Home >Java >javaTutorial >Xor of N numbers

Xor of N numbers

Patricia Arquette
Patricia ArquetteOriginal
2024-09-21 20:15:32309browse

Xor of N numbers

Given an integer number N, find the exor of the range 1 to N
exor of 1 ^ 2 ^ 3 ^4 ^.....N;

Brute force approach:
Tc:O(n)
Sc:O(1)

public int findExor(int N){

        //naive/brute force approach:
        int val  = 0;
        for(int i=1;i<5;i++){
            val =  val^ i;
        }
        return val;
    }

Optimal approach:
Tc:O(1)
Sc:O(1)

    public int getExor(int N){
        //better approach

        /**
         * one thing to observe is 
         * 1 = 001  = 1
         * 1 ^2 = 001 ^ 010 = 011=       3
         * 1^2^3 = 011 ^ 011 = 0=        0
         * 1^2^3^4 = 000^100 = 100=      4
         * 1^2^3^4^5 = 100^101 = 001=    1
         * 1^2^3^4^5^6 = 001^110 =111=   7
         * 1^2^3^4^5^6^7 = 111^111=000=  0
         * 
         * what we can observer is : 
         * 
         * N%4==0 then result is: N
         * N%4 ==1 then result is: 1
         * N%4 ==2 then result is: N+1
         * N%4==3 then result is: 0
         * 
         * */
         if(N%4==0) return N;
         else if(N%4 ==1) return 1;
         else if(N%4==2) return N+1;
         else return 0;

    }

What if we have to find the exor between ranges like L and R
example find an exor between numbers 4 and 7 i.e. 4^5^6^7.

For solving this we can leverage the same optimal solution above getExor()

first we will get exor till L-1 i.e getExor(L-1) = 1 ^ 2 ^ 3 (since L-1 = 3)......equation(1)

then we will find getExor(R) = 1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6 ^ 7 ----equation(2)

the finally,

Result  = equation(1) ^ equation(2)
        = (1 ^ 2 ^ 3) ^ (1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6 ^ 7)
        = (4^5^6^7)

public int findExorOfRange(int L, int R){
        return getExor(L-1) ^ getExor(R);
    }

public int getExor(int N){
        //better approach

        /**
         * one thing to observe is 
         * 1 = 001  = 1
         * 1 ^2 = 001 ^ 010 = 011=       3
         * 1^2^3 = 011 ^ 011 = 0=        0
         * 1^2^3^4 = 000^100 = 100=      4
         * 1^2^3^4^5 = 100^101 = 001=    1
         * 1^2^3^4^5^6 = 001^110 =111=   7
         * 1^2^3^4^5^6^7 = 111^111=000=  0
         * 
         * what we can observer is : 
         * 
         * N%4==0 then result is: N
         * N%4 ==1 then result is: 1
         * N%4 ==2 then result is: N+1
         * N%4==3 then result is: 0
         * 
         * */
         if(N%4==0) return N;
         else if(N%4 ==1) return 1;
         else if(N%4==2) return N+1;
         else return 0;

    }

The above is the detailed content of Xor of N numbers. For more information, please follow other related articles on the PHP Chinese website!

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn