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