# 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 â¤ 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 â¤ R, then Y = R/( 3 * 2 mx-2)
• We understand L â¤ Y â¤ X â¤ R, a variety of sets with S 1 â¤ 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 â¤ X is X– Y.
• Overall sets of optimum size = ( Y-L +1) * mx + X-Y.

Below is the application of the above method:

## C++

 ` #include << bits/stdc++. h>>` ` utilizing` ` namespace` ` sexually transmitted disease;` ` ` ` int` ` countSets(`` int` ` L, `` int` ` R)` ` {` ` ` ` ` ` ` ` `` int` ` mx = `` flooring``( log2( R/ L)) + 1;` ` ` ` ` ` ` ` ` ` `` if` `( mx == 1) {` ` `` return` `( R - L + 1);` ` ``} ` ` ` ` ` ` `` int` ` X = R/ `` pow``( 2, mx - 1);` ` `` int` ` Y = R/ (`` pow``( 2, mx - 2) * 3);` ` ` ` ` ` ` ` `` if` `( Y < < L) {` ` `` return` `( X - L + 1);` ` ``} ` ` ` ` ` ` `` int` ` ans = (Y - L + 1) * mx + (X - Y);` ` ` ` `` return` ` ans;` `} ` ` ` ` int` ` primary()` ` {` ` `` int` ` L = 3, R = 19;` ` ` ` ` ` `` cout << < < countSets( L, R) << < < endl;` `} `

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