Python3

# Python3 program for above approach
 
# This function sets the values of
# *x and *y to non-repeating elements
# in an array arr[] of size n
 
 
def UniqueNumbers2(arr, n):
 
    sums = 0
 
    for i in range(0, n):
 
        # Xor  all the elements of the array
        # all the elements occurring twice will
        # cancel out each other remaining
        # two unique numbers will be xored
        sums = (sums ^ arr[i])
 
    # Bitwise & the sum with it's 2's Complement
    # Bitwise & will give us the sum containing
    # only the rightmost set bit
    sums = (sums & -sums)
 
    # sum1 and sum2 will contains 2 unique
    # elements elements initialized with 0 box
    # number xored with 0 is number itself
    sum1 = 0
    sum2 = 0
 
    # Traversing the array again
    for i in range(0, len(arr)):
 
        # Bitwise & the arr[i] with the sum
        # Two possibilities either result == 0
        # or result > 0
        if (arr[i] & sums) > 0:
 
            # If result > 0 then arr[i] xored
            # with the sum1
            sum1 = (sum1 ^ arr[i])
 
        else:
 
            # If result == 0 then arr[i]
            # xored with sum2
            sum2 = (sum2 ^ arr[i])
 
    # Print the the two unique numbers
    print("The non-repeating elements are ",
          sum1, " and ", sum2)
 
 
# Driver Code
if __name__ == "__main__":
 
    arr = [2, 3, 7, 9, 11, 2, 3, 11]
    n = len(arr)
 
    UniqueNumbers2(arr, n)
 
# This code is contributed by akhilsaini

C#

// C# program for above approach
using System;
 
class GFG {
 
    // This function sets the values of
    // *x and *y to non-repeating elements
    // in an array arr[] of size n
    static void UniqueNumbers2(int[] arr, int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++) {
 
            // Xor  all the elements of the array
            // all the elements occurring twice will
            // cancel out each other remaining
            // two unique numbers will be xored
            sum = (sum ^ arr[i]);
        }
 
        // Bitwise & the sum with it's 2's Complement
        // Bitwise & will give us the sum containing
        // only the rightmost set bit
        sum = (sum & -sum);
 
        // sum1 and sum2 will contains 2 unique
        // elements elements initialized with 0 box
        // number xored with 0 is number itself
        int sum1 = 0;
        int sum2 = 0;
 
        // Traversing the array again
        for (int i = 0; i < arr.Length; i++) {
 
            // Bitwise & the arr[i] with the sum
            // Two possibilities either result == 0
            // or result > 0
            if ((arr[i] & sum) > 0) {
 
                // If result > 0 then arr[i] xored
                // with the sum1
                sum1 = (sum1 ^ arr[i]);
            }
            else {
 
                // If result == 0 then arr[i]
                // xored with sum2
                sum2 = (sum2 ^ arr[i]);
            }
        }
 
        // Print the the two unique numbers
        Console.WriteLine("The non-repeating "
                          + "elements are " + sum1 + " and "
                          + sum2);
    }
 
    // Driver Code
    static public void Main()
    {
        int[] arr = { 2, 3, 7, 9, 11, 2, 3, 11 };
        int n = arr.Length;
 
        UniqueNumbers2(arr, n);
    }
}
 
// This code is contributed by akhilsaini

Javascript

<script>
    // Javascript program for above approach
     
    // This function sets the values of
    // *x and *y to non-repeating elements
    // in an array arr[] of size n
    function UniqueNumbers2(arr, n)
    {
        let sum = 0;
        for(let i = 0; i < n; i++)
        {
 
            // Xor  all the elements of the array
            // all the elements occurring twice will
            // cancel out each other remaining
            // two unique numbers will be xored
            sum = (sum ^ arr[i]);
        }
 
        // Bitwise & the sum with it's 2's Complement
        // Bitwise & will give us the sum containing
        // only the rightmost set bit
        sum = (sum & -sum);
 
        // sum1 and sum2 will contains 2 unique
        // elements elements initialized with 0 box
        // number xored with 0 is number itself
        let sum1 = 0;
        let sum2 = 0;
 
        // Traversing the array again
        for(let i = 0; i < arr.length; i++)
        {
 
            // Bitwise & the arr[i] with the sum
            // Two possibilities either result == 0
            // or result > 0
            if ((arr[i] & sum) > 0)
            {
 
                // If result > 0 then arr[i] xored
                // with the sum1
                sum1 = (sum1 ^ arr[i]);
            }
            else
            {
 
                // If result == 0 then arr[i]
                // xored with sum2
                sum2 = (sum2 ^ arr[i]);
            }
        }
 
