# Maximizing XOR

### Problem Statement :

```Given two integers, l and r, find the maximal value of a xor b, written a+b, where a and b satisfy the following condition:
l <= a <= b <= r

For example, if l=11 and r=12, then
11 + 11 = 0
11 + 12 = 7
12 + 12 = 0

Our maximum value is 7.

Function Description

Complete the maximizingXor function in the editor below. It must return an integer representing the maximum value calculated.

maximizingXor has the following parameter(s):

l: an integer, the lower bound, inclusive
r: an integer, the upper bound, inclusive
Input Format

The first line contains the integer l.
The second line contains the integer r.

Constraints

1 <= l <= r <= 10^3

Output Format

Return the maximal value of the xor operations for all permutations of the integers from l to r, inclusive.```

### Solution :

```                            ```Solution in C :

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 a=0;
for (int i=l;i<=r;i++)
for (int j=i;j<=r;j++)
a=max(a,(i^j));
return a;

}

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

int _r;
cin >> _r;

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

return 0;
}

In Java :

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

public class Solution {
/*
* Complete the function below.
*/

static int maxXor(int l, int r) {
int max = 0;
for(int i = l;i <= r;i++) {
for(int j = l;j <= r;j++) {
int a = i ^ j;
max = Math.max(a, max);
}
}
return max;
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int res;
int _l;
_l = Integer.parseInt(in.nextLine());

int _r;
_r = Integer.parseInt(in.nextLine());

res = maxXor(_l, _r);
System.out.println(res);

}
}

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 max = 0,i,j;
for(i=l;i<r;i++)
for(j=i+1;j<=r;j++)
max = max<(i^j)?i^j:max;
return max;
}
int main() {
int res;
int _l;
scanf("%d", &_l);

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

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

return 0;
}

In Python3 :

l = int( input() )
r = int( input() )
m = 0
for i in range(l, r+1):
for j in range( i+1, r+1 ):
m = max( m, i^j )
print(m)```
```

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