You have just gained a subscriber. Out of all videos, this is so far the most comprehensible explanation. Thank you kind sir!!
@nikoo28 Жыл бұрын
Welcome aboard! Your feedback and love is much appreciated. Keeps me motivated :D
@ZachDift-kc4nk2 ай бұрын
did this code work when you submitted it on leetcode?
@butteredtequilla90462 ай бұрын
@@ZachDift-kc4nk I can’t remember honestly it was a long time ago
@ngm-oe8ow11 ай бұрын
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.
@peasantfaye540311 ай бұрын
It's not like you sort an array n times in sum2, however. You sort an array once, then you iterate through it in O(n) time. You could say it's O(n + log(n)), but it's still linear time, so we simplify it to O(n).
@ngm-oe8ow11 ай бұрын
With the inbuilt sorting or any sorting algorithm it would take O(nlog(n)) not O(log(n)) time. so O(nlog(n)+n)) simplifies to O(nlog(n)). so it's not linear time. it would take linear time if you use hashing.
@peasantfaye540311 ай бұрын
True, my bad.@@ngm-oe8ow
@bobaGogo10 ай бұрын
@@peasantfaye5403 you can only solve 2 sum in O(n) with a hashmap. If you do the 2 pointer solution you will get a O(1) space complexity, but O(n*log(n)) time complexity.
@GaganSingh-zz9el7 ай бұрын
why nlogn time complexity in 2sum using 2 pointer @@bobaGogo
@user-hp9kj8qt1hАй бұрын
Girl did 2 sum and 3 sum today!! Keep going, faang is waiting for me
@nikoo28Ай бұрын
You can do it!
@anoopghildiyal64133 ай бұрын
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??
@SRC_ARENA6 күн бұрын
Using a HashMap, the complexity for TwoSum will be O(N). Without sorting only it can be done.
@RakshithVrishab-ht8vk Жыл бұрын
Simple ,yet optimized , thanks Nikhil!
@architagarwal73795 ай бұрын
Time complexity is O[n^2logn]. We have to use hashmap apprach of 2 sum if the array is not sorted by default
@nikoo284 ай бұрын
O(n^2) will be dominant
@imdarkseid2343Ай бұрын
you did well soldier,nice approach simple and elegant but the duplicates numbers in arrays has to skipped to increase the faster execution
@checkraiser1002 ай бұрын
Damn, this video made the problem super easy
@hinocenciopaulo2 ай бұрын
Best approach.
@sumitkaranjkar3819Күн бұрын
what about edge cases like all +ve or all -ve or all 0 where you have to return {}, {}, {0,0,0} respectively
@user-do6rh1ev4k Жыл бұрын
The code takes 672 Runtime.Whether it is optimal
@shabanafirdose4671 Жыл бұрын
Too good , Thank you for sharing your knowledge
@nikoo28 Жыл бұрын
glad i could help you!!
@samridhshubham810911 ай бұрын
Very Excellent solution, was stuck in this prob for hrs
@nikoo2810 ай бұрын
glad I could help
@oknokokАй бұрын
Nice explanation, setup and video quality 📸
@RN-jo8zt8 ай бұрын
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?
@plutomessi21 Жыл бұрын
thank you bhaiya , i was stuck in this since yesterday
@karthikmalode-ir5twАй бұрын
Great solution ..understandable
@supershinobi7892 Жыл бұрын
your explanation is awesome thank you brother i was not able to solve this question before your video Now i solved.
@nikoo28 Жыл бұрын
You are most welcome
@rukhsanakhan95422 ай бұрын
dil se pyaar aapko sir
@ghayoorhussain8930 Жыл бұрын
Crystal Clear Explanation Sir
@architagarwal73795 ай бұрын
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
@sysybaba42011 ай бұрын
great explanation, congratulations on putting in this effort!!!
@nikoo2810 ай бұрын
Glad you enjoyed it!
@venkateshdhanasekaran539826 күн бұрын
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?
@nikoo2826 күн бұрын
I am using a HashSet, that takes care of duplicates
@kchemutai34832 ай бұрын
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.
@mayankkumargiri12892 ай бұрын
13:06 If you're not taking any extra space at all, the space complexity should be O(1)
@nikoo28Ай бұрын
i misspoke, you are taking the space of the HashSet which has a size (n). Hence O(n). Thanks for the correction.
@user-pv8pk4lh5i Жыл бұрын
Amazing explanation 🔥
@nikoo28 Жыл бұрын
Thank you 🙌 😄
@AbhishekKumar-hi8oj11 ай бұрын
very nice explanation, before that i went through couple of video to understand properly but this time I understood . Thanks.
@nikoo2811 ай бұрын
glad I could help
@gireeswar182 ай бұрын
Perfect Video !!!
@user-qy2fm3pu2bАй бұрын
great man
@AzharKhan-e9mАй бұрын
isme ek problem hai duplicate triplets ko lekr ... triplets double print horhe
@user-gn6ld5to8m7 ай бұрын
Thank you ... ❤
@purnimahianl5678 Жыл бұрын
You explain very well 👏
@shrutisrivastava55303 ай бұрын
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?
@user-lb7kn2bh5n Жыл бұрын
Nikhil We want 4sum leetcode solution.
