Given an integer N, the task is to find the count of natural Hexadecimal numbers with N digits.
Examples:
Input: N = 1
Output: 15Input: N = 2
Output: 240
Approach: It can be observed that for the values of N = 1, 2, 3, …, a series will be formed as 15, 240, 3840, 61440, 983040, 15728640, … which is a GP series whose common ratio is 16 and a = 15.
Hence the nth term will be 15 * pow(16, n – 1).
So, the count of n-digit natural hexadecimal numbers will be 15 * pow(16, n – 1).
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to return the count of n-digit // natural hexadecimal numbersint count(int n){ return 15 * pow(16, n - 1);}// Driver codeint main(){ int n = 2; cout << count(n); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG {// Function to return the count of n-digit // natural hexadecimal numbersstatic int count(int n){ return (int) (15 * Math.pow(16, n - 1));}// Driver codepublic static void main(String args[]) { int n = 2; System.out.println(count(n));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the above approach # Function to return the count of n-digit # natural hexadecimal numbers def count(n) : return 15 * pow(16, n - 1); # Driver code if __name__ == "__main__" : n = 2; print(count(n)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG { // Function to return the count of n-digit // natural hexadecimal numbers static int count(int n) { return (int) (15 * Math.Pow(16, n - 1)); } // Driver code public static void Main(String []args) { int n = 2; Console.WriteLine(count(n)); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript implementation of the above approach// Function to return the count of n-digit // natural hexadecimal numbersfunction count(n){ return 15 * Math.pow(16, n - 1);}// Driver codevar n = 2;document.write(count(n));</script> |
Output
240
Time Complexity: O(log n)
Auxiliary space: O(1), If the recursive stack is considered then it would be O(log(n))
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