How to return different values for different sips (with.Java)

In this article, we learned about How to return different values based on sips (with.Java)

We’re going to learn as we go along by solving coding test problems, reflecting on the problems we solved, and exploring other ways to solve them. Let’s start with the problem.

Problem

Given a positive integer n as a parameter, write a solution function that returns the sum of all positive integers less than or equal to n if n is odd, or the sum of the squares of all positive integers less than or equal to n if n is even.

Example input and output

n: 7 result: 16

This means that n is odd by 7. All positive odd numbers below 7 are 1, 3, 5, and 7, and their sum, 1 + 3 + 5 + 7 = 16, should be returned.

My solution to the problem

import java.util.*;

class Solution {
    public int solution(int n) {
        int answer = 0;
        int[] newArray = new int[n/2];
        if(n % 2 == 0){
            Arrays.setAll(newArray, even -> (int)Math.pow((even+1)*2,2));
            answer = Arrays.stream(newArray).sum();
        } else {
            Arrays.setAll(newArray, odd -> (odd+2)*2-1);
            answer = Arrays.stream(newArray).sum()+1;
        }
        } return answer;
    }
}

Solution

Specify the size of the array by dividing newArray in half. To determine even and odd numbers, we utilize the Arrays.setAll function to reset the values of the elements in newArray. For even numbers, we need to square the values, so we do that with the Math.pow() function. We also cast Math.pow() to int since its default type is double. We then reassign the sum of the elements to the answer variable with Arrays.stream(arr).sum(). For odd numbers, I implemented similar logic as for even numbers, and for sum()+1, I added it because I didn’t want the array to contain any 1s.

I solved the problem with an array as shown above, but there’s actually a simpler way. That’s right, you can do the math through a loop instead of controlling it with an array!

How to do math through a loop

class Solution {
    public long solution(int n) {
        long answer = 0;

        if (n % 2 == 1) {
            for (int i = 1; i <= n; i += 2) {
                answer += i;
            }
        } else {
            for (int i = 2; i <= n; i += 2) {
                answer += (long) i * i;
            }
        }
        } return answer;
    }
}