Given an array of numbers. For each number in array, find the multiples that exist in the array.

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Given an array of numbers. For each number in array, find the multiples that exist in the array.
Eg
Array - [2, 3, 6, 10]
Output- 
Multiples of 2 - 6,10
Multiples of 3 - 6
Multiples of 6 - No multiples
Multiples of 10 - No multiples

However I am getting

Multiples of each element of array is:
Multiples of2Is No multiples
Multiples of2is:6
Multiples of2is:10
Multiples of2is:0
Multiples of3is:6
Multiples of3Is No multiples
Multiples of3is:0
Multiples of6Is No multiples
Multiples of6is:0
Multiples of10is:0

What I have tried:

import java.util.*;
import java.io.*;
import java.lang.*;
public class multiplesarrays
{
    public static void main(String[] args)
    { 
        Scanner sc=new Scanner(System.in);
        
        System.out.println("Enter the no: of elements in array");
        int n=sc.nextInt();
        int arr[]=new int[100];
        for(int i=0; i<n; i++)
        {
            arr[i]=sc.nextInt();
        }
        System.out.println("Multiples of each element of array is:");
       
       for(int i=0; i<n; i++)
       {
      
      
       for(int j=i+1; j<=n; j++)
       {
           
       if(arr[j]%arr[i]!=0)
       {
           System.out.println("Multiples of"+arr[i]+ "Is No multiples");
    
       }
       else if(arr[j]%arr[i]==0)
       { 
           System.out.println("Multiples of"+arr[i]+"is:"+arr[j]);
         
       }
          
       }
       }
       
       
    }
}

Posted 23-Jun-23 18:22pm

Abhinav Kishore M

Updated 23-Jun-23 19:13pm

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Patrice T · Accepted Answer · 2023-06-23T19:13:00

Quote:
However I am getting

As programmer, your job is also to find the bugs in your code.
- If you look carefully at the input, there is no duplicates and values are sorted. If you don't enforce those conditions, output can not be correct.

Quote:
Java
Multiples of2is:0

- This tells you that you are testing 2 against a value that is not in the input.
- How many times "Multiples of 2 - " is printed in correct result, how many times is it printed with your code ?
- When do you know the there is no multiple ? When is your code printing it ?

Ask yourself those question, the answers should lead your to spot problems and then craft corrections.

[edit]
This changed code may help you to understand what is going on.

Java

import java.util.*;
import java.io.*;
import java.lang.*;
public class multiplesarrays
{
    public static void main(String[] args)
    { 
        Scanner sc=new Scanner(System.in);
        
        System.out.println("Enter the no: of elements in array");
        int n=sc.nextInt();
        int arr[]=new int[100];
        for(int i=0; i<n; i++)
        {
            arr[i]=sc.nextInt();
        }
        System.out.println("Multiples of each element of array is:");
       
       for(int i=0; i<n; i++)
       {
       System.out.println("i= "+i);
      
       for(int j=i+1; j<=n; j++)
       {
       System.out.println("j= "+j);

       if(arr[j]%arr[i]!=0)
       {
           System.out.println("Multiples of"+arr[i]+ "Is No multiples");
    
       }
       else if(arr[j]%arr[i]==0)
       { 
           System.out.println("Multiples of"+arr[i]+"is:"+arr[j]);
         
       }
          
       }
       }
       
       
    }
}

OriginalGriff · Accepted Answer · 2023-06-23T18:30:00

Simple solution: Sort the array. Then all duplicates are next to each other, and can be located, counted, and printed in a single pass.
Hint: Java has a sort method for arrays built in ...

But do yourself a favour and indent your code - it's a lot more readable that way.

[edit]
Just for kicks, I had 2 minutes to spare, so I tried this exercise in C# ...

C#

        private Dictionary<int, IEnumerable<int>> GetMultiples(int[] arr)
            {
            Array.Sort(arr);
            Dictionary<int, IEnumerable<int>> counts = new Dictionary<int, IEnumerable<int>>();
            for (int i = 0; i < arr.Length; i++)
                {
                int val = arr[i];
                counts[val] = arr.Where(x => x != val && x % val == 0);
                }
            return counts;
            }
...
            int[] arr = { 2, 3, 6, 10 };
            Dictionary<int, IEnumerable<int>> counts = GetMultiples(arr);
            foreach (int key in counts.Keys)
                {
                IEnumerable<int> values = counts[key];
                int count = values.Count();
                if (count == 0) Console.WriteLine($"{key} has no multiples");
                else
                    {
                    Console.WriteLine($"{key} has {count} multiples: {string.Join(",", values.Select(n => n.ToString()))}");
                    }
                }

Results

2 has 2 multiples: 6,10
3 has 1 multiples: 6
6 has no multiples
10 has no multiples

You can't use that code directly in Java, but at least it shows you it's possible ...
[/edit]

Given an array of numbers. For each number in array, find the multiples that exist in the array.

2 solutions

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