k-Rough Number or k-Jagged Number

A k-rough or k-jagged number is a number whose smallest prime factor is greater than or equal to the number ‘k’. Given numbers ‘n’ and ‘k’ as input, we are required to find whether ‘n; is a k-rough number or not.

Examples :

Input : n = 10, k = 2
Output : 10 is a 2-rough number
Explanation: The prime factors of 10 are 2 and 5 Hence its smallest prime factor is 2 which is greater than or equal to k, i.e 2

Input : n = 55, k = 7
Output : 55 is not a 7-rough number
Explanation: The prime factors of 55 are 5 and 11 Hence its smallest prime factor is 5 which is not greater than or equal to k, i.e 7

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

It may be inferred from above that every positive integer, except 1, is a 2-rough number since their smallest prime factor is either 2(for even positive integers) or greater than 2(for odd positive integers).

Simple Approach

First find all prime numbers up to the number ‘n’.
Next find the smallest prime factor of the number ‘n’ from its prime factorization.
Check if the smallest prime factor is greater than or equal to ‘k’ or not.

C++

// CPP to check if n is a k-rough number 
// or not 
#include <bits/stdc++.h> 
using namespace std; 
  
// Finds primes by Sieve of Eratosthenes 
// method 
vector<int> getPrimes(int n) 
{ 
    int i, j; 
    bool isPrime[n + 1]; 
    memset(isPrime, true, sizeof(isPrime)); 
    for (i = 2; i * i <= n; i++) 
    { 
  
        // If isPrime[i] is not changed, 
        // then it is prime 
        if (isPrime[i] == true) 
        { 
            // Update all multiples of p 
            for (j = i * 2; j <= n; j += i) 
                isPrime[j] = false; 
        } 
    } 
  
    // Forming array of the prime numbers found 
    vector<int> primes; 
    for (i = 2; i <= n; i++) 
        if (isPrime[i]) 
            primes.push_back(i); 
    return primes; 
} 
  
// Checking whether a number is k-rough or not 
bool isRough(int n, int k) 
{ 
    vector<int> primes = getPrimes(n); 
  
    // Finding minimum prime factor of n 
    int min_pf = n; 
    for (int i = 0; i < primes.size(); i++) 
        if (n % primes[i] == 0) 
            min_pf = primes[i]; 
  
    // Return true if minimum prime factor 
    // is greater than or equal to k. Else 
    // return false. 
    return (min_pf >= k); 
} 
  
// Driver Method 
int main() 
{ 
    int n = 75, k = 3; 
    if (isRough(n, k)) 
        cout << n << " is a "
            << k << "-rough number\n"; 
    else
        cout << n << " is not a "
            << k << "-rough number\n"; 
    return 0; 
} 

Java

// Java to check if n is  
// a k-rough number or not 
import java.io.*; 
import java.util.*; 
  
class GFG  
{ 
      
    // Finds primes by Sieve  
    // of Eratosthenes method 
    static ArrayList<Integer> getPrimes(int n) 
    { 
        int i, j; 
        int []isPrime = new int[n + 1]; 
          
        for(i = 0; i < n + 1; i++) 
            isPrime[i] = 1; 
          
        for (i = 2; i * i <= n; i++) 
        {      
            // If isPrime[i] is not  
            // changed, then it is prime 
            if (isPrime[i] == 1) 
            { 
                // Update all  
                // multiples of p 
                for (j = i * 2; j <= n; j += i) 
                    isPrime[j] = 0; 
            } 
        } 
      
        // Forming array of the  
        // prime numbers found 
        ArrayList<Integer> primes = new ArrayList<Integer>(); 
        for (i = 2; i <= n; i++) 
            if (isPrime[i] == 1) 
                primes.add(i); 
        return primes; 
    } 
      
    // Checking whether a  
    // number is k-rough or not 
    static boolean isRough(int n, int k) 
    { 
        ArrayList<Integer> primes = getPrimes(n); 
          
        // Finding minimum 
        // prime factor of n 
        int min_pf = n; 
        for (int i = 0; i < primes.size(); i++) 
            if (n % primes.get(i) == 0) 
                min_pf = primes.get(i); 
      
