Given an integer N, the task is to check if it is a icositetragonal number or not.
icositetragonal number:
s a class of figurate number. It has a 24-sided polygon called Icositetragon. The N-th Icositetragonal number count’s the number of dots and all others dots are surrounding with a common sharing corner and make a pattern
The first few icositetragonal numbers are 1, 24, 69, 136, 225, 336, …
Examples:
Input: N = 24
Output: Yes
Explanation:
Second icositetragonal number is 24.Input: N = 30
Output: No
Approach:
1. The Kth term of the icositetragonal number is given as
2. As we have to check that the given number can be expressed as a icositetragonal number or not. This can be checked as follows –
=>
=>
3. Finally, check the value computed using this formula is an integer, which means that N is a icositetragonal number.
Below is the implementation of the above approach:
C++
// C++ implementation to check that// a number is icositetragonal number or not#include <bits/stdc++.h>using namespace std;// Function to check that the// number is a icositetragonal numberbool isicositetragonal(int N){ float n = (10 + sqrt(44 * N + 100)) / 22; // Condition to check if the // number is a icositetragonal number return (n - (int)n) == 0;}// Driver Codeint main(){ int i = 24; // Function call if (isicositetragonal(i)) { cout << "Yes"; } else { cout << "No"; } return 0;} |
Java
// Java implementation to check that// a number is icositetragonal number or notclass GFG{// Function to check that the// number is a icositetragonal numberstatic boolean isicositetragonal(int N){ float n = (float)((10 + Math.sqrt(44 * N + 100)) / 22); // Condition to check if the // number is a icositetragonal number return (n - (int)n) == 0;}// Driver Codepublic static void main(String[] args){ int i = 24; // Function call if (isicositetragonal(i)) { System.out.print("Yes"); } else { System.out.print("No"); }}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation to check that # a number is icositetragonal number # or not import math# Function to check that the number# is a icositetragonal number def isicositetragonal(N): n = (10 + math.sqrt(44 * N + 100)) / 22 # Condition to check if the number # is a icositetragonal number return (n - int(n)) == 0# Driver Code i = 24# Function call if (isicositetragonal(i)): print("Yes") else: print("No") # This code is contributed by divyamohan123 |
C#
// C# implementation to check that// a number is icositetragonal number or notusing System;class GFG{// Function to check that the// number is a icositetragonal numberstatic bool isicositetragonal(int N){ float n = (float)((10 + Math.Sqrt(44 * N + 100)) / 22); // Condition to check if the // number is a icositetragonal number return (n - (int)n) == 0;}// Driver Codepublic static void Main(){ int i = 24; // Function call if (isicositetragonal(i)) { Console.Write("Yes"); } else { Console.Write("No"); }}}// This code is contributed by Akanksha_Rai |
Javascript
<script>// JavaScript implementation to check that// a number is icositetragonal number or not// Function to check that the// number is a icositetragonal numberfunction isicositetragonal(N){ var n = (10 + Math.sqrt(44 * N + 100)) / 22; // Condition to check if the // number is a icositetragonal number return (n - parseInt(n)) == 0;}// Driver Codevar i = 24;// Function callif (isicositetragonal(i)) { document.write("Yes");}else { document.write("No");} </script> |
Yes
Time Complexity: O(log N), it is time taken by inbuilt sqrt() function
Auxiliary Space: O(1)
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