Program to check if N is a Heptagonal Number

Given an integer N, the task is to check if N is a Heptagonal Number or not. If the number N is an Heptagonal Number then print “Yes” else print “No”.

Heptagonal Number represents Heptagon and belongs to a figurative number. Heptagonal has seven angles, seven vertices, and seven-sided polygon. The first few Heptagonal Numbers are 1, 7, 18, 34, 55, 81, …

Examples:

Input: N = 7 
Output: Yes 
Explanation:
Second heptagonal number is 7.

Input: N = 30 
Output: No  

Approach:

• The K^th term of the heptagonal number is given as
  
• As we have to check that the given number can be expressed as a Heptagonal Number or not. This can be checked as: 

=> 
=> 

• If the value of K calculated using the above formula is an integer, then N is a Heptagonal Number.
• Else N is not a Heptagonal Number.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if N is a
// Heptagonal number
bool isheptagonal(int N)
{
    float n
        = (3 + sqrt(40 * N + 9))
          / 10;

    // Condition to check if the
    // number is a heptagonal number
    return (n - (int)n) == 0;
}

// Driver Code
int main()
{
    // Given Number
    int N = 7;

    // Function call
    if (isheptagonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;

class GFG {

// Function to check if N
// is a heptagonal number
public static boolean isheptagonal(int N)
{
    double n = (3 + Math.sqrt(40 * N + 9)) / 10;

    // Condition to check if the number
    // is a heptagonal number
    return (n - (int)n) == 0;
}

// Driver code
public static void main(String[] args)
{

    // Given Number
    int N = 7;

    // Function call
    if (isheptagonal(N))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}

// This code is contributed by coder001

Python3

# Python3 program for the above approach
import math

# Function to check if N is a
# heptagonal number
def isheptagonal(N):

    n = (3 + math.sqrt(40 * N + 9)) / 10

    # Condition to check if the
    # number is a heptagonal number
    return (n - int(n)) == 0

# Driver Code
N = 7

# Function call
if (isheptagonal(N)):
    print("Yes")
else:
    print("No")

# This code is contributed by Shubham_Coder

C#

// C# program for the above approach
using System;

class GFG {

// Function to check if N
// is a heptagonal number
public static bool isheptagonal(int N)
{
    double n = (3 + Math.Sqrt(40 * N + 9)) / 10;

    // Condition to check if the number
    // is a heptagonal number
    return (n - (int)n) == 0;
}

// Driver code
public static void Main(String[] args)
{

    // Given Number
    int N = 7;

    // Function call
    if (isheptagonal(N))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}

// This code is contributed by Rohit_ranjan

Javascript

<script>

// JavaScript program for the above approach

// Function to check if N is a
// Heptagonal number
function isheptagonal(N)
{
    var n = (3 + Math.sqrt(40 * N + 9))
          / 10;

    // Condition to check if the
    // number is a heptagonal number
    return (n - parseInt(n)) == 0;
}

// Given Number
var N = 7;
// Function call
if (isheptagonal(N)) {
    document.write("Yes");
}
else {
    document.write("No");
}

</script>

Output

Yes

Time complexity: O(logN) because it is using inbuilt sqrt function
Auxiliary Space: O(1)

Approach: Binary Search

 Binary search to find if the given number N matches any Heptagonal Number. This approach is more efficient than brute 
 force as it reduces the search space by half at each iteration.

Below is code implementation of the above approach:

C++

#include <iostream>
using namespace std;

// Function to check if a number is a Heptagonal number
string isHeptagonal(int n) {
    int left = 1;
    int right = n;

    while (left <= right) {
        int mid = (left + right) / 2;

        // Calculate the Heptagonal Number using the formula
        int hn = mid * (5 * mid - 3) / 2;

        if (hn == n) {
            return "Yes"; // If the number is Heptagonal, return "Yes"
        } else if (hn < n) {
            left = mid + 1; // If the calculated Heptagonal number is less than the input, move to the right half
        } else {
            right = mid - 1; // If the calculated Heptagonal number is greater than the input, move to the left half
        }
    }

    return "No"; // If the number is not Heptagonal, return "No"
}

int main() {
    cout << isHeptagonal(7) << endl; // Check if 7 is a Heptagonal number
    return 0;
}

Java

public class GFG {
        static String isHeptagonal(int n) {
        int left = 1;
        int right = n;

        while (left <= right) {
            int mid = (left + right) / 2;

            // Calculate the Heptagonal Number using the formula
            int hn = mid * (5 * mid - 3) / 2;

            if (hn == n) {
                return "Yes";
            } else if (hn < n) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }

        return "No";
    }

    public static void main(String[] args) {
        System.out.println(isHeptagonal(7));
    }
}

Python

def is_heptagonal(n):
    left = 1
    right = n
    while left <= right:
        mid = (left + right) // 2
        # Calculate the Heptagonal Number using the formula
        hn = mid*(5*mid-3)//2
        if hn == n:
            return "Yes"
        elif hn < n:
            left = mid + 1
        else:
            right = mid - 1
    return "No"

# Test the function
print(is_heptagonal(7))

C#

using System;

class Program
{
    // Function to check if a number is a Heptagonal number
    static string IsHeptagonal(int n)
    {
        int left = 1;
        int right = n;

        while (left <= right)
        {
            int mid = (left + right) / 2;

            // Calculate the Heptagonal Number using the formula
            int hn = mid * (5 * mid - 3) / 2;

            if (hn == n)
            {
                return "Yes"; // If the number is Heptagonal, return "Yes"
            }
            else if (hn < n)
            {
                left = mid + 1; // If the calculated Heptagonal number is less than the input, move to the right half
            }
            else
            {
                right = mid - 1; // If the calculated Heptagonal number is greater than the input, move to the left half
            }
        }

        return "No"; // If the number is not Heptagonal, return "No"
    }

    static void Main()
    {
        Console.WriteLine(IsHeptagonal(7)); // Check if 7 is a Heptagonal number
    }
}

Javascript

function isHeptagonal(n) {
    let left = 1;
    let right = n;

    while (left <= right) {
        let mid = Math.floor((left + right) / 2);

        // Calculate the Heptagonal Number using the formula
        let hn = mid * (5 * mid - 3) / 2;

        if (hn === n) {
            return "Yes";
        } else if (hn < n) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }

    return "No";
}

// Test the function
console.log(isHeptagonal(7));

Output

Yes

Time Complexity: O(log n)
Auxiliary Space: O(1)