I solved it in Brute f tech Can you please tell me an efficient way

#1

I solved it in Brute f tech Can you please tell me an efficient way to solve this problem please help

#2

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

what is right shift equal to means???

#4

Right Shift :

Denoted as : >>

Eg: N>>i (N: first operand, i: second operand)

Takes two numbers, right shifts the bits of the first operand, the second operand decides the number of places to shift. In other words right shifting an integer “ x ” with an integer “ y ” denoted as ‘ (x>>y) ‘ is equivalent to dividing x with 2^y.

eg: lets take N=32 ; which is 100000 in Binary Form.

 Now, if “ N is right-shifted by 2”  i,e N=N>>2  then  N will become  N=N/(2^2). Thus,  N=32/(2^2)=8 which can be wriiten as  1000
#5

I know this one but in code you have written two<<=1 what does this mean<<=/>>=

#6

this means nothing but

x <<= 1 x = x<<1
also,
y <<= 1 y = y<<1
#7

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