Check if all Prime factors of number N are unique or not

Given a number N. The task is to check whether the given number N has unique prime factors or not. If yes then print YES else print NO.
Examples:

Input: N = 30 
Output: YES 
Explanation:
N = 30 = 2*3*5 
As all the prime factors of 30 are unique.
Input: N = 100 
Output: NO 
Explanation:
N = 100 = 2*2*5*5 
As all the prime factors of 100 are not unique because 2 and 5 are repeated twice. 

Approach:

1. Find all the prime factors of the given number N using Sieve Of Eratosthenes.
2. If the product of all the prime factors obtained is equaled to N then all prime factors are unique, so print YES.
3. Else print NO.

Below is the implementation of the above approach: 

CPP

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns the all the
// distinct prime factors in a vector
vector<int> primeFactors(int n)
{
    int i, j;
    vector<int> Prime;

    // If n is divisible by 2
    if (n % 2 == 0) {
        Prime.push_back(2);
    }

    // Divide n till all factors of 2
    while (n % 2 == 0) {
        n = n / 2;
    }

    // Check for the prime numbers other
    // than 2
    for (i = 3; i <= sqrt(n); i = i + 2) {

        // Store i in Prime[] i is a
        // factor of n
        if (n % i == 0) {
            Prime.push_back(i);
        }

        // Divide n till all factors of i
        while (n % i == 0) {
            n = n / i;
        }
    }

    // If n is greater than 2, then n is
    // prime number after n divided by
    // all factors
    if (n > 2) {
        Prime.push_back(n);
    }

    // Returns the vector Prime
    return Prime;
}

// Function that check whether N is the
// product of distinct prime factors
// or not
void checkDistinctPrime(int n)
{
    // Returns the vector to store
    // all the distinct prime factors
    vector<int> Prime = primeFactors(n);

    // To find the product of all
    // distinct prime factors
    int product = 1;

    // Find the product
    for (auto i : Prime) {
        product *= i;
    }

    // If product is equals to N,
    // print YES, else print NO
    if (product == n)
        cout << "YES";
    else
        cout << "NO";
}

// Driver Code
int main()
{
    int N = 30;
    checkDistinctPrime(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function that returns the all the
// distinct prime factors in a vector
static Vector<Integer> primeFactors(int n)
{
    int i, j;
    Vector<Integer> Prime = new Vector<Integer>();

    // If n is divisible by 2
    if (n % 2 == 0) {
        Prime.add(2);
    }

    // Divide n till all factors of 2
    while (n % 2 == 0) {
        n = n / 2;
    }

    // Check for the prime numbers other
    // than 2
    for (i = 3; i <= Math.sqrt(n); i = i + 2) {

        // Store i in Prime[] i is a
        // factor of n
        if (n % i == 0) {
            Prime.add(i);
        }

        // Divide n till all factors of i
        while (n % i == 0) {
            n = n / i;
        }
    }

    // If n is greater than 2, then n is
    // prime number after n divided by
    // all factors
    if (n > 2) {
        Prime.add(n);
    }

    // Returns the vector Prime
    return Prime;
}

// Function that check whether N is the
// product of distinct prime factors
// or not
static void checkDistinctPrime(int n)
{
    // Returns the vector to store
    // all the distinct prime factors
    Vector<Integer> Prime = primeFactors(n);

    // To find the product of all
    // distinct prime factors
    int product = 1;

    // Find the product
    for (int i : Prime) {
        product *= i;
    }

    // If product is equals to N,
    // print YES, else print NO
    if (product == n)
        System.out.print("YES");
    else
        System.out.print("NO");
}

// Driver Code
public static void main(String[] args)
{
    int N = 30;
    checkDistinctPrime(N);
}
}

// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach

# Function that returns the all the
# distinct prime factors in a vector
def primeFactors(n) :

    Prime = [];

