How do you detect and delete more than two duplicates in an Array?

Question

How do you detect and delete if there more than two duplicates in an array. In the following example, I have taken an array with the elements "10, 20, 30, 40, 40, 40, 50, 60, 70, 80 ,10". Here 40 is repeated thrice and 10 is repeated twice. I could write a program to detect two duplicates but I am not able to shrink 40(repeated thrice) to once. Note:- I want to do this without using any java collection

public class ArrayDuplicate {
public void run1()
{
    int[] a = {10, 20, 30, 40, 40, 40, 50, 60, 70, 80 ,10};
    int size=a.length;
    System.out.println("Array size before duplicate deletion "+size);
    for(int i =0;i<(size-1);i++)
    {
        for(int j=i+1;j<=(size-1);j++)
        {
            if(a[i]==a[j] &&i!=j)
            { 
                while(j<(size-1))
                {
                a[j]=a[j+1];
                j++;

                }
                size--;
            }

        }
    }
    System.out.print("The array after deleting the duplicates is ");
    for(int k=0;k<=(size-1);k++)
    {
        System.out.print(a[k]);  //40 is being printed twice
        if(k<(size-1))
        {
            System.out.print(",");
        }
        else
            System.out.print(".");
    }




}
public static void main(String[] args)
{
    ArrayDuplicate ob = new ArrayDuplicate();
    ob.run1();

}

}

What exact output do you expect? If you just want each element once, you could add the list to a set. — Tim Biegeleisen
– Tim Biegeleisen, Commented Aug 3, 2017 at 11:52
The output is 10,20,30,40,40,50,60,70,80. The 40 is getting repeated twice. I want the output to be 10,20,30,40,50,60,70,80 — Gaurav Thantry
– Gaurav Thantry, Commented Aug 3, 2017 at 11:56
@GauravThantry if you don't want array order then you can sort the array and then try your code. To sort array in asc order try this tutorial i randomly searched for sanfoundry.com/java-program-sort-array-ascending-order — Raju Sharma
– Raju Sharma, Commented Aug 3, 2017 at 11:59
Just as a note: Here at SO we have so many questions about "how to remove duplicates". I am sure that a quick research will help you solve the problem: Quick search. You can also use those Set techniques to reduce duplicates only to a specific amount. Therefore just use a Map<Element, Integer> where the keys are your elements and the values are counters that count how often you have seen the duplicate already. — Zabuzard
– Zabuzard, Commented Aug 3, 2017 at 12:04

albert_nil · Accepted Answer · 2017-08-03 12:33:43Z

3

You said that you only wanted to keep one element for those that are present 3 times or more in the list (not 2). If they were 2 or more, a TreeSet would be the only thing you would need.

You said that after being present 3 or more times, you only wanted to keep 1.

Here you are:

int[] input = whatever_your_input;

List<Integer> result = new ArrayList<>();
Map<Integer, Integer> valueCounter = new HashMap<>();

for (int i=0; i<input.length; i++) {
    int counter = valueCounter.get(input[i]);
    if (counter = null) {
        counter = 0;
    } 
    valueCounter.put(input[i], counter++);
    if (counter <3) {
        result.add(input[i]);
    }
 }

result will contain the output list.

Edit: Just to be clear, if you don't want any duplicate at all you only need this:

int[] input = whatever_your_input;

Set<Integer> result = new TreeSet<>();
for (int i=0; i<input.length; i++) {
     result.add(input[i]);
}

result will then contain your original list without any duplicate, and while keeping same sorting.

Edit2: Ok, it seems what OP wants is no duplicate at all. Also, no java Collection to be used. Here we go:

int[] input = whatever_your_input;

int[] tmpResult = new int[input.length];
int tmpLength = 0;

for (int i=0; i<input.length; i++) {
    boolean duplicated = false;
    for (int j=0; j<tmpLength; j++) {
        if (tmpResult[j] == input[i]) {
             duplicated = true;
             break;
        }
     }
     if (!duplicated) {
         tmpResult[tmpLength] = input[i];
         tmpLength++;
     }
}
int[] result = new int[tmpLength];
System.arraycopy(tmpResult, 0, result, 0, tmpLength);

result will contain values in same order and withot duplicates.

edited Aug 3, 2017 at 12:33

answered Aug 3, 2017 at 12:05

albert_nil

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6 Comments

Zabuzard Over a year ago

I am unsure if that really is what OP wants. For me the question sounds like "I am able to remove duplicate values if there are 2 of them but not if greater, how to do that?". But in any case your answer is useful in my opinion.

albert_nil Over a year ago

appreciate your feedback @Zabuza but the OP wrote "How do you detect and delete if there more than two duplicates in an array". In any case, since i felt like he only wanted to ensure no duplicates at all, i started by saying that then a TreeSet was enough, nothing more.

