solution.java

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

public class Solution {

    /* To maximize A xor B, we want A and B to differ as much as possible at every bit index.
    We first find the most significant bit that we can force to differ by looking at L and R.
    For all of the lesser significant bits in A and B, we can always ensure that they differ 
    and still have L <= A <= B <= R. Our final answer will be the number represented by all
    1s starting from the most significant bit that differs between A and B

    Example:
    L = 10111   
    R = 11100
        _X___  <-- that's most significant differing bit
        01111  <-- here's our final answer
    Notice that we never directly calculate the values of A and B
    **/
    static int maximizingXor(int L, int R) {
        int xored  = L ^ R;
        int significantBit = 31 - Integer.numberOfLeadingZeros(xored);
        int result = (1 << (significantBit + 1)) - 1;
        return result;
    }
    
    /* Driver Code */
    private static final Scanner scanner = new Scanner(System.in);
    public static void main(String[] args) throws IOException {
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int l = scanner.nextInt();
        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

        int r = scanner.nextInt();
        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

        int result = maximizingXor(l, r);

        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();

        bufferedWriter.close();

        scanner.close();
    }
}