I cannot find the problem with my code

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I am a novice programmer so I could be making a big mistake and I just do not see it but this code works for all input scenarios that I have tried and yet the verification still marks it as wrong, any hints would be greatly appreciated.

Use the keyboard to enter three integers. Display the number of positive numbers in the original set.

Here are some examples:
a) if you enter the numbers
-4 6 6
then we display
2

b) if you enter the numbers
-6 -6 -3
then we display
0

c) if you enter the numbers
0 1 2
then we display
2

Requirements:

The program should read the numbers from the keyboard.
The program must display a number on the screen.
The program should display the number of positive numbers in the original set.
If there are no positive numbers, the program should display "0".
Note that 0 is not a positive or negative number.

package en.codegym.task.jdk13.task04.task0428; import java.io.BufferedReader; import java.io.InputStreamReader; /* Positive number */ public class Solution { public static class number{ public static int count = 0; } public static void main(String[] args) throws Exception { //write your code here BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); String num1 = reader.readLine(); String num2 = reader.readLine(); String num3 = reader.readLine(); int aNum = Integer.parseInt(num1); int bNum = Integer.parseInt(num2); int cNum = Integer.parseInt(num3); int[] arr = new int[3]; arr[0] = aNum; arr[1] = bNum; arr[2] = cNum; /*if((aNum <= -1) && (bNum <= -1) && (cNum <= -1)){ System.out.println("0"); } else if((aNum >= 1) && (bNum >= 1) && (cNum >= 1)){ System.out.println("3"); } else{*/ for(int i = 0;i < arr.length;i++){ if((arr[i] >= 1) && (arr[i] != 0)){ number.count = number.count + 1; } else{ System.out.println(); } } System.out.println(number.count); } //} }

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Freeth

Level 6 , Netherlands

13 January 2022, 11:20

useful

Hi, It seems you are making things unnecessarily complicated on yourself. The question can basically be broken down into these components. - Extract three integers - Check if these integers are positive - Count the positive integers - Then display the amount You seem to be doing step 1 good but after that you might want to simplify the code this might not result in an elegant solution but it gets the job done and faccilitates easier understanding. I would suggest just using a 'positiveCount' variable which you print out with the right amount after you finished checking.

AmirMasoud Java Developer

13 January 2022, 06:25

you should also write code to a state that no positive number was found and print 0

Jay

Level 8 , United States

13 January 2022, 03:45

I took a different approach. I only use if statement to check the condition for each input. if the input number is greater then 0, and increment -> "counter variable". and print my statement code to the terminal after I checked all my if statement condition.

Lisa

Level 41

12 January 2022, 18:20

useful

That's a lot of different things you use here. Hmm... I'd first remove the inner class and its static variable and instead use a local counter var. CG sometimes doesn't like the usage of statics. So remove lines 12 -14 and instead add at line 31 int count = 0; change number.count to count (3 times). Then I'd remove lines 45 - 47. CG didn't ask you to output an empty line if the entered number is <= 0. Additional output often makes validation fail. Some more hints for better code line 42, that's a redundant condition if you want to make sure a number is > 0 then just check for number > 0 You do not necessarily need an array and you even do not need to store the number. The important info is just if the entered number is > 0, then increment the counter. That you just need to repeat three times like:

for (int i = 0; i < 3; i++)
    if (Integer.parseInt(reader.readLine()) > 0)
        count++;