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2023-06-09 18:52:05 +08:00
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commit 00282efd8b
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39
src/main/java/com/thealgorithms/maths/NonRepeatingElement.java
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@ -4,8 +4,8 @@ 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.
* Reason to use bitwise operator: It makes our program faster as we are operating on bits and not
* on actual numbers.
*/
public class NonRepeatingElement {
@ -15,17 +15,14 @@ public class NonRepeatingElement {
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
// 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"
);
"Enter " + n + " elements in the array. NOTE: Only 2 elements should not repeat");
for (i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
@ -36,40 +33,38 @@ public class NonRepeatingElement {
System.out.println("Unable to close scanner" + e);
}
//Find XOR of the 2 non repeating elements
// Find XOR of the 2 non repeating elements
for (i = 0; i < n; i++) {
res ^= arr[i];
}
//Finding the rightmost set bit
// 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
if ((res & arr[i]) > 0) { // Case 1 explained below
num1 ^= arr[i];
} else {
num2 ^= arr[i]; //Case 2 explained below
num2 ^= arr[i]; // Case 2 explained below
}
}
System.out.println(
"The two non repeating elements are " + num1 + " and " + num2
);
System.out.println("The two non repeating elements are " + num1 + " and " + num2);
}
/*
/*
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
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.
This is how we will get non repeating elements with the help of bitwise operators.
*/
}