Given a positive integer N, the task is to count the numbers that can be represented with N bits and whose 0th and Nth bits are set.
Input: N = 2
All possible 2-bit integers are 00, 01, 10 and 11.
Out of which only 11 has 0th and Nth bit set.
Input: N = 4
Approach: Out of the given N bits, only two bits need to be set i.e. the 0th and the Nth bit. So, setting these 2 bits as 1 we are left with the rest N – 2 bits every single of which can either be 0 or 1 and there are 2N – 2 ways of doing that.
Below is the implementation of the above approach:
- Count total set bits in all numbers from 1 to n | Set 2
- Count total unset bits in all the numbers from 1 to N
- Count set bits in the Kth number after segregating even and odd from N natural numbers
- Count unset bits in a range
- Count pairs with set bits sum equal to K
- Count pairs (A, B) such that A has X and B has Y number of set bits and A+B = C
- Count of divisors having more set bits than quotient on dividing N
- Count number of set bits in a range using bitset
- Find a number X such that (X XOR A) is minimum and the count of set bits in X and B are equal
- Minimum number N such that total set bits of all numbers from 1 to N is at-least X
- Number of ways to change the XOR of two numbers by swapping the bits
- Count numbers < = N whose difference with the count of primes upto them is > = K
- Count numbers which are divisible by all the numbers from 2 to 10
- Count numbers which can be constructed using two numbers
- Count numbers that don't contain 3
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