Given a range [L, R], we need to find two integers in this range such that their XOR is maximum among all possible choices of two integers. More Formally,

given [L, R], find max (A ^ B) where L <= A, B

Input : L = 8 R = 20 Output : 31 31 is XOR of 15 and 16. Input : L = 1 R = 3 Output : 3

A **simple solution** is to generate all pairs, find their XOR values and finally return the maximum XOR value.

An **efficient solution **is to consider pattern of binary values from L to R. We can see that first bit from L to R either changes from 0 to 1 or it stays 1 i.e. if we take the XOR of any two numbers for maximum value their first bit will be fixed which will be same as first bit of XOR of L and R itself.

After observing the technique to get first bit, we can see that if we XOR L and R, the most significant bit of this XOR will tell us the maximum value we can achieve i.e. let XOR of L and R is 1xxx where x can be 0 or 1 then maximum XOR value we can get is 1111 because from L to R we have all possible combination of xxx and it is always possible to choose these bits in such a way from two numbers such that their XOR becomes all 1. It is explained below with some examples,

Examples 1: L = 8 R = 20 L ^ R = (01000) ^ (10100) = (11100) Now as L ^ R is of form (1xxxx) we can get maximum XOR as (11111) by choosing A and B as 15 and 16 (01111 and 10000) Examples 2: L = 16 R = 20 L ^ R = (10000) ^ (10100) = (00100) Now as L ^ R is of form (0xx) we can get maximum xor as (111) by choosing A and B as 19 and 20 (10011 and 10100)

So the solution of this problem depends on the value of (L ^ R) only. We will calculate the L^R value first and then from most significant bit of this value, we will add all 1s to get the final result.

## C++

// C/C++ program to get maximum xor value // of two numbers in a range #include <bits/stdc++.h> using namespace std; // method to get maximum xor value in range [L, R] int maxXORInRange(int L, int R) { // get xor of limits int LXR = L ^ R; // loop to get msb position of L^R int msbPos = 0; while (LXR) { msbPos++; LXR >>= 1; } // construct result by adding 1, // msbPos times int maxXOR = 0; int two = 1; while (msbPos--) { maxXOR += two; two <<= 1; } return maxXOR; } // Driver code to test above methods int main() { int L = 8; int R = 20; cout << maxXORInRange(L, R) << endl; return 0; }

## Java

// Java program to get maximum xor value // of two numbers in a range class Xor { // method to get maximum xor value in range [L, R] static int maxXORInRange(int L, int R) { // get xor of limits int LXR = L ^ R; // loop to get msb position of L^R int msbPos = 0; while (LXR > 0) { msbPos++; LXR >>= 1; } // construct result by adding 1, // msbPos times int maxXOR = 0; int two = 1; while (msbPos-- >0) { maxXOR += two; two <<= 1; } return maxXOR; } // main function public static void main (String[] args) { int L = 8; int R = 20; System.out.println(maxXORInRange(L, R)); } }

Output:

31

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