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 2Input : 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
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.
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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