CodeGym/Help with Java Tasks/Can someone please point me to a mistake

Level 11

München

08.11.2020
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10comments

Can someone please point me to a mistake

Question about the task

Somehow average

Java Syntax, Level 4, Lesson 16

Resolved

The last condition is not met

package com.codegym.task.task04.task0441; /* Somehow average */ import java.io.*; public class Solution { public static void main(String[] args) throws Exception { //write your code here BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); int a = Integer.parseInt(reader.readLine()); int b = Integer.parseInt(reader.readLine()); int c = Integer.parseInt(reader.readLine()); if (a == b && b == c) { System.out.println(a); } else if (a >= b && a >= c) { if (b >= c) { System.out.println(b); } else System.out.println(c); } else if (b >= a && b >= c) { if (a >= c) { System.out.println(a); } else System.out.println(c); } else if (c >= a && c >= b) { if (a >= b) { System.out.println(a); } else System.out.println(b); } } }

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Misiu

Level 41 , Gdansk, Poland

8 November 2020, 13:12

In all instructions use only ">" operator (beside line 19). After line 20 add: if two numbers are equal ---> print one of them (three times this condition) Correct solution for some data doesn't mean that the solution is correct for all possible data.

Drazen Jankovic

Level 11 , München, Germany

8 November 2020, 14:40

Thanks a lot! Could you please help me find those data with whom the present solution wont work correctly? That would help me even more to figure out how does it code works with exceptions. Thanks a lot again!

Misiu

Level 41 , Gdansk, Poland

8 November 2020, 17:01

useful

I checked your solution with ">" only. There was no output for some data so I gave you my proposes. Now I've checked your solution with all kind of data and don't see any wrong outputs. Maybe thing is in line 31. IF is not neccesary. Here we already know that c is max.

Drazen Jankovic

Level 11 , München, Germany

8 November 2020, 17:52

Thank you! That was so confusing. I tried also with different data and outputs was good. That's way I was trying to point that out. There is no sintax error, logical error should return a bad result with some data... Right? And anyway, the task doesn't pass the test! You just saw that for yourself... 😀

Misiu

Level 41 , Gdansk, Poland

8 November 2020, 18:19

Right. But I still don't know - have you passed the task? If yes - what version?

Drazen Jankovic

Level 11 , München, Germany

8 November 2020, 18:55

I did pass the task, but not with this version... I used a Math.max after initial check if the one number is bigger from the other two... It was not the problem, task to pass, I wanted to figure how does it works, or why this specific version doesn't pass the test even if the results are ok... But it's not big deal anyway, maybe for me, because I'm still newbie in this... Anyway thanks a lot for your effort and help! 😀 P.S.: wish me luck in a new level... 😎

Misiu

Level 41 , Gdansk, Poland

8 November 2020, 21:02

One thing is a solution, the other what validator "thinks" about it ;) 5th level? Super. Go ahead!

Switch/Cypher

Level 25 , Bexleyheath, United Kingdom

8 November 2020, 11:40

You have a number of overlapping conditions for example:

a==b && b==c
a>=b && a>=c

are the same, if all the numbers are equal. Here is some code that achieves the task - I do some fancy stuff with arrays but the principles are all the same:

public static void main(String[] args) {
       Scanner keyboardInput = new Scanner(System.in);
       ArrayList<Integer> numbers = new ArrayList<>();

       while (numbers.size() < 3) {
           numbers.add(keyboardInput.nextInt());
       }
       keyboardInput.close();

       Collections.sort(numbers);

       if (numbers.get(0).equals(numbers.get(1))) {
           if (!numbers.get(1).equals(numbers.get(2)))
           System.out.println(numbers.get(0));
       }

       if (numbers.get(1).equals(numbers.get(2))) {
           if (!numbers.get(0).equals(numbers.get(1)))
               System.out.println(numbers.get(1));
           else System.out.println(numbers.get(0));
       }

       if (!numbers.get(0).equals(numbers.get(1)) && !numbers.get(1).equals(numbers.get(2))) {
           System.out.println(numbers.get(1));
       }


   }

Hit up google if you don't know what some of this does. It will help you in later tasks.

Drazen Jankovic

Level 11 , München, Germany

8 November 2020, 12:23

Tnx for a help! But that's not a point. You see, I've found many different solutions, but at this point, at lvl 4, we should use that what we have learned till now... And the second problem is that, if we test the program, and the result is as we expected, we expect that the code is correct.. But that is not a case here... 😀

Switch/Cypher

Level 25 , Bexleyheath, United Kingdom

8 November 2020, 14:43

Sure - even using things you already know you can clean your code up:

public static void main(String[] args) throws Exception {
       //write your code here
       BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
       int a = Integer.parseInt(reader.readLine());
       int b = Integer.parseInt(reader.readLine());
       int c = Integer.parseInt(reader.readLine());

       if (a == b && b == c) {
           System.out.println(a);
       } else if (a >= b && a >= c) {
           System.out.println(Math.max(b, c));
       } else if (b >= a && b >= c) {
           System.out.println(Math.max(a, c));
       } else {
           System.out.println(Math.max(a, b));
       }
   }

When I ran your code through IntelliJ it showed errors, so I suspect that's why it's not passing