Finding divisors (with.Java)

This is an article about the problem of “finding divisors.”

As I solve coding test problems, I look back on the problems I solved and look into different solution methods to learn more.

Let’s look at the problem first.

problem

When the integer n is given as a parameter, complete the solution function to return an array containing the divisors of n in ascending order.

Restrictions

• 1 ≤ n ≤ 10,000

Input/Output Example

My solution to the problem

class Solution {
     public int[] solution(int n) {
         int cnt = 0;
         for(int i = 1; i <= n; i++){
             if(n % i == 0){
                 cnt++;
             }
         }
         int[] answer = new int[cnt];
         int idx = 0;
         for(int i = 1; i <= n; i++){
             if(n % i == 0){
                 answer[idx++] = i;
             }
         }
         return answer;
     }
}

Solution explanation

• Declare the cnt variable and initialize it to 0. This variable is used to count the number of divisors.
• Execute a for statement that repeats from 1 to n. The variable i has values from 1 to n. If the remainder of dividing n by i is 0, that is, if i is a divisor of n, Increase the cnt value by 1.
• Declare an array answer of cnt length.
• Declare the idx variable and initialize it to 0. This variable points to an index in the answers array.
• Execute a for statement that repeats from 1 to n. The variable i has values from 1 to n. If the remainder of dividing n by i is 0, that is, if i is a divisor of n,
• Assign the value i to the idx index of the answer array and increase the idx value by 1.
• Returns an array of answers.