        // Return true if minimum  
        // prime factor is greater 
        // than or equal to k. Else 
        // return false. 
        return (min_pf >= k); 
    } 
      
    // Driver Code 
    public static void main(String args[]) 
    { 
        int n = 75, k = 3; 
        if (isRough(n, k)) 
            System.out.print(n + " is a " + k + 
                            "-rough number\n"); 
        else
            System.out.print(+ n + " is not a " + 
                          k + "-rough number\n"); 
    } 
} 
// This code is contributed by  
// Manish Shaw.(manishshaw1) 

C#

// C# to check if n is a k-rough number 
// or not 
using System; 
using System.Collections.Generic; 
using System.Linq; 
  
class GFG { 
      
    // Finds primes by Sieve of Eratosthenes 
    // method 
    static List<int> getPrimes(int n) 
    { 
        int i, j; 
        int []isPrime = new int[n + 1]; 
          
        for(i = 0; i < n + 1; i++) 
            isPrime[i] = 1; 
        // Array.Clear(isPrime, 1, isPrime.Length); 
          
        for (i = 2; i * i <= n; i++) 
        { 
      
            // If isPrime[i] is not changed, 
            // then it is prime 
            if (isPrime[i] == 1) 
            { 
                // Update all multiples of p 
                for (j = i * 2; j <= n; j += i) 
                    isPrime[j] = 0; 
            } 
        } 
      
        // Forming array of the prime numbers  
        // found 
        List<int> primes = new List<int>(); 
        for (i = 2; i <= n; i++) 
            if (isPrime[i] == 1) 
                primes.Add(i); 
        return primes; 
    } 
      
    // Checking whether a number is k-rough 
    // or not 
    static bool isRough(int n, int k) 
    { 
        List<int> primes = getPrimes(n); 
          
        // Finding minimum prime factor of n 
        int min_pf = n; 
        for (int i = 0; i < primes.Count; i++) 
            if (n % primes[i] == 0) 
                min_pf = primes[i]; 
      
        // Return true if minimum prime factor 
        // is greater than or equal to k. Else 
        // return false. 
        return (min_pf >= k); 
    } 
      
    // Driver Method 
    static void Main() 
    { 
        int n = 75, k = 3; 
        if (isRough(n, k)) 
            Console.Write(n + " is a " + k + 
                         "-rough number\n"); 
        else
            Console.Write(+ n + " is not a " 
                   + k + "-rough number\n"); 
    } 
} 
  
// This code is contributed by manishshaw. 

PHP

<?php 
// PHP to check if n is  
// a k-rough number or not 
  
// Finds primes by Sieve  
// of Eratosthenes method 
function getPrimes($n) 
{ 
    $isPrime = array_fill(0, ($n + 1), true); 
    for ($i = 2;  
         $i * $i <= $n; $i++) 
    { 
  
        // If isPrime[i] is  
        // not changed, 
        // then it is prime 
        if ($isPrime[$i] == true) 
        { 
            // Update all 
            // multiples of p 
            for ($j = $i * 2; 
                 $j <= $n; $j += $i) 
                $isPrime[$j] = false; 
        } 
    } 
  
    // Forming array of the 
    // prime numbers found 
    $primes = array(); 
    $k = 0; 
    for ($i = 2; $i <= $n; $i++) 
        if ($isPrime[$i]) 
            $primes[$k++]=$i; 
    return $primes; 
} 
  
// Checking whether a  
// number is k-rough or not 
function isRough($n, $k) 
{ 
    $primes = getPrimes($n); 
  
    // Finding minimum  
    // prime factor of n 
    $min_pf = $n; 
    for ($i = 0;  
         $i < count($primes); $i++) 
        if ($n % $primes[$i] == 0) 
            $min_pf = $primes[$i]; 
  
    // Return true if minimum  
    // prime factor is greater 
    // than or equal to k. Else 
    // return false. 
    return ($min_pf >= $k); 
} 
  
// Driver Code 
$n = 75;  
$k = 3; 
if (isRough($n, $k)) 
    echo $n . " is a " . 
         $k . "-rough number\n"; 
else
    echo $n . " is not a " . 
         $k . "-rough number\n"; 
           
// This code is contributed by mits 
?> 

Output:

75 is a 3-rough number

Efficient Solution :

The idea is based on Efficient program to print all prime factors of a given number.

