Is the logic efficient?

#1

just wanted to know if there is a better approach to the problem

#2

hello @granthdhir
yes there is better approach.

first get count of all bits in range [1…b] (say ans1)
the get count of all bits in range [1…a-1] (say ans2)
then ans1-ans2 will be the required answer.

now to get count of all the bits in range [1…n] refer this approach ->link

#3

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