Finding any duplicate element in array

Question

I am trying to find element that is duplicated in an array. Program is intended to take user input.

Here is the code:

   package sortingattempt;

import java.util.Scanner;

public class ArraySimilar {

    public static void main(String args[]){
        int[] a = new int[100];
        int[] b = new int[5];
        int Duplicate = 0;

        Scanner sc = new Scanner(System.in);
        System.out.println("Enter size of array");
        int size = sc.nextInt();

        //Scanner elem = new Scanner(System.in);
        System.out.println("Input elements in array" +size);
        for(int j = 0;j<size;j++){
            a[j] = sc.nextInt();
        }
        System.out.println(a.length);
        System.out.println("a[]" + a.toString());

    for (int i = 0; i < a.length;i++){
        b[0] = a[i];
        if (b[0] == a[i+1]){
            Duplicate = b[0];
            System.out.println(Duplicate);
        }
        i++;
        }
        System.out.println("No common variable");

    }

}

When I try to run it, its running properly till line where I ask user to give input. After entering input nothing is happening. Please indicate errors in program. Thank you.

Try inserting sc.nextLine(); before a[j] = sc.nextInt(); — Justin
– Justin, Commented Dec 7, 2013 at 1:18
No it did not work. I debugged program and found out that elements are not going in array. — Gaurav K.
– Gaurav K., Commented Dec 7, 2013 at 1:22

Paul Samsotha · Accepted Answer · 2013-12-07 02:10:55Z

I think what's happening is that you are not being prompted to enter a number with a System.out.println("Enter a number: "); So you think the program is not working, when really it is just waiting for you to keep entering numbers. You may want to prompt with System.out.println()s on each iteration:

System.out.println("Input elements in array" +size);
for(int j = 0;j<size;j++){
    System.out.println("Enter a number: ");
    a[j] = sc.nextInt();
}

Also, your initializng your arrays before hand. Shouldn't you wait to get the user input for size, then initialize them?

int[] a; 
int[] b;

Scanner sc = new Scanner(System.in);
System.out.println("Enter size of array");
int size = sc.nextInt();

a = new int[size];
b = new int[size];

Also, trying to print a.toString() will not give you your desired output. You have to iterate through it or do something like this

System.out.println("a[] " + Arrays.asList(a));

Edit: To Compare

You should just make the values equals in the first loop

for(int j = 0;j<size;j++){
    System.out.println("Enter a number: ");
    a[j] = sc.nextInt();
    b[j] = a[i];
}

Then in the bottom loop, compare in a nested loop

int dupCount = 0;
for (int i = 0; i < size; i++){
    for (int j = 0; j < size; j++){
        if (b[i] == a[j]){
            Duplicate = a[j];
            System.out.println(Duplicate);
            dupCount = 0;
        }
    }
}
if (dupCount == 0)
    System.out.println("No common variable");

Thanks. It is now taking elements of input as Array. but I am getting an error now: Exception in thread "main" a[][I@6bb1a986 java.lang.ArrayIndexOutOfBoundsException: 5 at sortingattempt.ArraySimilar.main(ArraySimilar.java:29)
Im assuming you entered 5 for the size? You can see the problem when you try access a[i + 1]. The max i in the loop is 4, so when you try and add 1 you're trying to access a[5] which doesnt exist in an array of size 5. max index would be 4. You need to fix that +1. I'm not exactly sure what you're trying to do, so I can't really offer a solution.
I am trying to put each element in a[] at b[0] position and compare it with other elements in a[].

notArefill · Accepted Answer · 2013-12-07 01:39:36Z

Another error is that you need to delete the i++ from this block of code

   Duplicate = b[0];
        System.out.println(Duplicate);
    }
    i++;  //delete that
    }

Also change that

  System.out.println("Input elements in array" +size);
for(int j = 0;j<size;j++){
    a[j] = sc.nextInt();
}
System.out.println(a.length);
System.out.println("a[]" + a.toString());//change that to a[j] and put it inside the loop.

I assume here that you want to print the elements of your array,so the a.toString won't do that.

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