XOR profit problem

#1

how to impliment it without using trie ds.also explain the algo.

#2

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 (1xx) 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.

Reference Code