@omkar._.k7 ай бұрын
Perfect
@melk481119 ай бұрын
Great work Nikhil
@hoddybhaba6704 Жыл бұрын
Easy to understand!!!
@jataman1235 ай бұрын
How does this solution ensures that we don't use one value multiple times?
@nikoo285 ай бұрын
because all 3 pointers point at different indexes
@mikedelta6588 ай бұрын
Thank you
@mrepic27973 күн бұрын
thnks
@sandeepkashyap7239 Жыл бұрын
great great explanation
@harshitha592911 ай бұрын
Good evening sir Thank you for such a keen explanation. Sir can you do leetcode problems on python 😊
@ZachDift-kc4nk2 ай бұрын
does this solution actually work in leetcode? i am getting an error when i submit (not run) the code.
@nikoo282 ай бұрын
yes it does, check out the complete implementation on the Github link available in video description
@meriyemelhajoui40839 ай бұрын
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
@nikoo288 ай бұрын
Sorting the array ensures that all smaller numbers are to the left abd larger to the right. Then you look at the sum obtained…if greater than target, then you need to pick a smaller number..so just move the right pointer by 1 place instead of traversing the entire array. Saves you a lot of time.
@hajarelalit7 ай бұрын
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 !
@asr.explores9 ай бұрын
well explained
@anurag_dutt6 ай бұрын
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.❤
@Prm9065 ай бұрын
to skip duplicates thats why he uses hashset
@rishikakoul63365 ай бұрын
CAN ANYONE EXPLAIN ME WHY HE DID ELSEIF(SUM
@sumitkaranjkar3819Күн бұрын
as he explained we need to adjust the index as per the sum ex- if sum
@kshitijpandey93768 ай бұрын
Understood the solution very well. Thank you
@sysybaba42011 ай бұрын
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?
@nikoo2810 ай бұрын
it is constant space.
@Mano_Vikas8 ай бұрын
You said, for viola in between at 5:22. What does that mean? Just curious
@nikoo288 ай бұрын
Means kind of ‘wow’ as an exclamation
@user-nu2tw8bi7sАй бұрын
but how to handl eduplicates
@nikoo28Ай бұрын
that is why a hashset is used
@jayprakashjaiswal8220 Жыл бұрын
Bro please bring more video on trees and graph
@nikoo28 Жыл бұрын
i am adding more and more videos every week. I am myself limited by resources and time. Hope you understand...if you have a particular topic/question in mind, let me know..and I can add it to my video list.
@user-do6rh1ev4k Жыл бұрын
@@nikoo28 Complete binary search problem series
@nikoo286 ай бұрын
The complete playlist on graphs is now available: kzfaq.info/sun/PLFdAYMIVJQHNFJQt2eWA9Sx3R5eF32WPn
@rawatbrothers0yt9685 ай бұрын
why are you looking like young Narendra Modi
@user-ph5ek8tg5l2 ай бұрын
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
@nikoo282 ай бұрын
That is why I sort the array. :)
@user-ph5ek8tg5l2 ай бұрын
@@nikoo28 Ok. got it. Since always the sorted List is added to the Set duplicate list addition is taken care of by the Set. But we can optimize the solution to avoid trying to add even those duplicate lists. int twoSum = A[left] + A[right]; if (twoSum == sum) { triplets.add(Arrays.asList(A[i], A[left], A[right])); /** * Only if we have found a solution for two values we can be sure we should move ahead of all their * duplicates. */ while (left < right && A[left + 1] == A[left]) { ++left; } ++left; while (left < right && A[right - 1] == A[right]) { --right; } --right; }
@rgowtham9445 Жыл бұрын
good 🤩
@NeerajManoj Жыл бұрын
nice bro
@TuringTested01 Жыл бұрын
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]
@gauratomar11 ай бұрын
In this way some cases would be lost as -1, 0,1 and 2,-1, -1 are also there both use -1 and both are different trplets
@vamshigud1512 ай бұрын
stock market bear bank side
@vinaypratapsingh3394 ай бұрын
if you sort the array, you will lost the indices.
@pradeepphulse48762 ай бұрын
But you need to return the values. Not the indexes.
@SuriyaT3001 Жыл бұрын
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 ..
@nikoo28 Жыл бұрын
check the code available on github in description. It does pass on leetcode.
@ngm-oe8ow11 ай бұрын
he is storing it in a HashSet so it won't have duplicates.
@user-fe8nh9kz7q5 ай бұрын
Your videos are awesome, thanks for all the details. Can we avoid having the duplicates, like adding memorization? Is that possible?
@user-ph5ek8tg5l2 ай бұрын
You are correct. His solution is wrong. it wont remove duplicates.
@parthbhayana63267 ай бұрын
Jo bhi bolo Hairfall toh bhot hogya 2 saalo me. 😂
@nikoo286 ай бұрын
can't escaping aging 😅
@joydeep_2 ай бұрын
Sorry to say, but not the best approach.
@nikoo28Ай бұрын
what would you suggest?
@ibrahim-abdallatif7 ай бұрын
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)
@nikoo286 ай бұрын
the solution I provided on my github profile does pass leetcode.
@satyaganesh71596 ай бұрын
Set uses equals and hashcode to compare elements in it, so list1.equals(list2) compares each element sequentially