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.


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++)
        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.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(;
        int res;
        int _l;
        _l = Integer.parseInt(in.nextLine());
        int _r;
        _r = Integer.parseInt(in.nextLine());
        res = maxXor(_l, _r);

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;
            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 )

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