CodeGym/Help with Java Tasks/Last condition not satisfied

Level 7

Delhi

06.01.2021
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11comments

Last condition not satisfied

Question about the task

Under discussion

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));
        String n1 = reader.readLine();
        int num1= Integer.parseInt(n1);
        String n2= reader.readLine();
        int num2 = Integer.parseInt(n2);
        String n3 = reader.readLine();
        int num3 = Integer.parseInt(n3);
        String n4 = reader.readLine();
        int num4 = Integer.parseInt(n4);
        Solution sl = new Solution();
        sl.Max(num1,num2,num3,num4);

    }
    void Max(int num1,int num2,int num3,int num4){
        if(num1>num2 && num1>num3 && num1>num4)
            System.out.println(num1);
        else if(num2>num1 && num2>num3 && num2>num4)
            System.out.println(num2);
        else if(num3>num1 && num3>num2 && num3>num4)
            System.out.println(num3);
        else if(num4>num1 && num4>num2 && num4>num3)
            System.out.println(num4);
        else if((num1==num2) && (num2==num3) && (num3==num4) && (num4==num1))
            System.out.println(num1);
    }
}

package com.codegym.task.task04.task0419; /* Maximum of four numbers */ 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)); String n1 = reader.readLine(); int num1= Integer.parseInt(n1); String n2= reader.readLine(); int num2 = Integer.parseInt(n2); String n3 = reader.readLine(); int num3 = Integer.parseInt(n3); String n4 = reader.readLine(); int num4 = Integer.parseInt(n4); Solution sl = new Solution(); sl.Max(num1,num2,num3,num4); } void Max(int num1,int num2,int num3,int num4){ if(num1>num2 && num1>num3 && num1>num4) System.out.println(num1); else if(num2>num1 && num2>num3 && num2>num4) System.out.println(num2); else if(num3>num1 && num3>num2 && num3>num4) System.out.println(num3); else if(num4>num1 && num4>num2 && num4>num3) System.out.println(num4); else if((num1==num2) && (num2==num3) && (num3==num4) && (num4==num1)) System.out.println(num1); } }

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Aditi

Level 7 , Delhi, India

8 January 2021, 08:26

Try replacing '>' by '>='.

Nouser

Level 36 , Germany

6 January 2021, 14:15

Try 1, 1, 2, 3 expected result: 3 actual output:

Dinesh

Level 7 , Delhi, India

7 January 2021, 05:15

Yes Its 3

remote87

Level 18 , Sofia, Bulgaria

6 January 2021, 14:10

LOL! This code is really hard to read. I would suggest to write a method that returns the max of only two numbers and than, when asking the user for 4 numbers, do something like this: max(max(num1, num2) , max(num3, num4)); This way it won't be so complex and so hard to read the code :)

BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int num1 = Integer.parseInt(reader.readLine();
        int num2 = Integer.parseInt(reader.readLine();
        int num3 = Integer.parseInt(reader.readLine();
        int num4 = Integer.parseInt(reader.readLine();
        //Solution sl = new Solution(); - redundant
       System.out.println(max(max(num1, num2), max(num3, num4)));

public static void max(int numOne, int numTwo){
        int maxOfTwo = 0;
        if(numOne > numTwo) maxOfTwo = numOne;
        else maxOfTwo = numTwo;
}

Not sure how I wrote it, because I use my phone

Dinesh

Level 7 , Delhi, India

7 January 2021, 04:43

Thank you so much for you kind guidance. Implemented as per your fruitful suggestions.

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 num1 = Integer.parseInt(reader.readLine());
        int num2 = Integer.parseInt(reader.readLine());
        int num3 = Integer.parseInt(reader.readLine());
        int num4 = Integer.parseInt(reader.readLine());

        System.out.println(max(max(num1, num2), max(num3, num4)));
    }

//how this method is able to compare 4 numbers ?
        public static void max(int numOne, int numTwo){
        int maxOfTwo = 0;
        if(numOne>numTwo) maxOfTwo = numOne;
        else maxOfTwo = numTwo;.


