Hello guys, today, you will learn how to solve another popular coding problem. You have given an array of objects, which could be an array of integers and or array of Strings or any object which implements the Comparable interface.

This also shows how

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In the first solution, we compare each element of the array to every other element. If it matches then its duplicate and if it doesn't, then there are no duplicates. This is also known as a

The time complexity of this problem is O(n^2) or quadratic. When you give this solution to your interviewer, he will surely ask you to come up with O(n) time complexity algorithm, which we will see next.

Here is the

In this program, instead of printing the duplicate elements, we have stored them in a Set and returned from the method, but if the interviewer doesn't ask you to return duplicates, then you can simply print them into the console as I have done in next solution.

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The second solution demonstrates how you can use a suitable data structure to come up with a better algorithm to solve the same problem. If you know, in Java, the Set interface doesn't allow duplicates, and it's based upon hash table data structure, so insertion takes O(1) time in the average case.

By using HashSet, a general-purpose Set implementation, we can find duplicates in O(n) time. All you need to do is iterate over an array using advanced for loop and insert every element into HashSet. Since it allows only unique elements, add() method will fail and return false when you try to add duplicates.

Bingo, you have to find the duplicate element, just print them off to console, as shown in the following program:

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Here is the Java program to combine both solutions, you can try running this solution on Eclipse IDE and see how it works. You can also write the JUnit test to see our solution work in all cases, especially corner cases like an empty array, array with null, etc.

That's all about

Data Structures and Algorithms: Deep Dive Using Java

Algorithms and Data Structures - Part 1 and 2

Grokking Dynamic Programming Patterns for Coding Interviews

The Coding Interview Bootcamp: Algorithms + Data Structures

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**How would you find duplicate elements from an array?****Can you solve this problem in O(n) complexity?**This is actually one of the frequently asked coding questions from Java interviews. There are multiple ways to solve this problem, and you will learn two popular ways here, first the brute force way, which involves comparing each element with every other element and other which uses a hash table-like data structure to reduce the time complexity of the problem from quadratic to linear, of course by trading off some space complexity.This also shows how

**by using a suitable data structure, you can come up with a better algorithm**to solve a problem. That's why a good knowledge of Data Structure and Algorithms are very important for all programmers.If you are new into programming world or want to refresh your knowledge about essential data structures like an array, string, linked list, hash table, binary tree, balanced tree, stack, queue, priority queue, etc then I suggest you go through a comprehensive data structure and algorithms course.

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__The logic of Solution 1 - finding duplicates in O(n^2)__

In the first solution, we compare each element of the array to every other element. If it matches then its duplicate and if it doesn't, then there are no duplicates. This is also known as a __The logic of Solution 1 - finding duplicates in O(n^2)__

**brute force algorithm**to find duplicate objects from Java array.The time complexity of this problem is O(n^2) or quadratic. When you give this solution to your interviewer, he will surely ask you to come up with O(n) time complexity algorithm, which we will see next.

Here is the

**code to find duplicate elements using a brute force algorithm**in Java:In this program, instead of printing the duplicate elements, we have stored them in a Set and returned from the method, but if the interviewer doesn't ask you to return duplicates, then you can simply print them into the console as I have done in next solution.

public static Set<Integer> findDuplicates(int[] input) { Set<Integer> duplicates = new HashSet<Integer>(); for (int i = 0; i < input.length; i++) { for (int j = 1; j < input.length; j++) { if (input[i] == input[j] && i != j) { // duplicate element found duplicates.add(input[i]); break; } } } return duplicates; }

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__How to Find duplicates in array in O(n) time Complexity__

The second solution demonstrates how you can use a suitable data structure to come up with a better algorithm to solve the same problem. If you know, in Java, the Set interface doesn't allow duplicates, and it's based upon hash table data structure, so insertion takes O(1) time in the average case.By using HashSet, a general-purpose Set implementation, we can find duplicates in O(n) time. All you need to do is iterate over an array using advanced for loop and insert every element into HashSet. Since it allows only unique elements, add() method will fail and return false when you try to add duplicates.

