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Maximum of four numbers

Java Syntax, Level 4, Lesson 6

Under discussion

I know there's a bit of code that could be compacted

package com.codegym.task.task04.task0419; /* Maximum of four numbers */ import java.io.*; public class Solution { public static void main(String[] args) throws Exception { BufferedReader reader = new BufferedReader (new InputStreamReader(System.in)); String aa = reader.readLine(); int a = Integer.parseInt(aa); String bb = reader.readLine(); int b = Integer.parseInt(bb); String cc = reader.readLine(); int c = Integer.parseInt (cc); String dd = reader.readLine(); int d = Integer.parseInt(dd); if ((a > b) && (b > c) && (c > d) && (a > c) && (b > a) && (d > a) && (a > d) && (b > d) && (c > a) && (c > b) && (d > b) && (d > c)) { System.out.println(a); } else if ((b > a) && (b > c) && (b > d)) { System.out.println(b); } else if ((c > a) && (c > b) && (c > d)) { System.out.println(c); } else if ((d > b) && (d > c) && (d > a)) { System.out.println(d); } } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

10 November 2020, 03:45

Everyone below gives good advice. Here is mine: When I approach a problem I try to break it down into easier/smaller sub problems. This program asks you to find the maximum of 4 entered numbers. I would ask you, as a much simpler problem, how would you find the maximum of 2 numbers? Most people would come up with this solution: if num1 is greater than num2 then num1 is the largest, else num2 is the largest Instead of outputting that result, why not save that result in a variable? You then can take that and compare it to the third number. Now that you are comparing the result of (num1 > num2) to num3, you have the same exact problem of only comparing 2 numbers, which you have already solved. So: if (result of num1 > num2) is greater than num3 then it is largest, else num3 is largest Now you can save that result in a variable so you can compare it to num4. Once again you are only comparing 2 numbers. When completed the result will be that you now have the largest of the four numbers It's simple and you can do this for a millions numbers if needed.

Switch/Cypher

Level 25 , Bexleyheath, United Kingdom

7 November 2020, 10:33

A simpler way without Collections would be to use a loop:

public static void main(String[] args) {
       try (BufferedReader keyboardInput = new BufferedReader(new InputStreamReader(System.in))) {

           int a = Integer.parseInt(keyboardInput.readLine());
           int b = Integer.parseInt(keyboardInput.readLine());
           int c = Integer.parseInt(keyboardInput.readLine());
           int d = Integer.parseInt(keyboardInput.readLine());
           int max = 0;

           keyboardInput.close();

           int[] numbers = new int[4];
           numbers[0] = a;
           numbers[1] = b;
           numbers[2] = c;
           numbers[3] = d;

           for (int i = 0; i < numbers.length; i++) {
               if (numbers[i] >= max){
                   max = numbers[i];
               }

           }

           System.out.println(max);


       } catch (Exception e) {
           e.printStackTrace();
       }


   }

Hopefully those two give you a flavour of other ways to achieve the same thing without a lot of 'its'

Misiu

Level 41 , Gdansk, Poland

7 November 2020, 12:13

Not working if all numbers are negative.

Switch/Cypher

Level 25 , Bexleyheath, United Kingdom

7 November 2020, 15:42

That's true. It was meant as an example of the kind of thing you might do, rather than a 'all cases' solution. I mean, I would use the solution below over this one anyway. Easily fixed though with:

int max = numbers[0];

Valuev Nutt

Level 4 , Chiba, Japan

10 November 2020, 00:53

Thank you so much for the explanation! I tried the code but it popped up with only two of the requirements complete

Switch/Cypher

Level 25 , Bexleyheath, United Kingdom

7 November 2020, 10:21

It can be compacted a lot - but probably with techniques you may not have learned yet. I'll show you for a taste:

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

       while (numbers.size() < 4){
           numbers.add(keyboardInput.nextInt());
       }
       keyboardInput.close();

       System.out.println(numbers.last());

TreeSet. automatically orders the numbers from lowest to highest.

Misiu

Level 41 , Gdansk, Poland

7 November 2020, 01:01

Line 26: (d > a) && (a > d) - always false Make line 26 in the way you created other IF's.

Valuev Nutt

Level 4 , Chiba, Japan

10 November 2020, 00:53

Not sure by what you mean, would you mind explaining a little further. Thank you!

Misiu

Level 41 , Gdansk, Poland

10 November 2020, 01:33

In lines 30 and 34 change > to >= Make line 26 as line 30 (or 34) but for "a" variable. Line 38 should be only: else print d