    }
}

Help please. What else I am not getting the right way. Thanks

Nouser

Level 36 , Germany

7 January 2021, 07:39

Your max method doesn't return a value. So when you call max(1, 2) you get no value at all. So the consecutive max call has no argument and fails.

public static int max(int numOne, int numTwo) {
    int maxOfTwo = 0;
    if(numOne>numTwo) maxOfTwo = numOne;
    else maxOfTwo = numTwo;

    return maxOfTwo;
    //or just
    //return numOne > numTwo ? numOne : numTwo;
}

remote87

Level 18 , Sofia, Bulgaria

7 January 2021, 09:25

Yep, as I said, I wrote it on my phone and I forgot to return the value from the method. Thank you, Nouser! :) As per your questin "//how this method is able to compare 4 numbers ?": It's really simple: return the max number of two (1, 2) -> 2, than do it again on the other pair of numbers max of two (3, 4) -> 4, than take the result of those two calls of the method and compare them so max of two (2, 4) -> 4. To understand it you can also do it with asignment:

int maxFirstPairs = max(num1, num2);
int maxSecondPairs = max(num3, num4);
int maxFinalPairs = max(maxFirstPairs, maxSecondPairs );

int maxFirstPairs = max(1, 2); //returns 2, thanks to Nouser
int maxSecondPairs = max(3, 4); //returns 4, thanks to Nouser
int maxFinalPairs = max(2, 4); //returns 4, thanks to Nouser

So, you call the method on the result of two inner calls of the same method to simplify your code :) Hope I helped and again, thank you Nouser :) Please rate our answers

Dinesh

Level 7 , Delhi, India

7 January 2021, 13:10

I was going through as per your suggested assignment. I got stuck in storing the values. Need help. Thank you.

import java.io.*;
public class Codegym{
    public static void main(String[]args) throws Exception{
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

        int num1 = Integer.parseInt(reader.readLine());
        int num2 = Integer.parseInt(reader.readLine());
        int num3 = Integer.parseInt(reader.readLine());
        int num4 = Integer.parseInt(reader.readLine());

        //How to Store the max value of these two for further comparison.
        System.out.println(max(num1, num2));
        System.out.println(max(num3, num4));

    }

    public static int max(int num1, int num2) {
        int maxOfTwo=0;
        if (num1>num2) maxOfTwo = num1;
            else maxOfTwo = num2;
        return maxOfTwo;
    }

}

remote87

Level 18 , Sofia, Bulgaria

7 January 2021, 15:00

//How to Store the max value of these two for further comparison.

 //compare 2 numbers and store the result in a variable
        int firstPair = max(num1, num2));
//print the first result of the comparison
        System.out.println(firstPair);

//compare the second 2 numbers and store the result
        int secondPair = (max(num3, num4));
//print the second result
        System.out.println(secondPair);

//if you want, you can store the result in a third variable, or you can print it directly
        int finalResult = max(firstPair, secondPair);
        System.out.println(finalResult);

         //print directly without storing the final result:
        //System.out.println(max(firstPair, secondPair));

Dinesh

Level 7 , Delhi, India

8 January 2021, 07:04

Thank You (remote87 & Nouser) for your kind efforts. Implemented and all conditions satisfied. Though I have one query regarding the max(). Please if possible make it clear as I am not getting it. as the last condition of the task says If there are several maximum numbers display any of them. So, 1. How the equality among maximum numbers being checked and displayed once? that is if I am giving input as 5 5 2 1 or 5 5 5 5 or 5 5 2 5 How only 5 as max value is being displayed. In following code we haven`t anything as such equality among numbers comparison.

public static int max(int numOne, int numTwo){
            int maxOfTwo = 0;

            if (numOne>numTwo) maxOfTwo = numOne;
                else maxOfTwo= numTwo;
                //return maxOfTwo;
                return numOne > numTwo ? numOne : numTwo;
        }

Though if it is inbuilt in max() function. I want its simple explanation. If Possible. Programmatically And The return statement using ternary(?) operator as suggested by Nouser

return numOne > numTwo ? numOne : numTwo;

doesn`t have any equality comparison in between numbers but still returning the result as required among different equal numbers as stated above. How? Please clarify. Thank You

Nouser

Level 36 , Germany

8 January 2021, 09:05

How only 5 as max value is being displayed. pretty easy if you look at your max method

if (a > b)
    return a;
else
    return b;

You have your mentioned a = 5 and b = 5. The check in line 1 fails (5 is not greater than 5) therefore it uses the else clause and returns b (5). You have one of the two 5s as max.

return a > b ? a : b;

is called ternary operator (google) and means nothing else than the above if else construct.