Bingo, you have to find the duplicate element, just print them off to console, as shown in the following program:

public static <T extends Comparable<T>> void getDuplicates(T[] array) { Set<T> dupes = new HashSet<T>(); for (T i : array) { if (!dupes.add(i)) { System.out.println("Duplicate element in array is : " + i); } } }This solution also demonstrates how you can use Generics to write type-safe code in Java. This method will work on any type of Java array, like Array with Integer, Array with String or any object which implements Comparable interface, but will not work with a primitive array because they are not objects in Java.

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__Java Program to find duplicate elements in Java using Generics__

Here is the Java program to combine both solutions, you can try running this solution on Eclipse IDE and see how it works. You can also write the JUnit test to see our solution work in all cases, especially corner cases like an empty array, array with null, etc.__Java Program to find duplicate elements in Java using Generics__

import java.util.Arrays; import java.util.HashSet; import java.util.Set; import static java.lang.System.*; /** * Java Program to find duplicate elements in an array. In this program, you * will learn two solution to find duplicate elements in integer array e.g. * brute force, by using HashSet data structure. * * @author java67 */ public class DuplicatesFromArray{ public static void main(String args[]) { int[] withDuplicates = { 1, 2, 3, 1, 2, 3, 4, 5, 3, 6 }; Set<Integer> duplicates = findDuplicates(withDuplicates); out.println("input array is : " + Arrays.toString(withDuplicates)); out.println("Duplicate elements found in array are : " + duplicates); // now calling our generic method to find duplicates String[] myArray = { "ab", "cd", "ab", "de", "cd" }; out.println("input string array is : " + Arrays.toString(myArray)); getDuplicates(myArray); } /** * Complexity of this solution is O(n^2) * * @param input * @return */ public static Set<Integer> findDuplicates(int[] input) { Set<Integer> duplicates = new HashSet<Integer>(); for (int i = 0; i < input.length; i++) { for (int j = 1; j < input.length; j++) { if (input[i] == input[j] && i != j) { // duplicate element found duplicates.add(input[i]); break; } } } return duplicates; } /** * Generic method to find duplicates in array. Complexity of this method is * O(n) because we are using HashSet data structure. * * @param array * @return */ public static <T extends Comparable<T>> void getDuplicates(T[] array) { Set<T> dupes = new HashSet<T>(); for (T i : array) { if (!dupes.add(i)) { System.out.println("Duplicate element in array is : " + i); } } } } Output : input array is : [1, 2, 3, 1, 2, 3, 4, 5, 3, 6] Duplicate elements found in array are : [1, 2, 3] input string array is : [ab, cd, ab, de, cd] Duplicate element in array is : ab Duplicate element in array is : cd

That's all about

**how to find duplicate elements in an array**. You have now learned two ways to solve this problem in Java. The first solution is the brute force algorithm, which is demonstrated by finding duplicate elements on integer array, but you can use the logic to find a duplicate on any kind of array. The second solution uses the HashSet data structure to reduce the time complexity from O(n^2) to O(n), and it also shows you can write generic methods to find duplicates on any object array.**Further Learning**Data Structures and Algorithms: Deep Dive Using Java

Algorithms and Data Structures - Part 1 and 2

Grokking Dynamic Programming Patterns for Coding Interviews

The Coding Interview Bootcamp: Algorithms + Data Structures

###
__Other Coding Problems for Practice__

__Other Coding Problems for Practice__

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- How to check if a year is a leap year in Java? (answer)
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- Top 10 Programming problems from Java Interviews? (article)
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- How do you swap two integers without using a temporary variable? (solution)
- Write a program to check if a number is the power of two or not? (solution)
- How to reverse String in Java without using StringBuffer? (solution)
- Write a program to code insertion sort algorithm in Java (program)

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How about checking the array for duplicates in string and convert the duplicates to uppercase??

ReplyDeleteThe should be inputted by a user not given.