HackerRank Xor sequence problem solution

In this HackerRank Xor sequence problem solution, we have given an array A and the left and right index L, R. we need to determine the XOR sum of the segment of A as A[L] XOR A[L + 1] XOR....A[R - 1] XOR A[R].

Problem solution in Python.

#!/bin/python3

from math import ceil, floor


Q = int(input().strip())
for a0 in range(Q):
    L,R = input().strip().split(' ')
    L,R = [int(L),int(R)]
    
    val = 0
    
    if R - L + 1 > 4:
        if L%4 == 0:
            val ^= L ^ 1 ^ (L + 3)
            L += 3

        elif L%4 == 1:
            val ^= 1 ^ (L + 2)
            L += 2

        elif L%4 == 2:
            val ^= L + 1
            L += 1
            
        if R%4 == 0:
            val ^= R
            R -= 1
            
        elif R%4 == 1:
            val ^= 1 ^ (R - 1)
            R -= 2
            
        elif R%4 == 2:
            val ^= (R + 1) ^ 1 ^ (R - 2)
            R -= 3
            
        if ((R - L)/4)% 2 == 1:
            val ^= 2
    
    else:    
        for i in range(L, R + 1):
            if i%4 == 1:
                val ^= 1
            elif i%4 == 2:
                val ^= i + 1
            elif i%4 == 0:
                val ^= i
            
    print(val)

Problem solution in Java.

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int Q = in.nextInt();
        for(int a0 = 0; a0 < Q; a0++){
            long L = in.nextLong();
            long R = in.nextLong();

            long result = 0;
            long lower = L + (3 - L % 4);
            long upper = R - R % 4;
            if (lower > upper) {
                for (long i = L; i <= R; ++i) result ^= xor(i);
            }
            else {
                for (long i = L; i <= lower; ++i) result ^= xor(i);
                for (long i = R; i >= upper; --i) result ^= xor(i);
            
                long mid = (upper-lower)/4;
                if (mid > 0) result ^= (mid % 2 == 0 ? 0 : 2);                
            }
            System.out.println(result);
        }
    }
    
    static long xor(long i) {
        if (i % 4 == 0) return i;
        else if (i % 4 == 1) return 1;
        else if (i % 4 == 2) return (i+1);
        return 0;
    }
}

Problem solution in C++.

# include <bits/stdc++.h>

# define fi first
# define se second
# define pb push_back


using namespace std;

typedef long long ll;

const int Maxn = 1e5 + 1, md = 1e9 + 7;
const double Eps = 0.01;

using namespace std;

ll n, l , r, a[Maxn];

ll X(ll x)
{
    ll y = x%4;
    if (y == 0)
        return x;
    if (y == 1)
        return 1;
    if (y == 2)
        return x + 1;
    return 0;
}
ll Y(ll x)
{
    ll y = x%8;
    if (x == 0)
        return 0;
    if (y == 0 || y == 1)
        return x;
    if (y == 6 || y == 7)
        return 0;
    if (y == 2 || y == 3)
        return 2;
    return x + 2;
}
int main() {

    cin >> n;

    while (n--)
    {
        cin >> l >> r;
        l = Y(l - 1);
        r = Y(r);
        l = l ^ r;
        cout << l << endl;
    }



}

Problem solution in C.

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
long long memoize[4];

unsigned long long arrayfinder(unsigned long long);
int main(){
    int Q;
    scanf("%d",&Q);
    for(int a0 = 0; a0 < Q; a0++){
        unsigned long long L; 
        unsigned long long R;
        unsigned long long total=0;
        scanf("%llu %llu",&L,&R);
        if((R-L)<17){
            for(int i= L;i<=R;i++){
                total^=arrayfinder(i);
            }
        }
        else{
            while(L%8!=7){
                total^=arrayfinder(L);
                L++;
            }
            while(R%8!=7){
                total^=arrayfinder(R);
                R--;
            }
        }
        printf("%llu\r\n",total);
    }
    return 0;
}
unsigned long long arrayfinder(unsigned long long x){
    

       if(x%4==0)return x;
       if(x%4==1)return 1;
       if(x%4==2)return x+1;
       if(x%4==3)return 0;
    
   
    return -1;
}