CODEWITHSUNDEEP
javaalgorithmxor
Viki
Viki

Reputation: 1

XORing of numbers within the given range

Here in this program of XORing of numbers within the given range, I can't understand the algorithm of how they are finding xor of numbers. It would be pretty good if someone explains how this algorithm works in finding the xor values within the range

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/*This program is used to find the xor of numbers within the range

u can enter multiple inputs.Here first enter the numbers of inputs u want to enter then in the next line enter the starting and ending range of the xoring 
numbers with space

for example : 

Input :
1
2 3

output:

odd


output displays whether the xor value  within the given range is odd r not

*/

class XORNEY {
    public static void main(String[] args) throws IOException {
        BufferedReader kb = new BufferedReader(new InputStreamReader(System.in));
        int testCases = Integer.parseInt(kb.readLine());
        StringBuilder output = new StringBuilder();

        while (testCases-- > 0) {
            //here without declaring the array dynamically how they are storing all the inputs
            String[] input = kb.readLine().trim().split(" ");  

            long L = Long.parseLong(input[0]);
            long H = Long.parseLong(input[1]);
            //i am more confused here abt why they are doing this
            long count = (H - L) / 2;   
            if ((L & 1) == 1 || (H & 1) == 1) count++;
            output.append((count & 1) == 1 ? "Odd\n" : "Even\n");
        }

        System.out.print(output);
    }
}

Upvotes: 0

Views: 104

Answers (1)

shakhawat
shakhawat

Reputation: 2727

here without declaring the array dynamically how they are storing all the inputs

They are just storing each test case line not all the inputs. Each line consist of two numbers separated by space, if you split it by space you get both the numbers on an array.

I am more confused here abt why they are doing this

I can give some ideas -

On a range of numbers, you will get binary numbers ending with an alternate of 0 and 1 sequence. For example -

  • 10 -> ***** 0
  • 11 -> ***** 1
  • 12 -> ***** 0
  • 13 -> ***** 1

for each pair of 0, 1 the XOR will be 1. Now you need to figure out if you are getting even number of pairs or odd number of pairs. Even number of pair will produce even number of 1 to XOR and give you a value ending with 0 that means the ultimate XOR value is an even number. And odd on the other way around.

If you have even number of elements in the range, you are quite sure that you have some pairs of (0, 1). But if it's odd, depending on the extra item being 0/1, your calculation needs to be adjusted.

Upvotes: 1

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