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 is 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 to use the Sieve of Eratosthenes to find all the prime numbers. 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 the 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 Sieve 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

Java

// Java implementation to check if
// a number is a prime power number
import java.io.*;
import java.util.*;
import java.lang.Math;

class GFG{

// Array to store the
// prime numbers
static ArrayList<Integer> primes = new ArrayList<Integer>();

// Function to mark the prime
// numbers using Sieve of
// Eratosthenes
public static void sieveOfEratosthenes(int n)
{

    // Create a boolean array "prime[0..n]" and initialize
    // all entries it as true. A value in prime[i] will
    // finally be false if i is Not a prime, else true.
    boolean prime[] = new boolean[n + 1];

    for(int i = 0; i < n; i++)
        prime[i] = true;

    for(int p = 2; p * p <= n; p++)
    {

        // If prime[p] is not changed,
        // then it is a prime
        if(prime[p] == true)
        {

            // Update all multiples of p
            for(int i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }

    // Print all prime numbers
    for(int i = 2; i <= n; i++)
    {
        if(prime[i] == true)
            primes.add(i);
    }
}

// Function to check if the
// number can be represented
// as a power of prime
public static int[] power_of_prime(int n)
{
    for(int ii = 0; ii < primes.size(); ii++)
    {
        int i = primes.get(ii);

        if (n % i == 0)
        {
            int c = 0;
            while(n % i == 0)
            {
                n /= i;
                c += 1;
            }

            if(n == 1)
                return new int[]{i, c};
            else
                return new int[]{-1, 1};
        }
    }
    return new int[]{-1, 1};
}

// Driver code
public static void main(String args[])
{
    int n = 49;
    int sq = (int)(Math.sqrt(n));
    sieveOfEratosthenes(sq + 1);

    // Function call
    int arr[] = power_of_prime(n);

    if (arr[0] > 1)
        System.out.println(arr[0] + " ^ " + arr[1]);
    else
        System.out.println("-1");
}
}

// This code is contributed by grand_master

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 Sieve 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)

C#

// C# implementation to check if
// a number is a prime power number
using System;
using System.Collections;

class GFG{

// Array to store the
// prime numbers
static ArrayList primes = new ArrayList();

// Function to mark the prime
// numbers using Sieve of
// Eratosthenes
public static void sieveOfEratosthenes(int n)
{

    // Create a boolean array "prime[0..n]"
    // and initialize all entries it as true.
    // A value in prime[i] will finally be
    // false if i is Not a prime, else true.
    bool []prime = new bool[n + 1];

    for(int i = 0; i < n; i++)
        prime[i] = true;

    for(int p = 2; p * p <= n; p++)
    {

        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true)
        {

            // Update all multiples of p
            for(int i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }

    // Print all prime numbers
    for(int i = 2; i <= n; i++)
    {
        if (prime[i] == true)
            primes.Add(i);
    }
}

// Function to check if the
// number can be represented
// as a power of prime
public static int[] power_of_prime(int n)
{
    for(int ii = 0; ii < primes.Count; ii++)
    {
        int i = (int)primes[ii];

        if (n % i == 0)
        {
            int c = 0;
            while (n % i == 0)
            {
                n /= i;
                c += 1;
            }

            if (n == 1)
                return new int[]{i, c};
            else
                return new int[]{-1, 1};
        }
    }
    return new int[]{-1, 1};
}

// Driver code
public static void Main(string []args)
{
    int n = 49;
    int sq = (int)(Math.Sqrt(n));
    sieveOfEratosthenes(sq + 1);

    // Function call
    int []arr = power_of_prime(n);

    if (arr[0] > 1)
        Console.Write(arr[0] + " ^ " +
                      arr[1]);
    else
        Console.Write("-1");
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// Javascript implementation to check if
// a number is a prime power number

// Array to store the
// prime numbers
let primes = [];

// Function to mark the prime
// numbers using Sieve of
// Eratosthenes
function sieveOfEratosthenes(n)
{

    // Create a boolean array "prime[0..n]"
    // and initialize all entries it as true.
    // A value in prime[i] will finally be
    // false if i is Not a prime, else true.
    let prime = Array.from({length: n+1}, (_, i) => 0);

    for(let i = 0; i < n; i++)
        prime[i] = true;

    for(let p = 2; p * p <= n; p++)
    {

        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true)
        {

            // Update all multiples of p
            for(let i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }

    // Print all prime numbers
    for(let i = 2; i <= n; i++)
    {
        if (prime[i] == true)
            primes.push(i);
    }
}

// Function to check if the
// number can be represented
// as a power of prime
function power_of_prime(n)
{
    for(let ii = 0; ii < primes.length; ii++)
    {
        let i = primes[ii];

        if (n % i == 0)
        {
            let c = 0;
            while (n % i == 0)
            {
                n /= i;
                c += 1;
            }

            if (n == 1)
                return [i, c];
            else
                return [-1, 1];
        }
    }
    return [-1, 1];
}



// Driver Code

        let n = 49;
    let sq = (Math.sqrt(n));
    sieveOfEratosthenes(sq + 1);

    // Function call
    let arr = power_of_prime(n);

    if (arr[0] > 1)
        document.write(arr[0] + " ^ " +
                      arr[1]);
    else
        document.write("-1");

</script>

Output:

7 ^ 2

Time Complexity: O(√n*log(log(√n)))+n)
Auxiliary Space: O(m) where m=1000001