Given a number N, check if it is Hexagonal or not. If it is then print “Yes” otherwise output “No”.
Examples:
Input: N = 6
Output: Yes
Explanation:
6 is the second hexagonal number.Input: N = 14
Output: No
Explanation:
The second hexagonal number is 6 while the third hexagonal number is 15. Hence 14 is not a hexagonal number.
Approach: To solve the problem mentioned above we have to note that the nth hexagonal Number is given by the formula: H(n) = n * (2n – 1). The formula indicates that the n-th hexagonal number depends quadratically on n. Therefore, find the positive integral root of N = H(n) equation.
Therefore, H(n) = nth hexagonal number and N is the given Number. Hence solve the following equation:
Here, H(n) = N
=> n * (2n – 1) = N
=> 2 * n * n – n – N = 0
The positive root of this equation is:
n = (1 + sqrt(8 N + 1)) / 4
After obtaining the value for n, check if it is an integer or not where n is an integer if n – floor(n) is 0.
Below is the implementation of the above approach:
C++
// C++ Program to check// if N is a Hexagonal Number#include <bits/stdc++.h>using namespace std;// Function to check// if number is hexagonalbool isHexagonal(int N){ float val = 8 * N + 1; float x = 1 + sqrt(val); // Calculate the value for n float n = (x) / 4; // Check if n - floor(n) // is equal to 0 if ((n - (int)n) == 0) return true; else return false;}// Driver codeint main(){ int N = 14; if (isHexagonal(N) == true) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
// Java program to check// if N is a hexagonal numberimport java.util.*;class GFG {// Function to check// if number is hexagonalstatic boolean isHexagonal(int N){ float val = 8 * N + 1; float x = 1 + (float)Math.sqrt(val); // Calculate the value for n float n = (x) / 4; // Check if n - floor(n) // is equal to 0 if ((n - (int)n) == 0) return true; else return false;}// Driver codepublic static void main(String[] args){ int N = 14; if (isHexagonal(N) == true) System.out.println("Yes"); else System.out.println("No");}}// This code is contributed by offbeat |
Python3
# Python3 program to check# if N is a hexagonal numberfrom math import sqrt# Function to check# if number is hexagonaldef isHexagonal(N): val = 8 * N + 1 x = 1 + sqrt(val) # Calculate the value for n n = x / 4 # Check if n - floor(n) # is equal to 0 if ((n - int(n)) == 0): return True else: return False# Driver codeif __name__ == '__main__': N = 14 if (isHexagonal(N) == True): print("Yes") else: print("No")# This code is contributed by BhupendraSingh |
C#
// C# program to check if // N is a hexagonal numberusing System;class GFG{// Function to check// if number is hexagonalstatic bool isHexagonal(int N){ float val = 8 * N + 1; float x = 1 + (float)Math.Sqrt(val); // Calculate the value for n float n = (x) / 4; // Check if n - floor(n) // is equal to 0 if ((n - (int)n) == 0) { return true; } else { return false; }}// Driver codepublic static void Main(string[] args){ int N = 14; if (isHexagonal(N) == true) Console.Write("Yes"); else Console.Write("No");}}// This code is contributed by rutvik_56 |
Javascript
<script>// Javascript program to check// if N is a hexagonal number// Function to check// if number is hexagonalfunction isHexagonal(N){ let val = 8 * N + 1; let x = 1 + Math.sqrt(val); // Calculate the value for n let n = (x) / 4; // Check if n - floor(n) // is equal to 0 if ((n - parseInt(n)) == 0) { return true; } else { return false; }}// Driver codelet N = 14;if (isHexagonal(N) == true){ document.write("Yes" + "</br>");}else{ document.write("No");}// This code is contributed by Ankita saini </script> |
No
Time Complexity: O(log(N)), for calculation square root.
Auxiliary Space: O(1)
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