If you are preparing for programming job interviews, then I also suggest you take a look at book, which contains 150 programming questions and solutions, good enough to do well on any programming job interviews e. These are your duplicate elements. Create one boolean duplicateElementsFound and assign it false. If the output of that list is supposed to be 4, then there is a very simple way to achieve that with a Set. How To Find Duplicate Elements In An Array In Java? This quick tutorial is going to show you how to clean up the duplicate elements from a List — first using plain Java, then Guava and finally a Java 8 Lambda-based solution.
Now, let's understand the logic behind each of those solutions in little more detail. Note: This approach is fast, but if the integer array contains large int values, then corresponding boolean array will take a lot of memory. Since we are comparing every element to every other element, this solution has quadratic time complexity i. Logic Used To Find Duplicate Characters In A String In Java : We use HashMap and Set to find the duplicate characters in a string. Sample Output : Enter number of elements : 5 Enter string 1 : 11 Enter string 2 : 22 Enter string 3 : 33 Enter string 4 : 44 Enter string 5 : 11 You have entered : 11 22 33 44 11 Duplicate elements using Brute Force method : 11 Duplicate elements using HashSet : 11 Enter number of elements : 4 Enter string 1 : aa Enter string 2 : bb Enter string 3 : cc Enter string 4 : dd You have entered : aa bb cc dd Duplicate elements using Brute Force method : No duplicate elements found.
Our third solution uses hash table data structure to make a table of elements and their count. Just make sure that name of your Java source file should be same as your public class e. This is in fact a very good solution because you can extend it to found count of duplicates as well. Now it will go inside if statement. Note: This approach requires extra overhead of sorting the array first. As pointed by Beginer , there can be addition and multiplication overflow problem with this approach.
As I said you can really test programming skill around an array of a Java programmer. In order to build map, you check if hash table contains the elements or not, if it is then increment the count otherwise insert element with count 1. Solution 2 : Second solution is even simpler than this. The approach used here is similar to method 2 of. This is in fact a very good solution because you can extend it to found count of duplicates as well.
Joshua Harris wrote: Kat Rollo wrote: So for example, in an ArrayList of strings which contains cat, cat, dog, horse, zebra, zebra, the answer should be two. That's all about how to find duplicate elements in an array in Java. And all elements occur once except two numbers which occur twice. So if an array contains 1 million elements, in worst case you would need an HashSet to store those 1 million elements. 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 problem from quadratic to linear, of course by trading off some space complexity. In Java, you can use Hash Set class to solve this problem.
No matter how many duplicates there are of a , it should only increase the duplicate count by 1. That makes the problem somewhat easy but what do you do if interviewer tells you that array contains duplicates and more than one numbers are missing? Detecting duplicate elements in Java array is another I like. You have now learned two ways to solve this problem in Java. It's been a long time since I have discussed any or , so I thought to revisit one of the most popular array based coding problem of finding missing numbers in given array. For example if your list contains: cat cat dog dog horse zebra zebra The output is 3 because cat, dog, and zebra are duplicated? But that appears not to be what you want.
You might have heard or seen this problem before on your programming job interview but there are a lot of different versions of increasing difficulty levels which interviewer normally use to confuse candidate and further test their ability to adapt to frequent changes. Call the method findDuplicateUsingHashSet to find duplicate elements using a Hash Set. There is a semicolon after the if condition, that means it is not doing anything. To find duplicate elements, we will count the frequency of each elements of array and store it in a Map. One of the most common way to find duplicates is by using brute force method, which compares each element of the array to every other element.
I hope this helps without giving too much away. Btw, if you need more such questions, can check the from Udemy. The performance of this method is very low if an array contains lots of elements. You can take advantage of this property to filter duplicate elements. If element is not added because the HashSet already contains the element, then add method returns false. Not being allowed to use Set does not require you to do a bad implementation.
This approach should be used only when the maximum possible element value is not too large. We can use the same approach to find all the missing numbers in the list. Im having a conflict with using text file as input, containing the same array elements. Duplicate elements using HashSet : No duplicate elements found. There are many methods through which you can find the duplicate elements in the given array. It also shows how use of a right data structure can improve performance of algorithm significantly. This method takes one String array as input.
So if an array contains 1 million elements, in worst case you would need an Hash Set to store those 1 million elements. . Print all the duplicate elements. If the gap is more than 1, then the current+1 number is missing. If add returns false it means that element is not allowed in the Set and that is your duplicate. The teacher has a register with names of all students, he goes through the list and mark absences on red.