        // Print the the two unique numbers
        document.write("The non-repeating " +
                          "elements are " + sum1 +
                          " and " + sum2);
    }
     
    let arr = [ 2, 3, 7, 9, 11, 2, 3, 11 ];
    let n = arr.length;
      
    UniqueNumbers2(arr, n);
 
// This code is contributed by vaibhavrabadiya117.
</script>

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

Method 3(Use Maps)

In this method, we simply count frequency of each element. The elements whose frequency is equal to 1 is the number which is non-repeating. The solution is explained below in the code-

C++

// C++ program for Find the two non-repeating elements in
// an array of repeating elements/ Unique Numbers 2
 
#include <bits/stdc++.h>
using namespace std;
 
/* This function prints the two non-repeating elements in an
 * array of repeating elements*/
 
void get2NonRepeatingNos(int arr[], int n)
{
    /*Create map and calculate frequency of array
       elements.*/
 
    map<int, int> m;
    for (int i = 0; i < n; i++) {
        m[arr[i]]++;
    }
 
    /*Traverse through the map and check if its second
      element that is the frequency is 1 or not. If this is
      1 than it is the non-repeating element print it.It is
      clearly mentioned in problem that all numbers except
      two are repeated once. So they will be printed*/
 
    cout << "The non-repeating elements are ";
    for (auto& x : m) {
        if (x.second == 1) {
            cout << x.first << " ";
        }
    }
}
 
/* Driver code */
int main()
{
    int arr[] = { 2, 3, 7, 9, 11, 2, 3, 11 };
    int n = sizeof(arr) / sizeof(arr[0]);
    get2NonRepeatingNos(arr, n);
}
 
// This code is contributed by Abhishek

Java

/*package whatever //do not write package name here */
 
//Java program to find 2 non repeating elements
//in array that has pairs of numbers
 
import java.util.*;
import java.io.*;
 
class GFG {
   
      //Method to print the 2 non repeating elements in an array
      public static void print2SingleNumbers(int[] nums){
       
          /*We use a TreeMap to store the elements
          in the sorted order*/
         TreeMap<Integer, Integer> map = new TreeMap<>();
       
          int n = nums.length;
       
          /*Iterate through the array and check if each
          element is present or not in the map. If the
        element is present, remove it from the array
        otherwise add it to the map*/
       
          for(int i = 0; i<n; i++){
            if(map.containsKey(nums[i]))
                  map.remove(nums[i]);
            else
                map.put(nums[i],1);
        }
       
          System.out.println("The non-repeating integers are " + map.firstKey() + " " + map.lastKey());
    }
      //Driver code
    public static void main (String[] args) {
        int[] nums = new int[]{2,11,3,11,7,3,9,2};
          print2SingleNumbers(nums);
    }
      //This code is contributed by Satya Anvesh R
}

Output

The non-repeating elements are 7 9

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

Method 4(Use Sets):

In this method, We check if the element already exists, if it exists we remove it else we add it to the set.

Approach:

Step 1: Take each element and check if it exists in the set or not. If it exists go to step-3. If it doesn’t exist go to step-2.

Step 2: Add the element to the set and go to step-4.

Step 3: Remove the element from the set and go to step-4.

Step 4: Print the elements of the set.

Implementation:

Java

/*package whatever //do not write package name here */
//Java program to find 2 non repeating elements
//in array that has pairs of numbers
 
import java.util.LinkedHashSet;
import java.util.Iterator;
import java.io.*;
 
class GFG {
   
      //Method to print the 2 non repeating elements in an array
      public static void print2SingleNumbers(int[] nums){
       
          // Create a Map Set to store the numbers
          LinkedHashSet<Integer> set = new LinkedHashSet<>();
       
          int n = nums.length;
       
          /*Iterate through the array and check if each
          element is present or not in the set. If the
        element is present, remove it from the array
        otherwise add it to the set*/
       
          for(int i = 0; i<n; i++){
              if(set.contains(nums[i]))
                  set.remove(nums[i]);
              else
                  set.add(nums[i]);
        }
           
          //Iterator is used to traverse through the set
          Iterator<Integer> i = set.iterator();
       
          /*Since there will only be 2 non-repeating elements
        we can directly print them*/
          System.out.println("The 2 non repeating numbers are : " + i.next() + " " + i.next());
    }
      //Driver code
    public static void main (String[] args) {
        int[] nums = new int[]{2, 3, 7, 9, 11, 2, 3, 11 };
          print2SingleNumbers(nums);
    }
      //This code contributed by Satya Anvesh R
}

Output

The 2 non repeating numbers are : 7 9

Time Complexity: O(n)

Auxiliary Space: O(n)

My Personal Notes

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