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hi, in this code I meet all conditions. when i put ex. 3, 3, 3, 1 it is only printing 3 once (one time) but program says not met

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)); int a = Integer.parseInt(reader.readLine()); int b = Integer.parseInt(reader.readLine()); int c = Integer.parseInt(reader.readLine()); int d = Integer.parseInt(reader.readLine()); if(a > b && a > c && a > d){ System.out.println(a); } else if(b > a && b > c && b > d){ System.out.println(b); } else if(c > b && c > a && c > d){ System.out.println(c); } else { System.out.println(d); } } }

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Shuy

Level 9 , Frankfurt, Germany

25 December 2019, 12:49

If you had the test numbers: a = 20 b=20 c=15 d=5 you would be displaying d. Hope that helps, otherwise just reply.

Rafał Szrajnert

Level 7 , Poddebice, Poland

25 December 2019, 13:37

thanks, i passed it: if(a >= b && a >= c && a >= d){ System.out.println(a); } else if(b >= a && b >= c && b >= d){ System.out.println(b); } else if(c >= b && c >= a && c >= d){ System.out.println(c); } else if (d >= a && d >= c && d >= b){ System.out.println(d); } but is there any way to make code smaller?

Shuy

Level 9 , Frankfurt, Germany

25 December 2019, 14:19

I think this will get covered a bit later, but basically you could write a method that gives you the bigger of 2 numbers, then youd simply call it twice, once with a,b and once with c, d and call it once again with the 2 return values of the first 2 calls.

Austeja

Level 10 , Kaunas, Lithuania

28 December 2019, 08:38

Shuy, I am trying to implement your solutions, looks very cool. Yet I can't figure out how to call the method that gives the bigger of the two return values of the previous two calls :/ How would you write it? Thanks!

Shuy

Level 9 , Frankfurt, Germany

6 January 2020, 08:42

Sry for the late reply and since youre already lvl 6 you probably already solved it. But just in case, you would only need a method to give you the bigger one of 2: public int bigger(int x, int y) { if (x >= y) return x; else return y; } and you could call it 3 times in one: bigger(bigger(x,y),bigger(z,v)) giving you the biggest of 4 numbers, since you pass the bigger value of x,y and the bigger value of z,v to the method.

Austeja

Level 10 , Kaunas, Lithuania

8 January 2020, 11:46

Thanks!

Jiesheng Yang

Level 8 , Sulzbach, Germany

1 April 2020, 14:28

yep, guess i can make it smaller: Scanner sc = new Scanner(System.in); int a = sc.nextInt(); int b = sc.nextInt(); int c = sc.nextInt(); int d = sc.nextInt(); int e = Math.max(Math.max(a,b),Math.max(c,d)); System.out.println(e);