Count of N-bit binary numbers without leading zeros
Given an integer N, the task is to find the count of N-bit binary numbers without leading zeros.
Examples:
Input: N = 2
Output: 2
10 and 11 are the only possible binary numbers.
Input: N = 4
Output: 8
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:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count // of possible numbers int count(int n) { return pow(2, n - 1); } // Driver code int main() { int n = 4; cout << count(n); return 0; } |
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Java
// Java implementation of the approach class GFG { // Function to return the count // of possible numbers static int count(int n) { return (int)Math.pow(2, n - 1); } // Driver code public static void main (String[] args) { int n = 4; System.out.println(count(n)); } } // This code is contributed by AnkitRai01 |
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Python3
# Python3 implementation of the approach # Function to return the count # of possible numbers def count(n): return pow(2, n - 1) # Driver code n = 4 print(count(n)) # This code is contributed by mohit kumar |
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C#
// C# implementation of the approach using System; class GFG { // Function to return the count // of possible numbers static int count(int n) { return (int)Math.Pow(2, n - 1); } // Driver code public static void Main (String[] args) { int n = 4; Console.WriteLine(count(n)); } } // This code is contributed by 29AjayKumar |
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Output:
8
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