# Java Program to Display All Prime Numbers from 1 to N

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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