Print all prime numbers less than or equal to N (original) (raw)
Last Updated : 11 Jul, 2025
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 TrueFunction 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
JavaScript
`
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
PHP
`
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))