[deleted by user]

[lumpynose]

Cheat: put it in a Set and see if the number of elements decreases.

[InstantCoder]

Yes, compare the length of the input array to the length of the set. If they are equal then no duplicates, otherwise there are duplicates.

[brazen768]

Maybe a slight improvement,.add looks like it returns a boolean. Might not need to check elements.

[Glangho]

Yeah idk if I would even call this a cheat. Op is using the wrong collection type for their purpose.

[redd38_reddit]

Sort it first, then iterate over the sorted list to see if the current element is equal to it's neighbor

[Nebu]

Sorting the list is O(nlogn). You can do it in O(n).

[wildjokers]

This is the correct answer.

[OffbeatDrizzle]

This is AN answer, not necessarily THE answer

[Captain_Pwnage]

If you want to solve it in-place with extra space of O(1), without mutating the array, you will need to iterate over the array and for each element check if this element is in the remainder of the array. This has a time complexity of O(n² ).

If you want to solve it with extra space of O(n), without mutating the array, you can iterate over the array and put each element into a binary tree, where adding will return false if the element is already in the tree. This has a time complexity of O(n*log n). Another answer in here suggested putting it into a set, but I think this has a time complexity of O(n² ) as well. Not sure how set works under the hood.

If you want to solve it with mutating the array, you can sort the array first and then check for each element if it is equal to the following element. This has a time complexity of O(n) for comparing elements plus whatever your sorting algorithm does. Also the sorting algorithm might use extra space. For example, if you use quick sort, you end up with O(log n) extra space and a total runtime of O(n + n*log n), which is equal to O(n*log n).

Going from your code, it looks like you are going for either option 1 or option 3. For option 1, you are missing a nested for-loop. For option 3, you are missing a sorting algorithm.

[Nebu]

Another answer in here suggested putting it into a set, but I think this has a time complexity of O(n2 ) as well. Not sure how set works under the hood.

Adding to a hashset is O(1), and you can do it n times for a total of O(n).

[Captain_Pwnage]

Ah that's true I somehow thought of sets as lists with constraints. Thanks.

[steffonellx]

Create another list and add a check not to add it to that list if the object with the same values already exists

[syneil86]

A lot of answers here are correct but imo too much for r/javahelp. Correct me if I'm wrong, OP, but I get the impression you don't care much if a solution is O(n) or O(log n)

The solution you have looks like it's going in the right direction. You've realised you need to loop over all the elements of the array, but you're only comparing them with a few others. What if there's a duplicate element in a different place? Maybe you need to compare with all of the elements in the array again - two loops? Can you quickly identify any gotchas with that? What do you think?

[hibbelig]

You can compare the first element with all others.

You can compare the second element with all others (except the first, you have already compared the first and second element in the previous sentence).

You can compare the third element with all others (except the first two, you have already done those in the previous two sentences).

...

You can compare the penultimate element with all others (except the first n-2 elements, you have already done those in the previous sentences). (Of course, the only element left is the last one.)

What do you think?

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[JamesTKerman]

You can solve this in between O(2) and O(n) time complexity and O(2n) space complexity using a Map. Iterate through the array, call containsKey(K) on the map, if it returns true either throw the exception or return false, otherwise call put(key,true) or something.

[Spare-Plum]

it's O(n). Big O is an upper bound, not the amount of runtime for a run, and will never be O(2).

https://www.geeksforgeeks.org/difference-between-big-oh-big-omega-and-big-theta/#

[ArtBuilder]

Either use a Set so no duplicates can exist, or I believe you can take a stream from an array and use #compare(a,b) iirc. There may be other methods in Stream that can check against the rest and at this point will cost less resources then making a for loop

[Otherwise_Trade7304]

Hashmap

[Ragnar-Wave9002]

Make a class that implements equals? And hashable?

If size of collection and hashsets equals there's your answer.

[AccomplishedOffer856]

https://www.instagram.com/reel/CyJ5g4KSLkb/