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Program to check if N is a Icosihexagonal Number

  • Last Updated : 23 Jun, 2021

Given an integer N, the task is to check if it is a Icosihexagonal number or not.
 

Icosihexagonal number is class of figurate number. It has 26 – sided polygon called Icosihexagon. The N-th Icosihexagonal number count’s the 26 number of dots and all other dots are surrounding with a common sharing corner and make a pattern. The first few Icosihexagonol numbers are 1, 26, 75, 148 … 
 

Examples: 
 

Input: N = 26 
Output: Yes 
Explanation: 
Second icosihexagonal number is 26.
Input: 30 
Output: No 
 

 



Approach: 
 

  1. The Kth term of the icosihexagonal number is given as 
    K^{th} Term = \frac{24*K^{2} - 22*K}{2}
     
  2. As we have to check that the given number can be expressed as a icosihexagonal number or not. This can be checked as follows – 
     

=> N = \frac{24*K^{2} - 22*K}{2}
=> K = \frac{22 + \sqrt{192*N + 484}}{48}
 

  1.  
  2. Finally, check the value of computed using this formulae is an integer, which means that N is a icosihexagonal number.

Below is the implementation of the above approach:
 

C++




// C++ program to check whether
// a number is a icosihexagonal number
// or not
 
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to check whether the
// number is a icosihexagonal number
bool isicosihexagonal(int N)
{
    float n
        = (22 + sqrt(192 * N + 484))
          / 48;
 
    // Condition to check if the
    // number is a icosihexagonal number
    return (n - (int)n) == 0;
}
 
// Driver code
int main()
{
    int i = 26;
 
    if (isicosihexagonal(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java




// Java program to check whether the
// number is a icosihexagonal number
// or not
class GFG{
 
// Function to check whether the
// number is a icosihexagonal number
static boolean isicosihexagonal(int N)
{
    float n = (float) ((22 + Math.sqrt(192 * N +
                                       484)) / 48);
     
    // Condition to check if the number
    // is a icosihexagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given number
    int N = 26;
     
    // Function call
    if (isicosihexagonal(N))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by shubham

Python3




# Python3 program to check whether
# a number is a icosihexagonal number
# or not
import numpy as np
 
# Function to check whether the
# number is a icosihexagonal number
def isicosihexagonal(N):
 
    n = (22 + np.sqrt(192 * N + 484)) / 48
 
    # Condition to check if the
    # number is a icosihexagonal number
    return (n - (int(n))) == 0
 
# Driver code
i = 26
 
if (isicosihexagonal(i)):
    print ("Yes")
else:
    print ("No")
 
# This code is contributed by PratikBasu

C#




// C# program to check whether the
// number is a icosihexagonal number
// or not
using System;
class GFG{
 
// Function to check whether the
// number is a icosihexagonal number
static bool isicosihexagonal(int N)
{
    float n = (float)((22 + Math.Sqrt(192 * N +
                                      484)) / 48);
     
    // Condition to check if the number
    // is a icosihexagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main()
{
     
    // Given number
    int N = 26;
     
    // Function call
    if (isicosihexagonal(N))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by Code_Mech

Javascript




<script>
// Javascript program to check whether
// a number is a icosihexagonal number
// or not
 
// Function to check whether the
// number is a icosihexagonal number
function isicosihexagonal(N)
{
    let n
        = (22 + Math.sqrt(192 * N + 484))
          / 48;
 
    // Condition to check if the
    // number is a icosihexagonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver code
let i = 26;
 
if (isicosihexagonal(i)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
// This code is contributed by rishavmahato348.
</script>
Output: 
Yes

 

Time Complexity: O(sqrt(n))

Auxiliary Space: O(1)

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