    # If n is divisible by 2
    if (n % 2 == 0) :
        Prime.append(2);

    # Divide n till all factors of 2
    while (n % 2 == 0) :
        n = n // 2;

    # Check for the prime numbers other
    # than 2
    for i in range(3, int(n ** (1/2)),2) :

        # Store i in Prime[] i is a
        # factor of n
        if (n % i == 0) :
            Prime.append(i);

        # Divide n till all factors of i
        while (n % i == 0) :
            n = n // i;

    # If n is greater than 2, then n is
    # prime number after n divided by
    # all factors
    if (n > 2) :
        Prime.append(n);

    # Returns the vector Prime
    return Prime;

# Function that check whether N is the
# product of distinct prime factors
# or not
def checkDistinctPrime(n) :

    # Returns the vector to store
    # all the distinct prime factors
    Prime = primeFactors(n);

    # To find the product of all
    # distinct prime factors
    product = 1;

    # Find the product
    for i in Prime :
        product *= i;

    # If product is equals to N,
    # print YES, else print NO
    if (product == n) :
        print("YES");
    else :
        print("NO");

# Driver Code
if __name__ == "__main__" :

    N = 30;
    checkDistinctPrime(N);

# This code is contributed by Yash_R

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function that returns the all the
// distinct prime factors in a vector
static List<int> primeFactors(int n)
{
    int i;
    List<int> Prime = new List<int>();

    // If n is divisible by 2
    if (n % 2 == 0) {
        Prime.Add(2);
    }

    // Divide n till all factors of 2
    while (n % 2 == 0) {
        n = n / 2;
    }

    // Check for the prime numbers other
    // than 2
    for (i = 3; i <= Math.Sqrt(n); i = i + 2) {

        // Store i in Prime[] i is a
        // factor of n
        if (n % i == 0) {
            Prime.Add(i);
        }

        // Divide n till all factors of i
        while (n % i == 0) {
            n = n / i;
        }
    }

    // If n is greater than 2, then n is
    // prime number after n divided by
    // all factors
    if (n > 2) {
        Prime.Add(n);
    }

    // Returns the vector Prime
    return Prime;
}

// Function that check whether N is the
// product of distinct prime factors
// or not
static void checkDistinctPrime(int n)
{
    // Returns the vector to store
    // all the distinct prime factors
    List<int> Prime = primeFactors(n);

    // To find the product of all
    // distinct prime factors
    int product = 1;

    // Find the product
    foreach (int i in Prime) {
        product *= i;
    }

    // If product is equals to N,
    // print YES, else print NO
    if (product == n)
        Console.Write("YES");
    else
        Console.Write("NO");
}

// Driver Code
public static void Main(String[] args)
{
    int N = 30;
    checkDistinctPrime(N);
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>

// JavaScript program for the above approach


// Function that returns the all the
// distinct prime factors in a vector
function primeFactors(n)
{
    let i, j;
    let Prime = new Array();

    // If n is divisible by 2
    if (n % 2 == 0) {
        Prime.push(2);
    }

    // Divide n till all factors of 2
    while (n % 2 == 0) {
        n = n / 2;
    }

    // Check for the prime numbers other
    // than 2
    for (i = 3; i <= Math.sqrt(n); i = i + 2) {

        // Store i in Prime[] i is a
        // factor of n
        if (n % i == 0) {
            Prime.push(i);
        }

        // Divide n till all factors of i
        while (n % i == 0) {
            n = n / i;
        }
    }

    // If n is greater than 2, then n is
    // prime number after n divided by
    // all factors
    if (n > 2) {
        Prime.push(n);
    }

    // Returns the vector Prime
    return Prime;
}

// Function that check whether N is the
// product of distinct prime factors
// or not
function checkDistinctPrime(n)
{
    // Returns the vector to store
    // all the distinct prime factors
    let Prime = primeFactors(n);

    // To find the product of all
    // distinct prime factors
    let product = 1;

    // Find the product
    for (let i of Prime) {
        product *= i;
    }

    // If product is equals to N,
    // print YES, else print NO
    if (product == n)
        document.write("YES");
    else
        document.write("NO");
}

// Driver Code

let N = 30;
checkDistinctPrime(N);


// This code is contributed by gfgking

</script>

Output:

YES

Time Complexity: O(N*log(log N)), where N is the given number.

Auxiliary Space: O(sqrt(n))