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:
- Firstly, consider the given number N as input.
- Use Sieve of Eratosthenes.
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