Check if the number is a Prime power number

Given an integer N, the task is to check if the number is a Prime power number. If yes, then print the number along with its power which is equal to N. Else print -1.

A prime power is a positive integer power of a single prime number.
For example: 7 = 7¹, 9 = 3² and 32 = 2⁵ are prime powers, while 6 = 2 × 3, 12 = 22 × 3 and 36 = 62 = 22 × 32 are not. (The number 1 is not counted as a prime power.)

Note: If there are no such prime number, print -1.
Examples:

Input: N = 49
Output: 7²
Explanation:
N can be represented as a power of prime number 7.
N = 49 = 7²

Input: N = 100
Output: -1
Explanation:
N cannot be represented as a power of any prime number.

Approach: The idea is use Seive of Eratosthenes to find all the prime number. Then, Iterate over all the prime numbers and check that if any prime number divides the given number N, if yes then divide it until it becomes 1 or not divisible by that prime number. Finally, check that the number is equal to 1, If yes then return prime number otherwise given number cannot be expressed as a prime number raised to some power.

Below is the implementation of the above approach:

C++

// C++ implementation to check if  
// a number is a prime power number 
#include<bits/stdc++.h> 
using namespace std; 
  
// Array to store the  
// prime numbers 
bool is_prime[1000001];  
vector<int> primes; 
  
// Function to mark the prime  
// numbers using Seive of  
// Eratosthenes 
void SieveOfEratosthenes(int n) 
{ 
    int p = 2; 
      
    for(int i = 0; i < n; i++) 
       is_prime[i] = true; 
         
    while (p * p <= n) 
    { 
          
        // If prime[p] is not  
        // changed, then it is a prime  
        if (is_prime[p] == true) 
        { 
              
            // Update all multiples of p  
            for(int i = p * p; i < n + 1; i += p) 
            { 
               is_prime[i] = false; 
            }  
        } 
        p += 1; 
    } 
    for(int i = 2; i < n + 1; i++) 
    { 
       if (is_prime[i]) 
           primes.push_back(i); 
    } 
} 
  
// Function to check if the  
// number can be represented  
// as a power of prime 
pair<int, int> power_of_prime(int n) 
{ 
    for(auto i : primes) 
    { 
         
       if (n % i == 0) 
       { 
           int c = 0; 
           while(n % i == 0) 
           { 
               n /= i; 
               c += 1;  
           } 
           if(n == 1) 
              return {i, c}; 
           else
              return {-1, 1}; 
       } 
    } 
} 
  
// Driver Code          
int main() 
{ 
    int n = 49; 
    SieveOfEratosthenes(int(sqrt(n)) + 1); 
      
    // Function Call 
    pair<int, int> p = power_of_prime(n); 
      
    if (p.first > 1) 
        cout << p.first << " ^ " 
             << p.second << endl; 
    else
        cout << -1 << endl; 
} 
  
// This code is contributed by Surendra_Gangwar 

Python3

# Python3 implementation to check  
# if a number is a prime power number 
  
from math import *
  
# Array to store the  
# prime numbers 
is_prime = [True for i in range(10**6 + 1)]  
primes =[] 
  
# Function to mark the prime  
# numbers using Seive of  
# Eratosthenes 
def SieveOfEratosthenes(n):  
    p = 2
    while (p * p <= n): 
        # If prime[p] is not  
        # changed, then it is a prime  
        if (is_prime[p] == True):  
            # Update all multiples of p  
            for i in range(p * p, n + 1, p):  
                is_prime[i] = False
        p += 1
    for i in range(2, n + 1): 
        if is_prime[i]: 
            primes.append(i) 
  
# Function to check if the  
# number can be represented  
# as a power of prime 
def power_of_prime(n): 
    for i in primes: 
        if n % i == 0: 
            c = 0
            while n % i == 0: 
                n//= i 
                c += 1
            if n == 1: 
                return (i, c) 
            else: 
                return (-1, 1) 
  
# Driver Code          
if __name__ == "__main__": 
    n = 49
    SieveOfEratosthenes(int(sqrt(n))+1) 
      
    # Function Call 
    num, power = power_of_prime(n) 
    if num > 1: 
        print(num, "^", power) 
    else: 
        print(-1) 

Output:

7 ^ 2

My Personal Notes

C++

Python3

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