Average Sum

[Kaspersky Ukshini]

Hey guys, what's up... I need help solving a problem as my homework (I got around 6hours left to complete it lol)...

Find the numbers which is closest to the average sum for a given array of N (1<=N<=50) of natural numbers. If there are two numbers who meet the requirement, return the smaller of the two.
For example for the array of: 1,2,3,4,5,6 the average sum is 3.5, so both 3 and 4 are the closest to that, but the program has to return 3, because it's smaller than 4.

The array can also contain duplicates.
First we type the number of elements in the array, then in each line we add the numbers. Name of the class: Array

**Note**: Create a data structure array and use it.

And this is what they have given to me, I just need to type the code:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Array<E> {


    public static int closestNumber( ) {
        //Your code here...
    }

    public static void main(String[] args) throws IOException {
        BufferedReader stdin = new BufferedReader( new InputStreamReader(System.in));
        String s = stdin.readLine();
        int N = Integer.parseInt(s);

        //Your code here...


        System.out.println(brojDoProsek(niza));
    }



}

[Campbell Ritchie]

Welcome to the Ranch

How do you work out such an average? How do you find the nearest whole number like that?
What have you written?

[Kaspersky Ukshini]

Campbell Ritchie wrote:Welcome to the Ranch

How do you work out such an average? How do you find the nearest whole number like that?
What have you written?

Hey, thanks a lot!

That's not my code, that's the peace of code it was given by the school, now we just have to fill it in... I didn't add anything yet...

[Joanne Neal]

Bashkim Ukshini wrote: I didn't add anything yet...

Okay. So lets go with what Campbell suggested and write some code to work out an average.
Suppose you had this array in your code

int numbers = new int { 2, 6, 37, 25, 98, 12, 45, 987 };

Can you write some code to work out the average of those numbers ?

[Kaspersky Ukshini]

Isn't that already written in this part:

public static void main(String[] args) throws IOException {  
        BufferedReader stdin = new BufferedReader( new InputStreamReader(System.in));  
        String s = stdin.readLine();  
        int N = Integer.parseInt(s);  

I don't quite get this to be honest, can you please explain shortly, I might get more confortable once I understand this part...

[Joanne Neal]

Bashkim Ukshini wrote:Isn't that already written in this part:.

It is indeed. I missed that.

Bashkim Ukshini wrote:I don't quite get this to be honest, can you please explain shortly, I might get more confortable once I understand this part...

BufferedReader stdin = new BufferedReader( new InputStreamReader(System.in));

System.in is what is known as standard input. Anything you type whilst your program is running will be added to this stream so that your program can access it. If you don't know what BufferedReader and InputStreamReader are, don't worry for the moment. Just accept that they improve the performance of reading what you type.

String s = stdin.readLine();  

This read a line of text that you type - i.e. everything you type until you hit the Enter/Return key. It stores the text in a String referred to by the s variable

int N = Integer.parseInt(s);

This converts the contents of the String into an integer variable

[Joanne Neal]

So, assuming you are comfortable with reading in the numbers of values you are going to use, first you need to create an array to hold those values.
You know how many values there are going to be, so can you write the code that creates that array ?
If not, you either need to check your class notes or check out this tutorial.

[Kaspersky Ukshini]

Should I use a Foor loop?
I also don't get this part: http://prntscr.com/4xq4gu
Why <E> ??

[Joanne Neal]

Bashkim Ukshini wrote:Should I use a Foor loop?

Not yet. For the moment I just want to see the line of code that declares the array variable and assigns an array of the correct size to it.

[Kaspersky Ukshini]

Okay, I've come up with this, I hope it's ok:

public static void main(String[] args) throws IOException{
		System.out.println("How long is the array?");
                BufferedReader stdin = new BufferedReader( new InputStreamReader(System.in)); 
		String s = stdin.readLine();
		int N = Integer.parseInt(s);
                

// This is the part for declaring the array, and reading from the keyboard

                int [] arr=new int [N];
                System.out.println("Start typing your array:");
                
		for(int i=0;i<N;i++){
                BufferedReader stdin1 = new BufferedReader( new InputStreamReader(System.in)); 
		String s1 = stdin1.readLine();
		arr[i] = Integer.parseInt(s1);
                }

[Joanne Neal]

Looks good. You don't need to create a new BufferedReader on line 14 - you can still use the one you created on line 3.
The next thing to do is to calculate the average of all those numbers you read in. Don't forget to think about the type of the variables you use.

