3 Ways To Notice Duplicate Elements Inwards An Array - Java

There are multiple ways to discovery duplicate elements inwards an array inwards Java in addition to nosotros volition come across 3 of them inwards this program. Solution in addition to logic shown inwards this article are generic in addition to applies to an array of whatsoever type e.g. String array or integer array or array of whatsoever object. One of the most mutual way to discovery duplicates is past times using creature strength method, which compares each chemical factor of the array to every other element. This solution has the fourth dimension complexity of O(n^2) and exclusively exists for the academic purpose. You shouldn't travel using this solution inwards the existent world. The measure way to discovery duplicate elements from an array is past times using HashSet information structure. If you lot remember, Set abstract information type doesn't allow duplicates. You tin conduct maintain payoff of this holding to filter duplicate elements. This solution has a fourth dimension complexity of O(n), every bit you lot exclusively demand to iterate over array once, exactly likewise has infinite complexity of O(n) as you lot demand to shop unique elements inwards the array.


Our third solution to discovery duplicate elements inwards an array is truly similar to our 2d solution exactly instead of using Set information construction nosotros volition utilisation hash tabular array information structure. This is a pretty proficient solution because you lot tin extend it to institute count of duplicates every bit well. In this solution, nosotros iterate over the array in addition to construct the map which stores array elements in addition to their count. Once the tabular array is build, you lot tin iterate over a hash tabular array in addition to impress out all the elements, who has a count greater than one, those are your duplicates.

How to discovery duplicates inwards Java array?

In the get-go paragraph, I conduct maintain given you lot a brief overview of 3 ways to discovery duplicate elements from Java array. Now, let's sympathise the logic behind each of those solutions inwards piffling to a greater extent than detail.

Solution 1 :

Our get-go solution is really simple. All nosotros are doing hither is to loop over an array in addition to comparison each chemical factor to every other element. For doing this, nosotros are using 2 loops, inner loop, in addition to outer loop. We are likewise making certain that nosotros are ignoring comparison of elements to itself past times checking for i != j before printing duplicates. Since nosotros are comparison every chemical factor to every other element, this solution has quadratic fourth dimension complexity i.e. O(n^2). This solution has the worst complexity inwards all 3 solutions.

 for (int i = 0; i < names.length; i++) {      for (int j = i + 1 ; j < names.length; j++) {           if (names[i].equals(names[j])) {                    // got the duplicate element           }      }  }

This interrogation is likewise really pop on programming interviews in addition to if you lot are preparing for them, I likewise propose you lot to solves problems from Cracking the Coding Interview: 150 Programming Questions in addition to Solutions. One of the best mass to gear upward for software developer interviews.

Solution 2 :

Second solution is fifty-fifty simpler than this. All you lot demand to know is that Set doesn't allow duplicates inwards Java. Which agency if you lot conduct maintain added an chemical factor into Set in addition to trying to insert duplicate chemical factor again, it volition non travel allowed. In Java, you lot tin utilisation HashSet shape to solve this problem. Just loop over array elements, insert them into HashSet using add() method in addition to banking concern lucifer provide value. If add() returns faux it agency that chemical factor is non allowed inwards the Set in addition to that is your duplicate. Here is the code sample to produce this :

 for (String name : names) {      if (set.add(name) == false) {         // your duplicate element      } }

Complexity of this solution is O(n) because you lot are exclusively going through array 1 time, exactly it likewise has infinite complexity of O(n) because of HashSet information structure, which contains your unique elements. So if an array contains 1 1 chiliad k elements, inwards worst illustration you lot would demand an HashSet to shop those 1 1 chiliad k elements.

Solution 3 :

Our tertiary solution takes payoff of some other useful information structure, hash table. All you lot demand to produce is loop through the array using enhanced for loop in addition to insert each chemical factor in addition to its count into hash table. You tin utilisation HashMap shape of JDK to solve this problem. It is the full general role hash tabular array implementation inwards Java. In lodge to construct table, you lot banking concern lucifer if hash tabular array contains the elements or not, if it is in addition to then growth the count otherwise insert chemical factor amongst count 1. Once you lot conduct maintain this tabular array ready, you lot tin iterate over hashtable in addition to impress all those keys which has values greater than one. These are your duplicate elements. This is inwards fact a really proficient solution because you lot tin extend it to institute count of duplicates every bit well. If you lot remember, I conduct maintain used this approach to find duplicate characters inwards String earlier. Here is how you lot code volition await similar :

// construct hash tabular array amongst count         for (String name : names) {             Integer count = nameAndCount.get(name);             if (count == null) {                 nameAndCount.put(name, 1);             } else {                 nameAndCount.put(name, ++count);             }         }          // Print duplicate elements from array inwards Java         Set<Entry<String, Integer>> entrySet = nameAndCount.entrySet();         for (Entry<String, Integer> entry : entrySet) {             if (entry.getValue() > 1) {                 System.out.printf("duplicate chemical factor '%s' in addition to count '%d' :", entry.getKey(), entry.getValue());             }         } 

Time complexity of this solution is O(2n) because nosotros are iterating over array twice in addition to infinite complexity is same every bit previous solution O(n). In worst illustration you lot would demand a hash tabular array amongst size of array itself.

