CodeGym/Help with Java Tasks/Can anyone help me???

Level 5

Chennai

07.01.2019
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10comments

Can anyone help me???

Question about the task

Three numbers

Java Syntax, Level 4, Lesson 6

Under discussion

Hopefully I attached my code snippet, Whats wrong in my code. Can anyone please give a solution with explanation ?

Use the keyboard to enter three integers. One of the numbers is unique. The other two are identical. Display the ordinal number of the number that is different from the others.

Example for 4 6 6:
1

Example for 6 6 3:
3

Requirements:

The program should read the numbers from the keyboard.
The program should use System.out.println() or System.out.print().
The program should display the ordinal number of the number that is different from the others.
If all the numbers are different, don't display anything.

package com.codegym.task.task04.task0424; /* Three numbers */ import java.io.*; import java.util.Scanner; public class Solution { public static int findIndex(int arr[], int t) { int j =0; while(j<arr.length) { if(arr[j] == t) return j+1; else return j = j +1; } return -1; } public static void main(String[] args) throws Exception { //write your code here Scanner sc = new Scanner(System.in); int a[] = new int[3]; for(int i=0;i<3;i++) { a[i] = sc.nextInt(); } int i =0; if(a[i] == a[i+1] && a[i] != a[i+2]) { System.out.println(findIndex(a,a[i+2])); } else if(a[i+1] == a[i+2] && a[i+1] != a[i]) { System.out.println(findIndex(a,a[i])); } else if(a[i+2] == a[i] && a[i+2] != a[i+1]) { System.out.println(findIndex(a,a[i+1])); } } }

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Ravi Ranjan

Level 7 , Muzaffarnagar, India

15 June 2019, 01:30

if(n1==n2){ System.out.println("3"); } else if(n1==n3){ System.out.println("2"); } else if(n3==n2){ System.out.println("1"); }

Okanlawon Oluwatobi Damilare

Level 9 , Banjul, Gambia

23 March 2019, 17:08

you don't need to build another function or use a for loop

Okanlawon Oluwatobi Damilare

Level 9 , Banjul, Gambia

23 March 2019, 17:08

you're taking this too far try int a,b,c; BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); try{ a = Integer.parseInt(reader.readLine()); b = Integer.parseInt(reader.readLine()); c = Integer.parseInt(reader.readLine()); if(a == b && b != c){ System.out.println(3); }else if(a == c && b != c){ System.out.println(2); }else if(b==c && a != b){ System.out.println(1); }else{} }catch(IOException e){ e.printStackTrace(); }

Roshan Singh

Level 12

17 February 2019, 16:48

You Can Try This Also..... if (num1!=num2 && num1!=num3 && num2==num3 ) { System.out.println("1"); } if (num2!=num1 && num2!=num3 && num1==num3 ) { System.out.println("2"); } if (num3!=num1 && num3!=num2 && num1==num2 ) { System.out.println("3"); }

Rathika

Level 5 , Chennai, India

8 January 2019, 11:03

@Guadalupe Gagnon, Thanks for your reply and opinion its really helpful.:)

Guadalupe Gagnon

Level 37 , Tampa, United States

8 January 2019, 14:19

Keep up the good work. I can tell you will be learning this stuff fast because of the things that you are trying.

Guadalupe Gagnon

Level 37 , Tampa, United States

7 January 2019, 14:47

useful

Learning is all about trying new things for greater understanding, but sometimes you need to stop and ask "why". Here are a few things that seem like they just make extra work: #1 why make an int i and start doing a{i+#]? Why not get rid of i and just use a[0], a[1], a[2]? #2 why make a findIndex method and then pass it the number by index? If the input is 6,6,3 and three is the correct one at index 2, your code says find me the index for the number at index 2....... The biggest thing to learn in computer programming is to keep things as simple as possible. You pretty much solved the problem but for one minor flaw, here is your solution with the fixes:

public class Solution {
    public static void main(String[] args) throws Exception {
        //write your code here
        Scanner sc = new Scanner(System.in);
        int a[] = new int[3];

        for(int i = 0; i < 3; i++)
        {
            a[i] = sc.nextInt();
        }


        if(a[0] == a[1] && a[0] != a[2])
        {
            System.out.println(2);
        }
        else if(a[1] == a[2] && a[1] != a[0])
        {
            System.out.println(0);
        }
        else if(a[2] == a[0] && a[2] != a[1])
        {
            System.out.println(1);
        }
    }
}

Angeget

Level 22 , Potsdam, Germany

8 January 2019, 08:09

I agree with you. Just keep it simple is the best way doing thing.

Andrew

Level 16 , Toronto, Canada

11 January 2019, 13:05

Hi, not sure if I am wrong here, but in the question the indexing starts at 1. I had essentially the same thing but the system.out.println's were all +1. Best, A

Guadalupe Gagnon

Level 37 , Tampa, United States

11 January 2019, 14:38

Hi Andrew, Array indexing always starts at 0 and ends at (length - 1). In the above example (original code, not my correction) the length is set to 3, so the indices would be 0, 1, and 2. If you set the length to 100 the indices would be 0-99. The for loop used above (on line 28) the numbers used are 0,1, and 2..... when 'i' was 3 the middle statement (i < 3) failed because 3 is not less than 3. Remember this in the future for 'for' loops, use the length [3 in this case] in the middle statement and you can't go wrong, Arrays have a length property you can use instead of hard coding the size in, arrayName.length (in above code it would be a.length). Message me your code, with the indices +1 so i can see what you did.