[Kaspersky Ukshini]

Okay then, I removed it, now that part looks like this:

for(int i=0;i<N;i++){
		String s1 = stdin.readLine();
		arr[i] = Integer.parseInt(s1);
                }

Now I have the array, I need to call the function inside main:

System.out.println(closestNo(arr));

and the function body is outside of many, and this is it:

public static int closestNo(int [] x){
            
            int sum=0;
            float average=0;
            for(int i=0;i<x.length;i++)
            {
                sum+=x[i];
            }
            average=sum/2;
            
	}

I just need to add the part that calculates the closest member to the "average"... Any ideas?

[Aj Prieto]

Bashkim Ukshini wrote:

public static int closestNo(int [] x){
            
            int sum=0;
            float average=0;
            for(int i=0;i<x.length;i++)
            {
                sum+=x[i];
            }
            average=sum/2;
            
	}

You should change the line on how you compute the average.

Remember, average = (sum of all elements) / (total number of elements)

[Kaspersky Ukshini]

Aj Prieto wrote:

Bashkim Ukshini wrote:

public static int closestNo(int [] x){
            
            int sum=0;
            float average=0;
            for(int i=0;i<x.length;i++)
            {
                sum+=x[i];
            }
            average=sum/2;
            
	}

You should change the line on how you compute the average.

Remember, average = (sum of all elements) / (total number of elements)

Ohh, right... it should be average=sum/x.length;
thank you.

how about finding the closest to the average? I found out how that is done, but what if I get two numbers with the equal distance to the average.. how to save them both? I need an idea.

[Joanne Neal]

Bashkim Ukshini wrote:what if I get two numbers with the equal distance to the average.. how to save them both? I need an idea.

You go back and read the specification

[Kaspersky Ukshini]

I believe I did it.. maybe it's complicated, but I hope it's correct.

This is the function that finds the closest number to the average:

public static int closestNo(int [] x){
            
            int sum=0;
            float average=0;
            //Frist we find the average sum:
            for(int i=0;i<x.length;i++)
            {
                sum+=x[i];
            }
            average=sum/x.length;
            
            //I assign the first number of the array as the closest number to the sum
           //And I store the absolute value of the difference between the first number and the sum to a variable,
           //so I can later compare that value with all the other members to see if I can find a smaller difference, hence a closer number to the average.

            float diff=(Math.abs((x[0]-sum))); 
            int number=x[0];

            // Now we find the closest number to the sum
        
       for(int i:x){
//If the is a number which difference with the sum, is smaller than the current one I have...
//store that number in the number variable, and store it's difference, so I can continu searching for a number with even smaller difference

           if((Math.abs(x[i]-sum))<diff)
           {   diff=(Math.abs(x[i]-sum));
               number=x[i];
//Now I made another for loop, to see if there is another number in the array which has the SAME distance with the curent one I have, if yes...
//I check if that number is smaller than my curent one, if yes, I store that number to 'number', and I go in the next one...
            for(int j:x)
            {
                if(Math.abs(x[j]-sum)==diff)
                {
                    if(x[j]<number)
                        number=x[j];
                        
                }
            }
           }
           
        }

What do you think guys?

[Joanne Neal]

Bashkim Ukshini wrote:What do you think guys?

The last bit seems a bit complicated. In the second loop you're checking the whole of the array for an equal number, but you've already checked all the numbers up to the current one in the first loop. In fact you don't even need the second loop.
What about this

for(int i:x){  
    int newdiff = Math.abs(x[i] - sum);
    if (newdiff < diff)  
    {
        diff= newdiff;  
        number=x[i];  
    } else (if (newdiff == diff) {
        if (x[i] < number)
            number = x[i];         
    }
}

[Kaspersky Ukshini]

the 2nd line should be -average, not -sum..

Anyways, your way is way way better than mine... I optimized it a bit more, and this is my final solution:

for(int i=0;i<x.length;i++){
          
           float newdiff=Math.abs(x[i]-average);
           if(newdiff<diff)
           {
              diff=newdiff;
              number=x[i];
           }
           else if (newdiff==diff && x[i]<number)
           {
               number=x[i];
           }
           
        }

Thank you so so so much guys!

[Campbell Ritchie]

Bashkim Ukshini wrote:Isn't that already written in this part:
. . .

Yes, and no.

What you could have done is separate calculating averages (more precisely arithmetical means) from the process of creating the array. So I think Joanne was correct to supply you with a ready made array.