Given an integer N, the task is to check if it is an Icositrigonal number or not.
Icositrigonal number is a class of figurate number. It has 23 – sided polygon called Icositrigon. The N-th Icositrigonal number count’s the 23 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few Icositrigonol numbers are 1, 23, 66, 130, 215, 321, 448 …
Examples:
Input: N = 23
Output: Yes
Explanation:
Second icositrigonal number is 23.Input: N = 30
Output: No
Approach:
1. The Kth term of the icositrigonal number is given as
2. As we have to check whether the given number can be expressed as an icositrigonal 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 an icositrigonal number.
Below is the implementation of the above approach:
C++
// C++ implementation to check that// a number is a icositrigonal number or not#include <bits/stdc++.h>using namespace std;// Function to check that the// number is a icositrigonal numberbool isicositrigonal(int N){ float n = (19 + sqrt(168 * N + 361)) / 42; // Condition to check if the // number is a icositrigonal number return (n - (int)n) == 0;}// Driver Codeint main(){ int i = 23; // Function call if (isicositrigonal(i)) { cout << "Yes"; } else { cout << "No"; } return 0;} |
Java
// Java implementation to check that// a number is a icositrigonal number or notimport java.util.*;class GFG{// Function to check that the// number is a icositrigonal numberstatic boolean isicositrigonal(int N){ float n = (float)(19 + Math.sqrt(168 * N + 361)) / 42; // Condition to check if the // number is a icositrigonal number return (n - (int)n) == 0;}// Driver Codepublic static void main(String args[]){ int i = 23; // Function call if (isicositrigonal(i)) { System.out.print("Yes"); } else { System.out.print("No"); }}}// This code is contributed by Akanksha_Rai |
Python3
# Python3 implementation to check that a # number is a icositrigonal number or not import math# Function to check that the number # is a icositrigonal number def isicositrigonal(N): n = (19 + math.sqrt(168 * N + 361)) / 42 # Condition to check if the number # is a icositrigonal number return (n - int(n)) == 0# Driver Code i = 23# Function call if (isicositrigonal(i)): print("Yes") else: print("No") # This code is contributed by divyamohan123 |
C#
// C# implementation to check that// a number is a icositrigonal number or notusing System;class GFG{// Function to check that the// number is a icositrigonal numberstatic bool isicositrigonal(int N){ float n = (float)(19 + Math.Sqrt(168 * N + 361)) / 42; // Condition to check if the // number is a icositrigonal number return (n - (int)n) == 0;}// Driver Codepublic static void Main(){ int i = 23; // Function call if (isicositrigonal(i)) { Console.Write("Yes"); } else { Console.Write("No"); }}}// This code is contributed by Nidhi_Biet |
Javascript
<script>// JavaScript implementation to check that// a number is a icositrigonal number or not// Function to check that the// number is a icositrigonal numberfunction isicositrigonal(N){ var n = (19 + Math.sqrt(168 * N + 361)) / 42; // Condition to check if the // number is a icositrigonal number return (n - parseInt(n)) == 0;}// Driver Codevar i = 23;// Function callif (isicositrigonal(i)) { document.write("Yes");}else { document.write("No");} </script> |
Yes
Time Complexity: O(log N) because sqrt() function is being used
Auxiliary Space: O(1)
