LeetCode Contains Duplicate Solution Explained - Java

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[WalkWithM]

I am beginner and I started leetcode literally three days ago and even some easy problems a bit hard for me but the more practice I do I feel more that I am becoming better. Thank you for your content

[shivashisharora]

HashSet approach can be simplified further. Just call the "add" method on the set for each element in the array and if "add" returns false (meaning the set already has the element) return true.

[prakashkrishna]

class Solution {
public boolean containsDuplicate(int[] nums) {
HashSet<Integer> data=new HashSet<Integer>();
for(int i=0;i<nums.length;i++){
if(data.add(nums[i])==false)
return true;
}
return false;
}
}

[vincentwang]

If the hashset method is supposed to be faster than the sorting method, then why is the the sorting method faster than the hashset method based on the runtime submissions?

[vikas]

It's TIMSORT, .sort() in Java. Thanks for valuable content

[Booyamakashi]

Its much easier in JS:
const unique = new Set(arr);
return unique.size !== arr.length;

[alxx]

How does that loop work when reaching the last number? Comparing to i + 1 is null right?

[PeterKoenen]

In python I just did this:
return len(set(nums)) != len(nums)
Is there a reason not to do it like this?

[giorgi]

How about nested loop? starting from i = 0, j = i + 1 etc

[KaisarAnvar]

“Too easy”… me sitting here for half an hour trying to solve this stuff 😭

[coolwhip-ucqt]

shouldn't it be for (int i = 0; i < nums.length ; i++){... for the second solution. cause say arr length=4, it checks 0 thru 3, 3 being index of fourth element?
P.S I'm a beginner

[edmund]

Why did hashset take longer (19ms) when it's O(n) compared to sorting O(nlogn) 7ms? Thanks for the simple solutions, I couldn't quite figure out how to do it without 2 for loops lol

[eboselumeenahoro]

Love the video, want to ask why do you add the element to the hash set if its not found

[hamsalekhavenkatesh]

Use XOR with index, will be quicker and efficient

[shivahere]

What if we add binary bits and capture %2 for the sum of bits in each place value.. finally of we get 0 no duplicates

[Ankushbindlish]

Sorting method can itself check for duplicate : insertion sort, bubble sort etc

[ericsodt]

why not put all nums in a Set and then compare the size of the Set to the size of the nums array. If they're the same, no dupes, else dupes.

[Shreyas-smjt]

Thanks for the video! Would really appreciate if you could increase the volume :)

[theinvestorengineer]

Love your work! How do you record your content? Both camera and screen?

[SK-ybbx]

The built in sort in Java is quicksort .