**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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