d

```
package com.thealgorithms.maths;
import java.util.Scanner;
/*
* Find the 2 elements which are non repeating in an array
* Reason to use bitwise operator: It makes our program faster as we are operating on bits and not
* on actual numbers.
*/
public final class NonRepeatingElement {
private NonRepeatingElement() {
}
public static void main(String[] args) {
try (Scanner sc = new Scanner(System.in)) {
int i, res = 0;
System.out.println("Enter the number of elements in the array");
int n = sc.nextInt();
if ((n & 1) == 1) {
// Not allowing odd number of elements as we are expecting 2 non repeating
// numbers
System.out.println("Array should contain even number of elements");
return;
}
int[] arr = new int[n];
System.out.println("Enter " + n + " elements in the array. NOTE: Only 2 elements should not repeat");
for (i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
try {
sc.close();
} catch (Exception e) {
System.out.println("Unable to close scanner" + e);
}
// Find XOR of the 2 non repeating elements
for (i = 0; i < n; i++) {
res ^= arr[i];
}
// Finding the rightmost set bit
res = res & (-res);
int num1 = 0, num2 = 0;
for (i = 0; i < n; i++) {
if ((res & arr[i]) > 0) { // Case 1 explained below
num1 ^= arr[i];
} else {
num2 ^= arr[i]; // Case 2 explained below
}
}
System.out.println("The two non repeating elements are " + num1 + " and " + num2);
sc.close();
}
}
/*
* Explanation of the code:
* let us assume we have an array [1,2,1,2,3,4]
* Property of XOR: num ^ num = 0.
* If we XOR all the elemnets of the array we will be left with 3 ^ 4 as 1 ^ 1
* and 2 ^ 2 would give
* 0. Our task is to find num1 and num2 from the result of 3 ^ 4 = 7. We need to
* find two's
* complement of 7 and find the rightmost set bit. i.e. (num & (-num)) Two's
* complement of 7 is 001
* and hence res = 1. There can be 2 options when we Bitise AND this res with
* all the elements in our
* array
* 1. Result will come non zero number
* 2. Result will be 0.
* In the first case we will XOR our element with the first number (which is
* initially 0)
* In the second case we will XOR our element with the second number(which is
* initially 0)
* This is how we will get non repeating elements with the help of bitwise
* operators.
*/
}
```