Given a Number 
. The task is to check if N is a Trojan Number or not.
Trojan Number is a number that is a strong number but not a perfect power. A number N is known as a strong number if, for every prime divisor or factor p of N, p2 is also a divisor. In other words, every prime factor appears at least twice.
All Trojan numbers are strong. However, not all strong numbers are Trojan numbers: only those that cannot be represented as mk, where m and k are positive integers greater than 1.
Examples:
Input : N = 108 Output : YES Input : N = 8 Output : NO
The idea is to store the count of each prime factor and check if the count is greater than 2 then it will be a Strong Number.
This part can easily be calculated by prime factorization through sieve.
The next step is to check if the given number cannot be expressed as xy. To check whether a number is perfect power or not refer to this article.
Below is the implementation of above problem:
C++
// CPP program to check if a number is// Trojan Number or not#include <bits/stdc++.h>using namespace std;// Function to check if a number// can be expressed as x^ybool isPerfectPower(int n){ if (n == 1) return true; // Try all numbers from 2 to sqrt(n) as base for (int x = 2; x <= sqrt(n); x++) { int y = 2; int p = pow(x, y); // Keep increasing y while power 'p' // is smaller than n. while (p <= n && p > 0) { if (p == n) return true; y++; p = pow(x, y); } } return false;}// Function to check if a number is Strongbool isStrongNumber(int n){ unordered_map<int, int> count; while (n % 2 == 0) { n = n / 2; count[2]++; } // count the number for each prime factor for (int i = 3; i <= sqrt(n); i += 2) { while (n % i == 0) { n = n / i; count[i]++; } } if (n > 2) count[n]++; int flag = 0; for (auto b : count) { // minimum number of prime divisors // should be 2 if (b.second == 1) { flag = 1; break; } } if (flag == 1) return false; else return true;}// Function to check if a number// is Trojan Numberbool isTrojan(int n){ if (!isPerfectPower(n) && isStrongNumber(n)) return true; else return false;}// Driver Codeint main(){ int n = 108; if (isTrojan(n)) cout << "YES"; else cout << "NO"; return 0;} |
Java
// Java program to check if a number is// Trojan Number or notimport java.util.*;class GFG { // Function to check if a number // can be expressed as x^y static boolean isPerfectPower(int n) { if (n == 1) { return true; } // Try all numbers from 2 to sqrt(n) as base for (int x = 2; x <= Math.sqrt(n); x++) { int y = 2; int p = (int) Math.pow(x, y); // Keep increasing y while power 'p' // is smaller than n. while (p <= n && p > 0) { if (p == n) { return true; } y++; p = (int) Math.pow(x, y); } } return false; } // Function to check if a number is Strong static boolean isStrongNumber(int n) { HashMap<Integer, Integer> count = new HashMap<Integer, Integer>(); while (n % 2 == 0) { n = n / 2; if (count.containsKey(2)) { count.put(2, count.get(2) + 1); } else { count.put(2, 1); } } // count the number for each prime factor for (int i = 3; i <= Math.sqrt(n); i += 2) { while (n % i == 0) { n = n / i; if (count.containsKey(i)) { count.put(i, count.get(i) + 1); } else { count.put(i, 1); } } } if (n > 2) { if (count.containsKey(n)) { count.put(n, count.get(n) + 1); } else { count.put(n, 1); } } int flag = 0; for (Map.Entry<Integer, Integer> b : count.entrySet()) { // minimum number of prime divisors // should be 2 if (b.getValue() == 1) { flag = 1; break; } } if (flag == 1) { return false; } else { return true; } } // Function to check if a number // is Trojan Number static boolean isTrojan(int n) { if (!isPerfectPower(n) && isStrongNumber(n)) { return true; } else { return false; } } // Driver Code public static void main(String[] args) { int n = 108; if (isTrojan(n)) { System.out.println("Yes"); } else { System.out.println("No"); } }} // This code is contributed by PrinciRaj1992 |
Python3
