Maximizing XOR - HackerRank || HackerRank Maximizing XOR Solution

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string.h>

int maximizingXOR(int m, int n)
{
    int max = 0;
    for (int i = m; i <= n; i++)
    {
        for (int j = i; j <= n; j++)
        {
            if ((i ^ j) > max)
            {
                max = (i ^ j);
            }
        }
    }

    return max;
}

int main()
{
    int l, r;
    scanf("%d %d", &l, &r);
    int result = maximizingXOR(l, r);
    printf("%d\n", result);

    return 0;
}

#include <bits/stdc++.h>

using namespace std;

int maximizingXOR(int m, int n)
{
    int max = 0;
    for (int i = m; i <= n; i++)
    {
        for (int j = i; j <= n; j++)
        {
            if ((i ^ j) > max)
            {
                max = (i ^ j);
            }
        }
    }

    return max;
}

int main()
{
    int l, r;
    cin >> l >> r;
    int result = maximizingXOR(l, r);
    cout << result << endl;

    return 0;
}

import java.util.Scanner;

public class Maximizing_XOR {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int l = input.nextInt();
        int r = input.nextInt();
        int max = 0;
        for (int i = l; i <= r; i++) {
            for (int j = i; j <= r; j++) {
                if ((i ^ j) > max) {
                    max = (i ^ j);
                }
            }
        }
        System.out.println(max);
    }
}