How to Find Duplicate Elements in an Array - Java Program | Java Interview Question and Answer

How to Find Duplicate Elements in an Array - Java Program | Java Interview Question and Answer | Test Automation Central
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[@akaki_khotcholava]

Time complexity is O(n**2). You can easily use HasMap and one for loop. I can provide JS code. And time complexity for this code will be O(n) function findDuplicates(arr) { const elementCount = {}; const duplicates = []; arr.forEach((element) => { // Initialize count for the element or increment existing count elementCount[element] = (elementCount[element] || 0) + 1; // Check if the element is a duplicate and hasn't been added to duplicates array yet if (elementCount[element] === 2) { duplicates.push(element); } }); return duplicates; }

[@JohnWickea]

lol your code is worse . check your space complexity xD. I can solve all this in O(n) lol.

[@akaki_khotcholava]

@@JohnWickea You must be a genius.

[@ladro_magro5737]

taking extra space is also not a good solution there is a much better way to handle this case with no extra space and O(n) time complexity

[@JohnWickea]

@@ladro_magro5737 let the kimd live in it's own universe, since he's not listening to anyone 😂

[@ahmedshaltout4437]

any operation on hashmap cost O(log N) time

[@ViktorTheRook]

U can sort array, iterate array and check array index i and i +1 until end of array… constant space and O(nlogn) because of sort…. Or use hash map for linear time but also linear space

[@cicartaya]

I'm just now learning about data structures and algorithms in C++. Correct me if I'm wrong, but I think this approach has a time complexity of O(n^2) because of the nested for loops, which is not that good if the arrays are really large. Is there a better/more efficient way to find duplicates in an array in Java?

[@sergeiivanov5739]

Sets...

[@electricity2703]

It takes combination(array.size(), 2) iteration

[@arsalanalam9622]

It doesn't work when you have multiple duplicate elements in same array

[@vedantjoshi2093]

No need for two for loops..it can be done in one

[@anubhav.codess]

Using hashmap and print the elements whose frequency is greater than 1 easy pezzy

[@theanimemanofindia8036]

Bro it is for the people who are only say core Java experience not collection

[@Pearl0812]

How to remove that??

[@crosswalker45]

Instead of brute force, try to implement other approach like hsshmap.

[@mySofCh786]

Don't use 2 for loop, don't use set/hashset

[@rgbgaming11]

We can simply use stream for this

[@prady1006]

In interview they will tell u stream is doing fine. Create your logic then you have to learn basic bro

[@jasonosunkoya]

Just use a hashmap,

[@nikhilraj1812]

Hey bro can we use two pointer for this question?👀

[@josejoy174]

why not just use "for each " element?..

[@eccomercebeast8911]

or you can use list arr.stream().filter(x -> x == 1).toList();

[@ChatGPT......]

😢wah bhi

[@akashsarjine]

Use distinct

[@rubenbarkhudaryan5606]

it doesnt work for 10000 elements arra

[@Thd369]

Good

[@dekhlohhgftgdrf6038]

What if all numbers were same let's say 1

[@Omkar_manjare]

Not optimized solution

[@nihanshupethe8983]

Bc 🔥🔥

[@shaikafrid2186]

It will fail for some edge cases

[@subhojitbiswas3876]

Ye s like if in the last if u have one more 1

[@bikramprajapati5358]

Data structure sikh pehle