Given an integer N, the task is to find the count of N-bit binary numbers without leading zeros.
Input: N = 2
10 and 11 are the only possible binary numbers.
Input: N = 4
Approach: Since the numbers cannot have leading zeros so the left-most bit has to be set to 1. Now for the rest of the N – 1 bits, there are two choices they can either be set to 0 or 1. So, the count of possible numbers will be 2N – 1.
Below is the implementation of the above approach:
- Count numbers having N 0's and and M 1's with no leading zeros
- Comparing leading zeros in binary representations of two numbers
- All possible numbers of N digits and base B without leading zeros
- Number of leading zeros in binary representation of a given number
- Count unique numbers that can be generated from N by adding one and removing trailing zeros
- Count number of trailing zeros in Binary representation of a number using Bitset
- Count numbers have all 1s together in binary representation
- Count of Binary Digit numbers smaller than N
- Count number of trailing zeros in (1^1)*(2^2)*(3^3)*(4^4)*..
- Count number of trailing zeros in product of array
- Minimum count of Full Binary Trees such that the count of leaves is N
- 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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