Given an integer N, the task is to check if N is a Concentric Hexagonal Numbers or not. If the number N is a Concentric Hexagonal Number then print “Yes” else print “No”.
Concentric Hexagonal Numbers are the number sequence forms a pattern with concentric hexagons, and the numbers denote the number of points required after the N-th stage of the pattern. The first few concentric hexagonal numbers are 0, 1, 6, 13, 24, 37, 54, 73, 96, 121 …
Examples:
Input: N = 6
Output: Yes
Explanation: Third Concentrichexagonal number is 6.Input: N = 20
Output: No
Approach:
- The Kth term of the Concentric hexagonal number is given as:
Kth Term = (3 * K * K) / 2
- As we have to check that the given number can be expressed as a Concentric hexagonal number or not. This can be checked as:
Here, Kth Term = N => (3 * K * K) / 2 = N => 3 * K * K – 2 * N = 0 The positive root of this equation is: K = sqrt((2 * N )/3)
- If the value of K calculated using the above formula is an integer, then N is a Concentric Hexagonal Number.
- Else the number N is not a ConcentricHexagonal Number.
C++
// C++ program to check if N is a// Concentric Hexagonal Number#include <bits/stdc++.h>using namespace std;// Function to check if the// number is a Concentric hexagonal numberbool isConcentrichexagonal(int N){ float n = sqrt((2 * N) / 3); // Condition to check if the // number is a Concentric // hexagonal number return (n - (int)n) == 0;}// Driver Codeint main(){ int N = 6; // Function call if (isConcentrichexagonal(N)) { cout << "Yes"; } else { cout << "No"; } return 0;} |
Java
// Java program to check if N is a// Concentric Hexagonal Numberclass GFG{// Function to check if the// number is a Concentric hexagonal numberstatic boolean isConcentrichexagonal(int N){ float n = (float) Math.sqrt((2 * N) / 3); // Condition to check if the // number is a Concentric // hexagonal number return (n - (int)n) == 0;}// Driver Codepublic static void main(String[] args){ int N = 6; // Function call if (isConcentrichexagonal(N)) { System.out.print("Yes"); } else { System.out.print("No"); }}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to check if N is a# concentric hexagonal numberimport math# Function to check if the number# is a concentric hexagonal numberdef isConcentrichexagonal(N): n = math.sqrt((2 * N) / 3) # Condition to check if the # number is a concentric # hexagonal number return (n - int(n)) == 0# Driver codeN = 6if isConcentrichexagonal(N): print("Yes")else: print("No")# This code is contributed by divyeshrabadiya07 |
C#
// C# program to check if N is a// concentric hexagonal numberusing System;class GFG{// Function to check if the number// is a concentric hexagonal numberstatic bool isConcentrichexagonal(int N){ float n = (float) Math.Sqrt((2 * N) / 3); // Condition to check if the // number is a concentric // hexagonal number return (n - (int)n) == 0;}// Driver Codepublic static void Main(){ int N = 6; // Function call if (isConcentrichexagonal(N)) { Console.Write("Yes"); } else { Console.Write("No"); }}}// This code is contributed by Code_Mech |
Javascript
// JS program to check if N is a // concentric hexagonal number // Function to check if the number// is a concentric hexagonal number function isConcentrichexagonal(N){ let n = Math.sqrt((2 * N) / 3) // Condition to check if the // number is a concentric // hexagonal number return (n - Math.floor(n)) == 0} // Driver codelet N = 6 if (isConcentrichexagonal(N)) console.log("Yes")else console.log("No") // This code is contributed by phasing17 |
Output
Yes
Time complexity: O(logN) for given n, as it is using inbuilt sqrt function
Auxiliary space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Share your thoughts in the comments
