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:
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- Count numbers in range [L, R] having K consecutive set bits
- Count numbers from range whose prime factors are only 2 and 3 using Arrays | Set 2
- Find Nth positive number whose digital root is X
- Nth number whose sum of digit is multiple of 10
- Nth positive number whose absolute difference of adjacent digits is at most 1
- Count numbers < = N whose difference with the count of primes upto them is > = K
- Count numbers in a given range whose count of prime factors is a Prime Number
- Count pairs (A, B) such that A has X and B has Y number of set bits and A+B = C
- Find a number X such that (X XOR A) is minimum and the count of set bits in X and B are equal
- Count of N digit Numbers whose sum of every K consecutive digits is equal | Set 2
- Count total unset bits in all the numbers from 1 to N
- Print all numbers whose set of prime factors is a subset of the set of the prime factors of X
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