XOR profit hint

#1

I don’t understand the solution. I’m not able to grasp what he’s trying to say. Please help

#2

@aiman.mumtaz
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,
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)

#3

So, we just need to check and convert the bit of MSB to 1?

#4

Yeah just xor L and R and get the MSB, after that you can automatically select all ones for lower bits because it will always be possible to choose such a combo.

#5

Workeed. Thanks a lot