Print all prime numbers less than or equal to N
Given a number N, the task is to print all prime numbers less than or equal to N.
Examples:
Input: 7 Output: 2, 3, 5, 7 Input: 13 Output: 2, 3, 5, 7, 11, 13
Naive Approach: Iterate from 2 to N, and check for prime. If it is a prime number, print the number.
Below is the implementation of the above approach:
C++
// C++ program to print all primes less than N #include <bits/stdc++.h> using namespace std; // function check whether a number is prime or not bool isPrime(int n) { // Corner case if (n <= 1) return false; // Check from 2 to n-1 for (int i = 2; i < n; i++) if (n % i == 0) return false; return true; } // Function to print primes void printPrime(int n) { for (int i = 2; i <= n; i++) if (isPrime(i)) cout << i << " "; } // Driver Code int main() { int n = 7; printPrime(n); } |
C
// C program to print all primes less than N #include <stdbool.h> #include <stdio.h> // function check whether a number is prime or not bool isPrime(int n) { // Corner case if (n <= 1) return false; // Check from 2 to n-1 for (int i = 2; i < n; i++) if (n % i == 0) return false; return true; } // Function to print primes void printPrime(int n) { for (int i = 2; i <= n; i++) if (isPrime(i)) printf("%d ", i); } // Driver Code int main() { int n = 7; printPrime(n); } // This code is contributed by Sania Kumari Gupta |
Java
// Java program to print // all primes less than N class GFG { // function check whether // a number is prime or not static boolean isPrime(int n) { // Corner case if (n <= 1) return false; // Check from 2 to n-1 for (int i = 2; i < n; i++) if (n % i == 0) return false; return true; } // Function to print primes static void printPrime(int n) { for (int i = 2; i <= n; i++) { if (isPrime(i)) System.out.print(i + " "); } } // Driver Code public static void main(String[] args) { int n = 7; printPrime(n); } } // This code is contributed // by ChitraNayal |
Python3
# Python3 program to print # all primes less than N # Function to check whether # a number is prime or not . def isPrime(n): # Corner case if n <= 1 : return False # check from 2 to n-1 for i in range(2, n): if n % i == 0: return False return True # Function to print primes def printPrime(n): for i in range(2, n + 1): if isPrime(i): print(i, end = " ") # Driver code if __name__ == "__main__" : n = 7 # function calling printPrime(n) # This code is contributed # by Ankit Rai |
C#
// C# program to print // all primes less than N using System; class GFG { // function check whether // a number is prime or not static bool isPrime(int n) { // Corner case if (n <= 1) return false; // Check from 2 to n-1 for (int i = 2; i < n; i++) if (n % i == 0) return false; return true; } // Function to print primes static void printPrime(int n) { for (int i = 2; i <= n; i++) { if (isPrime(i)) Console.Write(i + " "); } } // Driver Code public static void Main() { int n = 7; printPrime(n); } } // This code is contributed // by ChitraNayal |
PHP
<?php // PHP program to print // all primes less than N // function check whether // a number is prime or not function isPrime($n) { // Corner case if ($n <= 1) return false; // Check from 2 to n-1 for ($i = 2; $i < $n; $i++) if ($n % $i == 0) return false; return true; } // Function to print primes function printPrime($n) { for ($i = 2; $i <= $n; $i++) { if (isPrime($i)) echo $i . " "; } } // Driver Code $n = 7; printPrime($n); // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // Javascript program to print all primes // less than N // function check whether a number // is prime or not function isPrime(n) { // Corner case if (n <= 1) return false; // Check from 2 to n-1 for (let i = 2; i < n; i++) if (n % i == 0) return false; return true; } // Function to print primes function printPrime(n) { for (let i = 2; i <= n; i++) { if (isPrime(i)) document.write(i +" "); } } // Driver Code let n = 7; printPrime(n); // This code is contributed by Mayank Tyagi </script> |
Output:
2 3 5 7
Time Complexity: O(N * N)
Auxiliary Space: O(1)
A better approach is based on the fact that one of the divisors must be smaller than or equal to ?n. So we check for divisibility only till ?n.
C++
// C++ program to print all primes // less than N #include <bits/stdc++.h> using namespace std; // function check whether a number // is prime or not bool isPrime(int n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; } // Function to print primes void printPrime(int n) { for (int i = 2; i <= n; i++) { if (isPrime(i)) cout << i << " "; } } // Driver Code int main() { int n = 7; printPrime(n); } |
Java
// Java program to print // all primes less than N import java.io.*; class GFG { // function check // whether a number // is prime or not static boolean isPrime(int n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so // that we can skip // middle five numbers // in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; } // Function to print primes static void printPrime(int n) { for (int i = 2; i <= n; i++) { if (isPrime(i)) System.out.print(i + " "); } } // Driver Code public static void main (String[] args) { int n = 7; printPrime(n); } } // This code is contributed // by anuj_67. |
C#
// C# program to print // all primes less than N using System; class GFG { // function check // whether a number // is prime or not static bool isPrime(int n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so // that we can skip // middle five numbers // in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; } // Function to print primes static void printPrime(int n) { for (int i = 2; i <= n; i++) { if (isPrime(i)) Console.Write(i + " "); } } // Driver Code public static void Main () { int n = 7; printPrime(n); } } // This code is contributed // by ChitraNayal |
Python3
# function to check if the number is # prime or not def isPrime(n) : # Corner cases if (n <= 1) : return False if (n <= 3) : return True # This is checked so that we can skip # middle five numbers in below loop if (n % 2 == 0 or n % 3 == 0) : return False i = 5 while(i * i <= n) : if (n % i == 0 or n % (i + 2) == 0) : return False i = i + 6 return True # print all prime numbers # less than equal to N def printPrime(n): for i in range(2, n + 1): if isPrime(i): print (i, end =" ") n = 7 printPrime(n) |
Javascript
<script> // Javascript program to print all primes // less than N // function check whether a number // is prime or not function isPrime(n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (let i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; } // Function to print primes function printPrime(n) { for (let i = 2; i <= n; i++) { if (isPrime(i)) document.write(i + " "); } } // Driver Code let n = 7; printPrime(n); // This code is contributed by subhammahato348. </script> |
PHP
<?php // PHP program to print // all primes less than N // function check whether // a number is prime or not function isPrime($n) { // Corner cases if ($n <= 1) return false; if ($n <= 3) return true; // This is checked so that // we can skip middle five // numbers in below loop if ($n % 2 == 0 || $n % 3 == 0) return false; for ($i = 5; $i * $i <= $n; $i = $i + 6) if ($n % $i == 0 || $n % ($i + 2) == 0) return false; return true; } // Function to print primes function printPrime($n) { for ($i = 2; $i <= $n; $i++) { if (isPrime($i)) echo $i . " "; } } // Driver Code $n = 7; printPrime($n); // This code is contributed // by ChitraNayal ?> |
Output:
2 3 5 7
Time Complexity: O(N3/2)
Auxiliary Space: O(1)
The best solution is to use Sieve of Eratosthenes. The time complexity is O(N * loglog(N))
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