XOR Profit Problem

#1

Not able to grasp the concept of question properly, kindly explain. Also please refer a video if have so.

#2

Hey the approach you are using is not right.
See this, you will get the intution
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.

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.

image

#3

Ok, I am getting it gradually, but can you please explain that in second while lop we making the number as 11111(for maximum XOR formation) for example 1 and hence returning it giving output as -1, so how would we convert those them to two integers 19 and 20.

#4

You don’t have to convert them in integer again, you have to print the maximum xor value that we can get from numbers between L to R. So won’t be required 19 & 20 again.

#5

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