Given an **integer N**, the task is to check if it can be expressed as a product of exactly **K** prime divisors. **Examples:**

Input:N = 12, K = 3Output:YesExplanation:12 can be expressed as product of 2×2×3.Input:N = 14, K = 3Output:NoExplanation:14 can be only expressed as product of 2×7.

**Approach:**

To solve the problem mentioned above we are given the value N and we will find the **maximum number of values we can split N into**. We can represent prime factorization of **N** as where *p _{i}* are the prime factors of

**N**and

*a*are the exponents. We know that total number of divisors of

_{i}**N**is . Therefore, we can observe that we have to check whether it is possible to represent

**N**as product of

**K**numbers or not. If the maximum split is less than K then it is not possible to express it in exactly K prime divisors, else it is always possible.

## C++

`// CPP implementation to Check if a` `// number can be expressed as a` `// product of exactly K prime divisors` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// function to find K prime divisors` `void` `KPrimeDivisors(` `int` `N, ` `int` `K)` `{` ` ` `int` `maximum_split = 0;` ` ` `// count number of 2s that divide N` ` ` `while` `(N % 2 == 0) {` ` ` `maximum_split++;` ` ` `N /= 2;` ` ` `}` ` ` `// N must be odd at this point.` ` ` `// So we can skip one element` ` ` `for` `(` `int` `i = 3; i * i <= N; i = i + 2) {` ` ` `while` `(N % i == 0) {` ` ` `// divide the value of N` ` ` `N = N / i;` ` ` `// increment count` ` ` `maximum_split++;` ` ` `}` ` ` `}` ` ` `// Condition to handle the case when n` ` ` `// is a prime number greater than 2` ` ` `if` `(N > 2)` ` ` `maximum_split++;` ` ` `// check if maximum_split is less than K` ` ` `// then it not possible` ` ` `if` `(maximum_split < K) {` ` ` `printf` `(` `"No\n"` `);` ` ` `return` `;` ` ` `}` ` ` `printf` `(` `"Yes\n"` `);` `}` `/* Driver code */` `int` `main()` `{` ` ` `// initialise N and K` ` ` `int` `N = 12;` ` ` `int` `K = 3;` ` ` `KPrimeDivisors(N, K);` ` ` `return` `0;` `}` |

## Java

`// Java implementation to Check if a` `// number can be expressed as a` `// product of exactly K prime divisors` `class` `GFG {` ` ` ` ` `// function to find K prime divisors` ` ` `static` `void` `KPrimeDivisors(` `int` `N, ` `int` `K)` ` ` `{` ` ` `int` `maximum_split = ` `0` `;` ` ` ` ` `// count number of 2s that divide N` ` ` `while` `(N % ` `2` `== ` `0` `) {` ` ` `maximum_split++;` ` ` `N /= ` `2` `;` ` ` `}` ` ` ` ` `// N must be odd at this point.` ` ` `// So we can skip one element` ` ` `for` `(` `int` `i = ` `3` `; i * i <= N; i = i + ` `2` `) {` ` ` ` ` `while` `(N % i == ` `0` `) {` ` ` `// divide the value of N` ` ` `N = N / i;` ` ` ` ` `// increment count` ` ` `maximum_split++;` ` ` `}` ` ` `}` ` ` ` ` `// Condition to handle the case when n` ` ` `// is a prime number greater than 2` ` ` `if` `(N > ` `2` `)` ` ` `maximum_split++;` ` ` ` ` `// check if maximum_split is less than K` ` ` `// then it not possible` ` ` `if` `(maximum_split < K) {` ` ` `System.out.println(` `"No"` `);` ` ` `return` `;` ` ` `}` ` ` ` ` `System.out.println(` `"Yes"` `);` ` ` `}` ` ` ` ` `/* Driver code */` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `// initialise N and K` ` ` `int` `N = ` `12` `;` ` ` `int` `K = ` `3` `;` ` ` ` ` `KPrimeDivisors(N, K);` ` ` `}` `}` `// This code is contributed by Yash_R` |

