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Java Program to Display All Prime Numbers from 1 to N

Last Updated : 02 Jul, 2024
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For a given number N, the purpose is to find all the prime numbers from 1 to N.

Examples:  

Input: N = 11
Output: 2, 3, 5, 7, 11
Input: N = 7
Output: 2, 3, 5, 7

Approach 1:

  • Firstly, consider the given number N as input.
  • Then apply a for loop in order to iterate the numbers from 1 to N.
  • At last, check if each number is a prime number and if it's a prime number then print it using brute-force method.
Java
// Java program to find all the // prime numbers from 1 to N class gfg {  // Function to print all the  // prime numbers till N  static void prime_N(int N)  {  // Declaring the variables  int x, y, flg;  // Printing display message  System.out.println(  "All the Prime numbers within 1 and " + N  + " are:");  // Using for loop for traversing all  // the numbers from 1 to N  for (x = 1; x <= N; x++) {  // Omit 0 and 1 as they are  // neither prime nor composite  if (x == 1 || x == 0)  continue;  // Using flag variable to check  // if x is prime or not  flg = 1;  for (y = 2; y <= x / 2; ++y) {  if (x % y == 0) {  flg = 0;  break;  }  }  // If flag is 1 then x is prime but  // if flag is 0 then x is not prime  if (flg == 1)  System.out.print(x + " ");  }  }  // The Driver code  public static void main(String[] args)  {  int N = 45;  prime_N(N);  } } 

Output
All the Prime numbers within 1 and 45 are: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 

Time Complexity: O(N2)

Auxiliary Space: O(1)

Approach 2:

  • Firstly, consider the given number N as input.
  • Then apply a for loop in order to iterate the numbers from 1 to N.
  • At last, check if each number is a prime number and if it's a prime number then print it using the square root method.
Java
// Java program to find all the // prime numbers from 1 to N class gfg {  // Function to print all the  // prime numbers till N  static void prime_N(int N)  {  // Declaring the variables  int x, y, flg;  // Printing display message  System.out.println(  "All the Prime numbers within 1 and " + N  + " are:");  // Using for loop for traversing all  // the numbers from 1 to N  for (x = 2; x <= N; x++) {  // Using flag variable to check  // if x is prime or not  flg = 1;  for (y = 2; y * y <= x; y++) {  if (x % y == 0) {  flg = 0;  break;  }  }  // If flag is 1 then x is prime but  // if flag is 0 then x is not prime  if (flg == 1)  System.out.print(x + " ");  }  }  // The Driver code  public static void main(String[] args)  {  int N = 45;  prime_N(N);  } } 

 
 


Output
All the Prime numbers within 1 and 45 are: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 

Time Complexity: O(N3/2)


 

Approach 3:


 


 

Java
// Java program to print all // primes smaller than or equal to // n using Sieve of Eratosthenes class SieveOfEratosthenes {  void sieveOfEratosthenes(int n)  {  // Create a boolean array  // "prime[0..n]" and  // initialize all entries  // it as true. A value in  // prime[i] will finally be  // false if i is Not a  // prime, else true.  boolean prime[] = new boolean[n + 1];  for (int i = 0; i <= n; i++)  prime[i] = true;  for (int p = 2; p * p <= n; p++) {  // If prime[p] is not changed, then it is a  // prime  if (prime[p] == true) {  // Update all multiples of p  for (int i = p * p; i <= n; i += p)  prime[i] = false;  }  }  // Print all prime numbers  for (int i = 2; i <= n; i++) {  if (prime[i] == true)  System.out.print(i + " ");  }  }  // Driver Code  public static void main(String args[])  {  int N = 45;  System.out.println(  "All the Prime numbers within 1 and " + N  + " are:");  SieveOfEratosthenes g = new SieveOfEratosthenes();  g.sieveOfEratosthenes(N);  } } 

 
 


Output
All the Prime numbers within 1 and 45 are: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 


 

Time complexity : O(n*log(log(n))) 

Auxiliary space: O(n) as using extra space for array prime


 


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