# Python 3 program to check if a number # is Trojan Number or notfrom math import sqrt, pow# Function to check if a number# can be expressed as x^ydef isPerfectPower(n): if n == 1: return True # Try all numbers from 2 to # sqrt(n) as base for x in range(2, int(sqrt(n)) + 1): y = 2 p = pow(x, y) # Keep increasing y while power # 'p' is smaller than n. while p <= n and p > 0: if p == n: return True y += 1 p = pow(x, y) return False# Function to check if a number # is Strongdef isStrongNumber(n): count = {i:0 for i in range(n)} while n % 2 == 0: n = n // 2 count[2] += 1 # count the number for each # prime factor for i in range(3,int(sqrt(n)) + 1, 2): while n % i == 0: n = n // i count[i] += 1 if n > 2: count[n] += 1 flag = 0 for key,value in count.items(): # minimum number of prime # divisors should be 2 if value == 1: flag = 1 break if flag == 1: return False return True# Function to check if a number# is Trojan Numberdef isTrojan(n): return isPerfectPower(n) == False and isStrongNumber(n) # Driver Codeif __name__ == '__main__': n = 108 if (isTrojan(n)): print("YES") else: print("NO")# This code is contributed by# Surendra_Gangwar |
C#
// C# program to check if a number is// Trojan Number or notusing System;using System.Collections.Generic; class GFG { // Function to check if a number // can be expressed as x^y static bool isPerfectPower(int n) { if (n == 1) { return true; } // Try all numbers from 2 to sqrt(n) as base for (int x = 2; x <= Math.Sqrt(n); x++) { int y = 2; int p = (int) Math.Pow(x, y); // Keep increasing y while power 'p' // is smaller than n. while (p <= n && p > 0) { if (p == n) { return true; } y++; p = (int) Math.Pow(x, y); } } return false; } // Function to check if a number is Strong static bool isStrongNumber(int n) { Dictionary<int, int> count = new Dictionary<int, int>(); while (n % 2 == 0) { n = n / 2; if (count.ContainsKey(2)) { count[2] = count[2] + 1; } else { count.Add(2, 1); } } // count the number for each prime factor for (int i = 3; i <= Math.Sqrt(n); i += 2) { while (n % i == 0) { n = n / i; if (count.ContainsKey(i)) { count[i] = count[i] + 1; } else { count.Add(i, 1); } } } if (n > 2) { if (count.ContainsKey(n)) { count[n] = count[n] + 1; } else { count.Add(n, 1); } } int flag = 0; foreach(KeyValuePair<int, int> b in count) { // minimum number of prime divisors // should be 2 if (b.Value == 1) { flag = 1; break; } } if (flag == 1) { return false; } else { return true; } } // Function to check if a number // is Trojan Number static bool isTrojan(int n) { if (!isPerfectPower(n) && isStrongNumber(n)) { return true; } else { return false; } } // Driver Code public static void Main(String[] args) { int n = 108; if (isTrojan(n)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }}// This code is contributed by Princi Singh |
Javascript
<script>// javascript program to check if a number is// Trojan Number or not // Function to check if a number // can be expressed as x^y function isPerfectPower(n) { if (n == 1) { return true; } // Try all numbers from 2 to sqrt(n) as base for (var x = 2; x <= Math.sqrt(n); x++) { var y = 2; var p = parseInt( Math.pow(x, y)); // Keep increasing y while power 'p' // is smaller than n. while (p <= n && p > 0) { if (p == n) { return true; } y++; p = parseInt( Math.pow(x, y)); } } return false; } // Function to check if a number is Strong function isStrongNumber(n) { var count = new Map(); while (n % 2 == 0) { n = n / 2; if (count.has(2)) { count.set(2, count.get(2) + 1); } else { count.set(2, 1); } } // count the number for each prime factor for (var i = 3; i <= Math.sqrt(n); i += 2) { while (n % i == 0) { n = n / i; if (count.has(i)) { count.set(i, count.get(i) + 1); } else { count.set(i, 1); } } } if (n > 2) { if (count.has(n)) { count.set(n, count.get(n) + 1); } else { count.set(n, 1); } } var flag = 0; const iterator = count[Symbol.iterator](); let itr = iterator.next() for (let i = 0; i < count.size; i++) { console.log(itr.value, itr.done) if (itr.value == 1) { flag = 1; break; } itr = iterator.next() } if (flag == 1) { return false; } else { return true; } } // Function to check if a number // is Trojan Number function isTrojan(n) { if (!isPerfectPower(n) && isStrongNumber(n)) { return true; } else { return false; } } // Driver Code var n = 108; if (isTrojan(n)) { document.write("Yes"); } else { document.write("No"); }// This code contributed by gauravrajput1 </script> |
Output:
YES
Auxiliary Space: O(n)
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