You have just gained a subscriber. Out of all videos, this is so far the most comprehensible explanation. Thank you kind sir!!

If you sort the array the complexity becomes O(nLog(n)) in the 2 sum part and you said the complexity becomes o(n), but for the 3sum sorting is okay because you are reducing it to o(n^2). The explanation was quite good and understandable thanks.

Girl did 2 sum and 3 sum today!! Keep going, faang is waiting for me

At 5:30 how the time complexity is O(n)?? you have sorted the array so it would be O(nlogn) already for the sorting so how it would be O(n) for total algorithm??

Simple ,yet optimized , thanks Nikhil!

Time complexity is O[n^2logn]. We have to use hashmap apprach of 2 sum if the array is not sorted by default

you did well soldier,nice approach simple and elegant but the duplicates numbers in arrays has to skipped to increase the faster execution

Damn, this video made the problem super easy

Best approach.

what about edge cases like all +ve or all -ve or all 0 where you have to return {}, {}, {0,0,0} respectively

The code takes 672 Runtime.Whether it is optimal

Too good , Thank you for sharing your knowledge

Very Excellent solution, was stuck in this prob for hrs

Nice explanation, setup and video quality 📸

i got it that nums.length - 2 in the loop condition is to ensure that there are at least two elements to the right of the current index i even i can find triplet with for (int i = 0; i < nums.length ; i++) { insted of for (int i = 0; i < nums.length -2; i++) { so is there any reason we are writing -2?

thank you bhaiya , i was stuck in this since yesterday

Great solution ..understandable

your explanation is awesome thank you brother i was not able to solve this question before your video Now i solved.

dil se pyaar aapko sir

Crystal Clear Explanation Sir

Nice explanation but there are 2 cases over here. if the array is sorted already or the array is not sorted. if the array is sorted we can go with the approach explained by nukhil but if its not sorted the better use hashmap approach which gives O[n] TC and O[N] SC

great explanation, congratulations on putting in this effort!!!

After seeing lot many videos, finally I found the crystal clear approach.. Thank you so much brother. One question, how you handle the duplicate element in this?

Great explanation, I just wanted a clarification on the time complexity, i think we left out the time complexity for sorting. What is the time complexity for sorting, otherwise the rest is O(n^2) as explained.

13:06 If you're not taking any extra space at all, the space complexity should be O(1)

Amazing explanation 🔥

very nice explanation, before that i went through couple of video to understand properly but this time I understood . Thanks.

Perfect Video !!!

great man

isme ek problem hai duplicate triplets ko lekr ... triplets double print horhe

Thank you ... ❤

You explain very well 👏

The solution was very simple to understand thankyou so much. But i had a doubt as I implemented this on leetcode, the time it took for all the test cases was: 457ms , and 9ms solutions were also available. So I have to study the more optimum approach or it is fine for technical round or interview round?

Nikhil We want 4sum leetcode solution.

Perfect

Great work Nikhil

Easy to understand!!!

How does this solution ensures that we don't use one value multiple times?

Thank you

thnks

great great explanation

Good evening sir Thank you for such a keen explanation. Sir can you do leetcode problems on python 😊

does this solution actually work in leetcode? i am getting an error when i submit (not run) the code.

I dont see what we should sort the array ? if we gonna loop over the loop and everytime fix and try to found a sum that s equal to 0

Nice solution, I have now subscribed to your channel, very good explanation and solution. I want to clear Toptal interview, please guide me in some way if possible. Thanks !

well explained

Bhaiya, the explanation is so good but we have to skip the duplicate triplets. So, we have to do this If (i>0 && nums[i]==nums[i-1]){ Continue; } Duplicate triplets are the question [-4,-1,-1,0,1,2] are See the [-1(1 idx),0,1] and [-1(2 idx),0,1].. Thank you.❤

CAN ANYONE EXPLAIN ME WHY HE DID ELSEIF(SUM

Understood the solution very well. Thank you

you are using a set to arrive at a solution right? so how do you say that you are not using any extra space? or is it just constant space?

You said, for viola in between at 5:22. What does that mean? Just curious

but how to handl eduplicates

Bro please bring more video on trees and graph

why are you looking like young Narendra Modi

Bro, Your solution is wrong where you are assuming that HashSet will remove duplicate Lists. {1,2,3} & {3,2,1} are two different lists. They wont be considered duplicate by the HashSet as List equals method wont return equal for both. Set result = new HashSet(); result.add(Arrays.asList(1, 2, 3)); result.add(Arrays.asList(3, 2, 1)); System.out.println(result.size()); // Prints 2 NOT 1

good 🤩

nice bro

I think one more optimization you can do is this: Keeping a boolean array of "done" numbers and marking the numbers with which triplets are already made, because there could be duplicates of each number and for each of them you dont have to find the triplets because triplets would be unique, for example: if there is an array of 3000 integers all containing zero, your code would go to every zero and find the triplets using all zeros whereas the answer would just be [0,0,0]

stock market bear bank side

if you sort the array, you will lost the indices.

This is not right .. pls check it out it gives duplicate output but in question they asked only unique subsets.. while dry run of your code pls execute in leet code itself ..

Jo bhi bolo Hairfall toh bhot hogya 2 saalo me. 😂

Sorry to say, but not the best approach.

Thanks for the clear explanation, but please note that this solution allows duplicate triplets in the result, which is not correct and won't pass leetcode submission (I tried it myself), here is the correct solution after some changes: >>> EDIT: When I tried it I was using a LinkedList instead of a HashSet as shown in the video(using HashSet won't allow duplicates indeed and hence the presented solution is correct), anyway here is the correct way to solve it with LinkedList. class Solution { public List threeSum(int[] nums) { Arrays.sort(nums); List result = new LinkedList(); for (int i = 0; i < nums.length -2; i++) { if(i == 0 || (i > 0 && nums[i] != nums[i-1])) { int left = i + 1; int right = nums.length - 1; while(left < right) { int sum = nums[i] + nums[left] + nums[right]; if (sum == 0) { result.add(Arrays.asList(nums[i], nums[left], nums[right])); while(left < right && nums[left+1] == nums[left]) left++; while(left < right && nums[right-1] == nums[right]) right--; left++; right--; } else if(sum < 0) { left++; } else { right--; } } } } return result; } } // TC: O(n^2), SC: O(n)