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 Nclass 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 Nclass 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 Eratosthenesclass 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