Gaurav Thantry Over a year ago

@albert_nil Can't we get this by using only Arrays ? I could have used the 'List' class as it doesn't allow any duplicates, but I wanted to do this by using only arrays.

albert_nil Over a year ago

for sure it can be done, but then please edit your question to reflect you wanted avoid the use of any java Collection

albert_nil Over a year ago

btw, List allow duplicates, it's Set the one that avoids that

|

Yugerten · Accepted Answer · 2017-08-03 12:09:26Z

2

int[] a = {10, 20, 30, 40, 40, 40, 50, 60, 70, 80 ,10};
Set<Integer> set = new HashSet<Integer>();
for (int i : a) {
   set.add(i);
}
a = set.toArray(new int[set.size()]);

Hope that help you!

answered Aug 3, 2017 at 12:09

Yugerten

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3 Comments

Zabuzard Over a year ago

While this answer is indeed helpful I think you could improve the quality by explaining the approach. You could explain the advantages of Sets and why they help to solve the problem.

albert_nil Over a year ago

this would remove the sorting (but not explicitely stated by OP that he wants sorting to be retained).

Yugerten Over a year ago

the idea is: the Java Set collection dont support duplication

Raju Sharma · Accepted Answer · 2017-08-03 21:38:31Z

I know why the 40 is getting repeated twice.In your loop, when i = 3 and j = 4, a[i] = a[j],so in while-loop,the array moves left,
a[4] = a[5],a[5] = a[6],a[6] = a[7], a[7] = a[8]...
then,
a[] = {10, 20, 30, 40, 40, 50, 60, 70, 80 ,10} ,size = 9,j= 9.
continues the for-loop,i = 4,j =5,but when i = 3, a[3] = a[4] = 40,so the loop can't shrink 40(repeated thrice) to once. I modified your code as shown below.

  public class ArrayDuplicate {
    public void run1()
    {
        int[] a = {10, 20, 30, 40, 40, 40, 50, 40, 60, 70, 80 ,10};
        int size=a.length;
        System.out.println("Array size before duplicate deletion "+size);
        for(int i =0;i<(size-1);i++)
        {
            for(int j=i+1;j<=(size-1);j++)
            {
                if(a[i]==a[j] &&i!=j)
                {      
                    int temp = j;
                    while(temp<(size-1))
                    {
                        a[temp]=a[temp+1];
                        temp++;
                    }
                    size--;
                    j=i;  
                }

            }
        }
        System.out.print("The array after deleting the duplicates is ");
        for(int k=0;k<=(size-1);k++)
        {
            System.out.print(a[k]);  //40 is being printed twice
            if(k<(size-1))
            {
                System.out.print(",");
            }
            else
                System.out.print(".");
        }
    }
    public static void main(String[] args)
    {
        ArrayDuplicate ob = new ArrayDuplicate();
        ob.run1();

    }
}

Raju Sharma · Accepted Answer · 2017-08-03 12:45:30Z

I think the OP doesnt want to use Collections to remove duplicates. So i just modified his code to look as below.

for (int i = 0; i < (size - 1); i++) {
                  for (int j = i + 1; j <= (size - 1); j++) {
                        if (a[i] == a[j] && i != j) {
                              while (j < (size - 1)) {
                                    a[j] = a[j + 1];
                                    j++;

                              }
                              i--;//recheck in inner for loop is performed (i--) , to check adjacent duplicate element.
                              size--;
                        }

                  }
            }

Here an recheck in inner for loop is performed (i--) , to check adjacent duplicate element.

Joginder Kumar · Accepted Answer · 2017-08-03 12:46:26Z

1

Hope below code will do.

public static void main(String[] args) {

    int[] a = { 10, 20, 30, 40, 40, 40, 50, 60, 70, 80, 10 };

    Map<Integer, Integer> m = new HashMap<Integer, Integer>();
    ArrayList<Integer> al = new ArrayList<Integer>();

    for (int i = 0; i < a.length; i++) {
        int count = 1;

        if (m.containsKey(a[i])) {

            m.put(a[i], m.get(a[i]) + 1);
        }

        else {
            m.put(a[i], count);
        }

    }

    for (int i = 0; i < a.length; i++) {

        if (m.get(a[i]) > 2) {

        } else {
            al.add(a[i]);
        }

    }
    System.out.println(al);

}

answered Aug 3, 2017 at 12:46

Joginder Kumar

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Comments

tbsalling · Accepted Answer · 2017-08-03 12:33:49Z

0

Implementations of Set cannot hold duplicates, so by inserting all values into a Set all the deplicates would be removed.

Using Java 8 Streams and Collections this could be implemented like:

// Remove all duplicates by inserting into a Set
Set<Integer> noDupes = Arrays.stream(a).boxed().collect(Collectors.toCollection(TreeSet::new));

// Set to primitive array
a = noDupes.stream().mapToInt(Integer::intValue).toArray();

then a would hold the original values minus duplicates.

answered Aug 3, 2017 at 12:33

tbsalling

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