## Python3

`# Python implementation to Check if a` `# number can be expressed as a` `# product of exactly K prime divisors` `import` `math as mt` `# function to find K prime divisors` `def` `KPrimeDivisors(n, k):` ` ` ` ` `# To count maximum split of N` ` ` `maximum_split ` `=` `0` ` ` ` ` `# count number of 2s that divide N` ` ` `while` `n ` `%` `2` `=` `=` `0` `:` ` ` `maximum_split` `+` `=` `1` ` ` `n ` `=` `n ` `/` `/` `2` ` ` ` ` `# n must be odd at this point` ` ` `# so we skip one element` ` ` `for` `i ` `in` `range` `(` `3` `, mt.ceil(mt.sqrt(n)), ` `2` `):` ` ` `while` `n ` `%` `i ` `=` `=` `0` `:` ` ` `n ` `=` `n ` `/` `i;` ` ` `maximum_split` `+` `=` `1` ` ` ` ` `# Condition to handle the case when n` ` ` `# is a prime number greater than 2` ` ` `if` `n > ` `2` `:` ` ` `maximum_split` `+` `=` `1` ` ` ` ` `# check if maximum_split is less than K` ` ` `# then it not possible` ` ` `if` `maximum_split < k:` ` ` `print` `(` `"No"` `)` ` ` `return` ` ` `print` `(` `"Yes"` `)` ` ` ` ` `# Driver code` `N ` `=` `12` `K ` `=` `3` `KPrimeDivisors(N, K)` |

## C#

`// C# implementation to Check if a` `// number can be expressed as a` `// product of exactly K prime divisors` `using` `System;` `class` `GFG {` ` ` ` ` `// function to find K prime divisors` ` ` `static` `void` `KPrimeDivisors(` `int` `N, ` `int` `K)` ` ` `{` ` ` `int` `maximum_split = 0;` ` ` ` ` `// count number of 2s that divide N` ` ` `while` `(N % 2 == 0) {` ` ` `maximum_split++;` ` ` `N /= 2;` ` ` `}` ` ` ` ` `// N must be odd at this point.` ` ` `// So we can skip one element` ` ` `for` `(` `int` `i = 3; i * i <= N; i = i + 2) {` ` ` ` ` `while` `(N % i == 0) {` ` ` `// divide the value of N` ` ` `N = N / i;` ` ` ` ` `// increment count` ` ` `maximum_split++;` ` ` `}` ` ` `}` ` ` ` ` `// Condition to handle the case when n` ` ` `// is a prime number greater than 2` ` ` `if` `(N > 2)` ` ` `maximum_split++;` ` ` ` ` `// check if maximum_split is less than K` ` ` `// then it not possible` ` ` `if` `(maximum_split < K) {` ` ` `Console.WriteLine(` `"No"` `);` ` ` `return` `;` ` ` `}` ` ` ` ` `Console.WriteLine(` `"Yes"` `);` ` ` `}` ` ` ` ` `/* Driver code */` ` ` `public` `static` `void` `Main(String[] args)` ` ` `{` ` ` `// initialise N and K` ` ` `int` `N = 12;` ` ` `int` `K = 3;` ` ` ` ` `KPrimeDivisors(N, K);` ` ` `}` `}` `// This code is contributed by 29AjayKumar` |

## Javascript

`<script>` `// javascript implementation to Check if a` `// number can be expressed as a` `// product of exactly K prime divisors ` `// function to find K prime divisors` ` ` `function` `KPrimeDivisors(N , K)` ` ` `{` ` ` `var` `maximum_split = 0;` ` ` `// count number of 2s that divide N` ` ` `while` `(N % 2 == 0)` ` ` `{` ` ` `maximum_split++;` ` ` `N /= 2;` ` ` `}` ` ` `// N must be odd at this point.` ` ` `// So we can skip one element` ` ` `for` `(i = 3; i * i <= N; i = i + 2)` ` ` `{` ` ` `while` `(N % i == 0)` ` ` `{` ` ` ` ` `// divide the value of N` ` ` `N = N / i;` ` ` `// increment count` ` ` `maximum_split++;` ` ` `}` ` ` `}` ` ` `// Condition to handle the case when n` ` ` `// is a prime number greater than 2` ` ` `if` `(N > 2)` ` ` `maximum_split++;` ` ` `// check if maximum_split is less than K` ` ` `// then it not possible` ` ` `if` `(maximum_split < K)` ` ` `{` ` ` `document.write(` `"No"` `);` ` ` `return` `;` ` ` `}` ` ` `document.write(` `"Yes"` `);` ` ` `}` ` ` `/* Driver code */` ` ` ` ` `// initialise N and K` ` ` `var` `N = 12;` ` ` `var` `K = 3;` ` ` `KPrimeDivisors(N, K);` `// This code is contributed by gauravrajput1.` `</script>` |

**Output:**

Yes

**Time Complexity:** O(sqrt(N))

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