Java Program to discovery duplicate elements inwards array

Here is our 3 solutions packed into a Java plan to discovery duplicate elements inwards array. You tin run this illustration from ascendancy occupation or Eclipse IDE, whatever suits you. Just brand certain that lift of your Java rootage file should travel same every bit your populace shape e.g. "DuplicatesInArray". I conduct maintain left chip of exercise for you, of class if you lot would similar to do. Can you lot refactor this code into methods, you lot tin produce that easily past times using extract method characteristic of IDE similar Eclipse in addition to Netbans in addition to write unit of measurement examine to banking concern lucifer the logic of each approach. This would hand you lot some practise inwards refactoring code in addition to writing JUnit tests, 2 of import attributes of a professional person programmer.

package dto;  import java.util.HashMap; import java.util.HashSet; import java.util.Map; import java.util.Map.Entry; import java.util.Set;  /**  * Java Program to discovery duplicate elements inwards an array. There are 2 immediately  * frontwards solution of this occupation first, creature strength way in addition to 2d past times using  * HashSet information structure. Influenza A virus subtype H5N1 tertiary solution, similar to 2d 1 is past times using  * hash tabular array information construction e.g. HashMap to shop count of each chemical factor in addition to  * impress chemical factor amongst count 1.  *   * @author java67  */  public class DuplicatesInArray{      public static void main(String args[]) {          String[] names = { "Java", "JavaScript", "Python", "C", "Ruby", "Java" };          // First solution : finding duplicates using creature strength method         System.out.println("Finding duplicate elements inwards array using creature strength method");         for (int i = 0; i < names.length; i++) {             for (int j = i + 1; j < names.length; j++) {                 if (names[i].equals(names[j]) ) {                    // got the duplicate element                 }             }         }          // Second solution : utilisation HashSet information construction to discovery duplicates         System.out.println("Duplicate elements from array using HashSet information structure");         Set<String> shop = new HashSet<>();          for (String lift : names) {             if (store.add(name) == false) {                 System.out.println("found a duplicate chemical factor inwards array : "                         + name);             }         }          // Third solution : using Hash tabular array information construction to discovery duplicates         System.out.println("Duplicate elements from array using hash table");         Map<String, Integer> nameAndCount = new HashMap<>();          // construct hash tabular array amongst count         for (String lift : names) {             Integer count = nameAndCount.get(name);             if (count == null) {                 nameAndCount.put(name, 1);             } else {                 nameAndCount.put(name, ++count);             }         }          // Print duplicate elements from array inwards Java         Set<Entry<String, Integer>> entrySet = nameAndCount.entrySet();         for (Entry<String, Integer> entry : entrySet) {              if (entry.getValue() > 1) {                 System.out.println("Duplicate chemical factor from array : "                         + entry.getKey());             }         }     } } Output : Finding duplicate elements inwards array using creature strength method Duplicate elements from array using HashSet information construction institute a duplicate chemical factor inwards array : Java Duplicate elements from array using hash tabular array Duplicate chemical factor from array : Java

From the output, you lot tin come across that the exclusively duplicate chemical factor from our String array, which is "Java" has been institute past times all of our 3 solutions.


That's all about how to discovery duplicate elements inwards an array inwards Java. In this tutorial, you lot conduct maintain learned 3 ways to solve this problem. The creature strength way require you lot to compare each chemical factor from array to another, thus has quadratic fourth dimension complexity. You tin optimize functioning past times using HashSet information structure, which doesn't allow duplicates. So a duplicate chemical factor is the 1 for which add() method of HashSet provide false. Our tertiary solution uses hash tabular array information construction to brand a tabular array of elements in addition to their count. Once you lot construct that table, iterate over it in addition to impress elements whose count is greater than one. This is a really proficient coding occupation in addition to oftentimes asked inwards Java Interview. It likewise shows how utilisation of a correct information construction tin better functioning of algorithm significantly.

If you lot conduct maintain similar this coding problem, you lot may likewise similar to solve next coding problems from Java Interviews :

• How to discovery all pairs on integer array whose amount is equal to given number? [solution]
• How to take away duplicates from an array inwards Java? [solution]
• How to form an array inwards house using QuickSort algorithm? [solution]
• Write a plan to discovery piece of work past times 2 numbers from an integer array? [solution]
• How produce you lot take away duplicates from array inwards place? [solution]
• How produce you lot contrary array inwards house inwards Java? [solution]
• Write a plan to discovery missing publish inwards integer array of 1 to 100? [solution]
• How to banking concern lucifer if array contains a publish inwards Java? [solution]
• How to discovery maximum in addition to minimum publish inwards unsorted array? [solution]

Further Learning
The Coding Interview Bootcamp: Algorithms + Data Structures
Data Structures in addition to Algorithms: Deep Dive Using Java
Algorithms in addition to Data Structures - Part 1 in addition to 2

Recommended books to Prepare for Software Engineer Interviews

If you lot are solving these coding problems to gear upward for software engineer chore interviews, you lot tin likewise conduct maintain a await at next books. They incorporate wealth of noesis in addition to several oftentimes asked coding problems from Java in addition to C++ interviews :

• Programming Interviews Exposed: Secrets to Landing Your Next Job (book)
• Coding Puzzles: Thinking inwards code By codingtmd (book)
• Cracking the Coding Interview: 150 Programming Questions in addition to Solutions (book)

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