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

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++ proogram for above approach#include using namespace std; /* This function sets the values of *x and *y to non-repeating elementsin 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; }

Output:

`The non-repeating elements are 7 and 9`

Time Complexity: O(n)
Auxiliary Space: O(1)

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