If n is divisible by 2 (smallest prime number), then we return true if k is smaller than or equal to 2. Else we return false.
Then we one by one try all odd numbers. As soon as we find an odd number that divides n, we compare it with k and return true if the odd number is greater than or equal to k, else false. This solution works because if a prime number does not divide n, then its multiples will also not divide.

C++

// CPP program to check if given n is 
// k-rough or not. 
# include <bits/stdc++.h> 
using namespace std; 
  
// Returns true if n is k rough else false 
bool isKRough(int n, int k) 
{ 
    // If n is even, then smallest prime 
    // factor becomes 2. 
    if (n % 2 == 0) 
        return (k <= 2); 
  
    // n must be odd at this point.  So we 
    // can skip one element (Note i = i +2) 
    for (int i = 3; i*i <= n; i = i+2) 
        if (n%i == 0) 
            return (i >= k); 
  
   return (n >= k); 
} 
  
/* Driver program to test above function */
int main() 
{ 
    int n = 75, k = 3; 
    if (isKRough(n, k)) 
        cout << n << " is a "
             << k << "-rough number\n"; 
    else
        cout << n << " is not a "
             << k << "-rough number\n"; 
    return 0; 
} 

Java

// Java program to check if given n is 
// k-rough or not. 
  
class GFG { 
      
// Returns true if n is k rough else false 
static boolean isKRough(int n, int k) 
{ 
    // If n is even, then smallest  
    // prime factor becomes 2. 
    if (n % 2 == 0) 
        return (k <= 2); 
  
    // n must be odd at this point. So we 
    // can skip one element (Note i = i + 2) 
    for (int i = 3; i*i <= n; i = i + 2) 
        if (n % i == 0) 
            return (i >= k); 
  
return (n >= k); 
} 
  
/* Driver program to test above function */
public static void main(String[] args) 
{ 
    int n = 75, k = 3; 
    if (isKRough(n, k)) 
        System.out.println(n + " is a " + 
                     k + "-rough number"); 
    else
        System.out.println(n + " is not a " + 
                        k + "-rough number"); 
} 
} 
  
// This code is contributed by Smitha 

C#

// C# program to check if given n is 
// k-rough or not. 
using System; 
  
class GFG { 
      
// Returns true if n is k rough else false 
static bool isKRough(int n, int k) 
{ 
    // If n is even, then smallest prime 
    // factor becomes 2. 
    if (n % 2 == 0) 
        return (k <= 2); 
  
    // n must be odd at this point. So we 
    // can skip one element (Note i = i + 2) 
    for (int i = 3; i*i <= n; i = i + 2) 
        if (n % i == 0) 
            return (i >= k); 
  
return (n >= k); 
} 
  
/* Driver program to test above function */
public static void Main() 
{ 
    int n = 75, k = 3; 
    if (isKRough(n, k)) 
        Console.Write(n + " is a " + 
             k + "-rough number\n"); 
    else
        Console.Write(n + " is not a " + 
                 k + "-rough number\n"); 
} 
} 
  
// This code is contributed by Smitha 

PHP

<?php 
// PHP program to check if  
// given n is k-rough or not. 
  
// Returns true if n is k  
// rough else false 
function isKRough($n, $k) 
{ 
    // If n is even, then smallest  
    // prime factor becomes 2. 
    if ($n % 2 == 0) 
        return ($k <= 2); 
  
    // n must be odd at this point.  
    // So we can skip one element 
    // (Note i = i +2) 
    for ($i = 3; $i * $i <= $n; $i = $i + 2) 
        if ($n % $i == 0) 
            return ($i >= $k); 
  
return ($n >= $k); 
} 
  
// Driver Code 
$n = 75; 
$k = 3; 
if (isKRough($n, $k)) 
    echo $n . " is a " . $k .  
           "-rough number\n"; 
else
    echo $n . " is not a " . $k .  
               "-rough number\n";  
      
// This code is contributed by Sam007 
?> 

Output :

75 is a 3-rough number

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SaagnikAdhikary

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Improved By : Smitha Dinesh Semwal, Sam007, manishshaw1, Mithun Kumar

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