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:
- The Kth term of the icosihexagonal number is given as
- As we have to check that the given number can be expressed as a icosihexagonal number or not. This can be checked as follows –
=>
=>
-
- 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++ 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 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 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# 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 |
<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)