Factors of a numbers

Brute force solution:

public class factors{
    public static void main(String[] args) {
        int n = 20;
        findFactors(n);
    }
    static void findFactors(int n){
        for(int i = 1; i <= n; i++){
            if(n%i == 0){
                System.out.print(i + " ");
            }
        } 
    }
}

Time complexity: O(n)

Space complexity: O(1)

Somewhat optimized:

public class factors{
    public static void main(String[] args) {
        int n = 20;
        findFactors(n);
    }
    static void findFactors(int n){
        for(int i = 1; i <= n/2; i++){
            if(n%i == 0){
                System.out.print(i + " ");
            }
        }
        System.out.print(n + " "); 
    }
}

Time complexity: O(n)

Space complexity: O(1)

Optimized solution:

public class factors{
    public static void main(String[] args) {
        int n = 36;
        findFactors(n);
    }
    static void findFactors(int n){
//create a list to store the 2nd half factors
        ArrayList<Integer> list = new ArrayList<>();

//run the loop root of n times
        for(int i = 1; i*i <= n; i++){
            if(n%i == 0){

//if n/i == i it means that it is the root of n so, to avoid repetition of 
//root print it just 1 time
                if(n/i == i){
                    System.out.print(i + " ");
                }
                else{
                System.out.print(i + " ");
                list.add(n/i);
                }
            }
        }
        //print the second half of factors
        for(int i = list.size()-1; i >= 0; i--){
            System.out.print(list.get(i) + " ");
        }
      
    }
}

Time complexity: O(√n)

Space complexity: O(√n)

Thank you for reading. If you have any queries then, please let me know in the comment section. I will surely be responsive toward it.