Count the possible integer sets of optimum size with the offered condition

Offered 2 integers L and R, the job is to discover the count of the sets of a optimum size such that each component in the set is in between L and R (inclusive), and for any 2 components in the set among them is divisible by the other.

Examples:

Input: L = 3, R = 19
Output: 4
Description: There will be 4 possible sets– {3, 6, 12}, {3, 6, 18}, {3, 9, 18}, {4, 8, 16}

Input: L = 4, R = 8
Output: 1

Method: This can be resolved with the following concept:

• Let {S ₁, S ₂, S ₃.. S _mx} be a set of optimum sizes pleasing the offered conditions. Let M _i = S _{i +1}/ S _i.
• It is instinctive that for S _mx to be the minimum we require to pick S ₁ and M _i for all i as low as possible, the minimum worth of S ₁ can be L and the minimum worth of M _i can be 2
• Nevertheless, we can pick among the M _i to be 3 so that S _mx will be ( 3/2) times the preliminary worth of S _mx (which must be less than R), if we pick any worth of M _i to be more than 3, the size of the set would not be optimal as there can constantly be a brand-new component S _{mx +1}= 2 * S _mx such size of the set would end up being mx +1.

Follow the listed below actions to execute the concept:

• Very first compute the worth of mx, i.e the optimum possible size of the set. This can be determined presuming all the worths of M _i are 2, and the worth of S ₁ is L, then mx = flooring( log2( r/l)) + 1
• Determine the optimum worth of S ₁ such that a set of size mx pleasing the offered conditions is possible. Let’s call it X, We understand 2 ^mx-1 * X ≤ R, then X = R/2 ^mx-1
• Determine the optimum worth of S ₁ such that a set of size mx pleasing the offered conditions is possible and among the worths of M _i can be 3 rather of 2, let us call it Y We understand that 3 * 2 ^mx-2 * Y ≤ R, then Y = R/( 3 * 2 ^mx-2)
• We understand L ≤ Y ≤ X ≤ R, a variety of sets with S ₁ ≤ Y are ( Y-L +1) * mx, note that we increased by mx as any of the M _i in these sets can be 3. A variety of sets with S ₁>> Y and S ₁ ≤ X is X– Y.
• Overall sets of optimum size = ( Y-L +1) * mx + X-Y.

Below is the application of the above method:

Time Intricacy: O( 1 )
Auxilairy Area: O( 1 )