# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 equals 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 ` `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 greter 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 primeFactors(``int` `n) ` `{ ` `    ``int` `i, j; ` `    ``Vector Prime = ``new` `Vector(); ` `  `  `    ``// 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 greter 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 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 greter 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 greter 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 `

Output:

```YES
```

Time Complexity: O(N*log(log N)), where N is the given number. My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : sapnasingh4991, Yash_R