# HackerRank Maximizing XOR problem solution

In this HackerRank Maximizing XOR problem solution we have given to integers L and R. we need to find the maximal value of A xor B where l <= a <= b <= r.

## Problem solution in Python.

```#!/usr/bin/python3
def maxXor(l, r):
res = 0
for i in range(l, r+1):
for j in range(i, r+1):
if i ^ j > res:
res = i ^ j
return res

if __name__ == '__main__':
l = int(input())
r = int(input())
print(maxXor(l, r))```

## Problem solution in Java.

```import java.io.*;
import java.util.*;

public class Solution {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int L = in.nextInt();
int R = in.nextInt();
int ans = Integer.MIN_VALUE;
for (int A = L; A <= R; A++)
for (int B = A; B <= R; B++)
ans = Math.max(ans, A ^ B);
System.out.println(ans);
}
}```

## Problem solution in C++.

```#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <cstdlib>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;
/*
* Complete the function below.
*/
int maxXor(int l, int r) {
int m=0,i,j;
for(i=l;i<=r-1;i++)
{ for(j=i+1;j<=r;j++)
m=max(m,i^j);
}
return m;
}

int main() {
int res;
int _l;
cin >> _l;

int _r;
cin >> _r;

res = maxXor(_l, _r);
cout << res;

return 0;
}```

## Problem solution in C.

```#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <assert.h>
/*
* Complete the function below.
*/
int maxXor(int l, int r) {
int x, y, z, ans = 0;
for (x = l; x <= r; x++)
for (y = x + 1; y <= r; y++) {
z = x ^ y;
if (z > ans) ans = z;
}
return ans;
}
int main() {
int res;
int _l;
scanf("%d", &_l);

int _r;
scanf("%d", &_r);

res = maxXor(_l, _r);
printf("%d", res);

return 0;
}```