Given an integer n, the task is to count the numbers having only 1 set bit in the range [0, n].
Input: n = 7
000, 001, 010, 011, 100, 101, 110 and 111 are the binary representation of all the numbers upto 7.
And there are only 3 numbers ( 001, 010 and 100 ) having only 1 set bit.
Input: n = 3
Approach: If k bits are required to represent n then there are k numbers possible as 1 can be positioned at k different positions each time.
Below is the implementation of the above approach
Time Complexity: O(log n)
- Count Odd and Even numbers in a range from L to R
- Count of numbers from range [L, R] whose sum of digits is Y
- Count the numbers divisible by 'M' in a given range
- Count factorial numbers in a given range
- Count numbers in range 1 to N which are divisible by X but not by Y
- Numbers in range [L, R] such that the count of their divisors is both even and prime
- Count of common multiples of two numbers in a range
- Count all the numbers in a range with smallest factor as K
- Count of all even numbers in the range [L, R] whose sum of digits is divisible by 3
- Count numbers from range whose prime factors are only 2 and 3
- Count of Numbers in Range where the number does not contain more than K non zero digits
- Count numbers with unit digit k in given range
- Count numbers in range L-R that are divisible by all of its non-zero digits
- Count of Numbers in a Range where digit d occurs exactly K times
- Count numbers in a range having GCD of powers of prime factors equal to 1
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Improved By : jit_t