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

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.

**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 }

## 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.*/ // Create a Map Set to store the numbers set<int> s; for (int i = 0; i < n; i++) { /*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*/ if (s.find(arr[i]) != s.end()) s.erase(arr[i]); else s.insert(arr[i]); } cout << "The 2 non repeating numbers are : "; for (auto it : s) cout << it << " "; cout << endl; } /* 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 Aditya kumar

**Time Complexity: O(n)**

** Auxiliary Space: O(n)**