How to find a duplicate string in a string array

Question

I have the following code where i want to retrieve only those values which are duplicate but could not find the right result.I don't know where i'm wrong.

public class TestDummy {
    public static void main(String args[]){
       String arr[] ={"lady", "bird", "is","bird","lady","cook"};
       int len = arr.length;
       System.out.println("Size "+len);
       for(int i=0 ; i<=len;i++){
           for(int j=1 ; j< len-1;j++){
            if(arr[i]==arr[j]){
              System.out.println("Duplicate "+arr[i]);
          } 
       }  
           }
    }


}

Did you try adding them to a set?. It is more efficient than your current approach which takes` O(n^2)` — TheLostMind
– TheLostMind, Commented Jun 11, 2015 at 6:17
I don't want to use set. Can you help me to know where is the error in code ? — Little bird
– Little bird, Commented Jun 11, 2015 at 6:19
Look at this stackoverflow.com/a/1937083/1737819. From the map of words with their occurrences choose those that have occurencies > 1. :) — Developer Marius Žilėnas
– Developer Marius Žilėnas, Commented Jun 11, 2015 at 6:24
What will happen when i=1 and j=1? How can you prevent i and j from having the same value, and comparing one word to itself? — James Newton
– James Newton, Commented Jun 11, 2015 at 6:32
first for has an error for(int i=0 ; i<**=**len;i++){. Can you spot it? — Developer Marius Žilėnas
– Developer Marius Žilėnas, Commented Jun 11, 2015 at 6:47

Peter Pan · Accepted Answer · 2015-06-11 06:45:32Z

3

    String arr[] ={"lady", "bird", "is","bird","lady","cook"};
    Map<String, Integer> map = new HashMap<>();
    for(String str: arr) {
        if(map.containsKey(str)) {
            map.put(str, map.get(str)+1);
        } else{
            map.put(str, 1);
        }
    }
    for(String str: map.keySet()) {
        if(map.get(str) > 1) {
            System.out.println("Duplicate: "+ str+" count:"+map.get(str));
        }
    }

output:

Duplicate: bird count:2
Duplicate: lady count:2

answered Jun 11, 2015 at 6:45

Peter Pan

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Comments

TheLostMind · Accepted Answer · 2015-06-11 06:22:06Z

1

You have to change your code to :

public static void main(String args[]) {
        String arr[] = { "lady", "bird", "is", "bird", "lady", "cook" };
        int len = arr.length;
        System.out.println("Size " + len);
        for (int i = 0; i < len; i++) { // not <= but only <
            for (int j = i + 1; j < len; j++) {  // start from i+1 and go upto last element
                if (arr[i].equals(arr[j])) { // use equals()
                    System.out.println("Duplicate " + arr[i]);
                }
            }
        }
    }

O/P :

Size 6
Duplicate lady
Duplicate bird

answered Jun 11, 2015 at 6:22

TheLostMind

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2 Comments

Little bird Over a year ago

Can i get number counts as well?

TheLostMind Over a year ago

@Littlebird - For that you will have to use a Map / array to keep count

Uma Kanth · Accepted Answer · 2015-06-11 06:19:46Z

0

Firsty Strings are to be compared with .equals() not ==

if(arr[i].equals(arr[j]))
{
   System.out.println("Duplicate "+arr[i]);
}

And I would recommend you to use sets as they dont allow duplicate values to be entered into it, or a list and check if it already exists using a .contains() method.

List<String> list = new ArrayList();

for(int i = 0; i < arr.length; i++)
   if(list.contains(arr[i])
     System.out.println("Duplicate" + arr[i]);
   else
     list.add(arr[i]);

answered Jun 11, 2015 at 6:19

Uma Kanth

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Comments

Vihar · Accepted Answer · 2015-06-11 06:20:44Z

0

you can use a simpler approach like the following

Set<String> strings=new Hashset<String>();
for(int i=0;i<len;i++){
    if(strings.add(arr[i])==false){
        System.out.println(arr[i]+" is a duplicate");
    }
}

and in your existing code do it like arr[i].equals(arr[j]) to see if values are equal(if they are equal they are duplicate)

== checks for reference equivalence, while equals() method checks for value equivalence, so you should be using equals method whenever you need to check equivalence of two objects

hope it helps!

Good luck!

answered Jun 11, 2015 at 6:20

Vihar

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Comments

Codebender · Accepted Answer · 2015-06-11 06:21:39Z

0

There are a few issues.

   for(int i=0 ; i<len;i++){
       for(int j=i +1 ; j< len;j++){
        if(arr[i].equals(arr[j])){
          System.out.println("Duplicate "+arr[i]);
      } 
   }  
       }

Note that: I have change j=1 to j=i and == to .equals and <= to < and len -1 to len

answered Jun 11, 2015 at 6:21

Codebender

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Comments

GreenEyedAndy · Accepted Answer · 2015-06-11 06:25:13Z

0

This gives you a list with the duplicates and their count:

var duplicates =
from word in arr
group word by word into g
where g.Count() > 1
select new { g.Key, Count = g.Count() };

answered Jun 11, 2015 at 6:25

GreenEyedAndy

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Comments

Developer Marius Žilėnas · Accepted Answer · 2015-06-11 06:54:18Z

0

Create a Map of your words and occurencies.

import java.util.*;

public class TestDummy {
    public static void main(String args[]) {
        String arr[] = {
            "lady", "bird", "is", "bird", "lady", "cook"
        };
        Map<String, Integer> dictionary = new TreeMap<>();

        int len = arr.length;
        System.out.println("Size " + len);
        for (int i = 0; i < len; i++) {
            if (dictionary.containsKey(arr[i])) {
                dictionary.put(arr[i], dictionary.get(arr[i]) + 1);
                System.out.format("Duplicate %s%n", arr[i]);
            } else {
                dictionary.put(arr[i], 1);
            }
        }
    }
}



**Output**
    Size 6
    Duplicate bird
    Duplicate lady

edited Jun 11, 2015 at 6:54

answered Jun 11, 2015 at 6:44

Developer Marius Žilėnas

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Comments

Mehdi · Accepted Answer · 2015-06-11 08:10:15Z

0

public static String[] removeDuplicates(String[] array){                                           
    return new HashSet<String>(Arrays.asList(array)).toArray(new String[0]);       
}

answered Jun 11, 2015 at 8:10

Mehdi

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Comments

Asheesh Kumar · Accepted Answer · 2015-06-11 11:51:36Z

0

above solutions does work
but just to make code clean you could use Linq.js

answered Jun 11, 2015 at 11:51

Asheesh Kumar

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