first time ever saw the face of legend thank you for your tutorials !!

@techdose4u

2 жыл бұрын

Welcome 😀

best explanation on youtube....Everyone on youtube just expalins brutforce approach but no one shows the code like you...HATS OFF

I might not have thought about the complexity of chained elements of hashmap in the interviews. Great!

@techdose4u

3 жыл бұрын

Now you know tha analysis :)

@spetsnaz_2

3 жыл бұрын

@@techdose4u yea :D

@techdose4u

3 жыл бұрын

:)

Lovely indeed ! thankyou sir for this.

you are amazing. thank you for the detailed explanation and clear thought and explanation process.

@techdose4u

2 жыл бұрын

Welcome :)

Bro, Aren't you getting overflow error when u submit the code in leetcode if we go with method 2?

sir i like prefer these type of videos , as in from the rest of your videos that i saw , this one has your face , and i dont know why but these type of videos where i can see the instructors face , help me have better understanding

@techdose4u

Жыл бұрын

Noted :)

Thank you!!! This was a great explanation!

in second approach , when sum= target u hv written, left=left+1, what if there is duplicate element in while loop also, how to check for dupliacte element in the innermost (while) loop

Thank you so much buddy. Your video is really clear and very good explained... in details! thank you

@techdose4u

3 жыл бұрын

Welcome :)

Hello sir, your explanation is great. I want to know which software are you using to write on the screen. And also what is your set up, like are you using a tablet and writing on it using stylus or you are writing on the laptop using the mouse cursor?

can someone please suggest best way to implement hashing function for implementing 3rd approach in java as time complexity of custom HashSet depends on how best we can override the hashCode method

thank you so much

First, thanks for all the awesome videos. There could be another approach where it would take O(N^3) time if we first sort the array and then take O(N^2) two sums and iter over the original array to find K - ( i th two sum ) exist in the array and that should take O(N) time for each such operation. So the total complexity should be O(N^3)

@apoorvbedmutha457

Жыл бұрын

this approach gives TLE on leetcode

Thanks for the explanation! In my opinion the easiest one was the solution with hashtable.

This is an amazing explanation. Thank you.

@techdose4u

2 жыл бұрын

Welcome 😊

In the second approach I think we don't need to use a set to handle the duplicates as we can simply compare the last added quadruple with the current one as you did in the 3rd approach.

Thanks for taking the time to explain the ideas. However, the first continue optimization doesn't seem to be correct. It would not let the 4 pointers work in tandem. Anyway, the set will take care of the duplicates. I have yet to implement the 2 sum one.

Thank you sir. how much time it takes you to make a one video from code explanation to code writing?

@techdose4u

3 жыл бұрын

3 days :)

i found that it can done in o(n^3) .... Loop through i and j and doing a two pointer approach

@techdose4u

3 жыл бұрын

👍🏼

A question plz. At 5:14, you said that the big O for set operation is O(lg N). is that true ? Time Complexity of HashSet Operations: The underlying data structure for HashSet is hashtable. So amortize (average or usual case) time complexity for add, remove and look-up (contains method) operation of HashSet takes O(1) time. For HashMap.containsKey(), it should be O(1) too. right ?

@techdose4u

3 жыл бұрын

Avg is O(1) but WC is O(logN)

@techdose4u

3 жыл бұрын

@@user-oy4kf5wr8l no need to make things complex. Just say O(1). If interviewer asks in detail then explain :)

@techdose4u

3 жыл бұрын

@@user-oy4kf5wr8l I am not smart 😂

Thank you Tech dose.

@techdose4u

3 жыл бұрын

Welcome :)

Great explanation sir..thanks for the explanation.

@techdose4u

3 жыл бұрын

Welcome :)

awesome!!!!!!!!!!!!!!

Really! Amazing video..... One more added in liked video.

@techdose4u

2 жыл бұрын

Great 🔥

Sir can you explain why we have kept i

@techdose4u

3 жыл бұрын

No. Because a set of 4 elements should have unique indexed elements.

in method 2 if sum == tar why we are only moving only lefft++ , why not r++ or both can anyone explain

great explanation !

@techdose4u

3 жыл бұрын

Thanks

Insertion in set is considered O(1) amortized, not O(log(n)).

@matrixtoogood5601

2 жыл бұрын

In C++, you cannot insert a vector in the unordered_set without providing a hash function. Insertion in unordered_set is O(1) but insertion in set is O(logN)

if make set at last in method 2 then time complexity will be O(n3 +nlogN)=O(n3) , am I right?

Reminder ON :)

@techdose4u

3 жыл бұрын

Great :)

How come insertion in set is LogN. Isn't it amortized O(1)?

@techdose4u

3 жыл бұрын

Avg O(1)

Great explanation

@techdose4u

2 жыл бұрын

Thanks ☺️

Few days earlier same problem has been asked in DAILYHUNT. I couldn't solve😭

@techdose4u

3 жыл бұрын

Ohhh....no problem. Maybe this will help in future.

It will be nice if you could mention in which playlist you have a certain video in description of every video

@techdose4u

2 жыл бұрын

👍🏼

finally wait is over.

@techdose4u

3 жыл бұрын

I am currently very busy so couldn't really post frequently. I will try to post atleast 1 per week.

2nd approach can be done without using set, which will make our time complexity O(n^3) instead of O(n^3)log(n) with this check while(start while(start This will make sure we don't check 2 sum for same values. Since array is sorted all same values will be adjacent. So skip the same values until we find new value Code - class Solution { public: vector fourSum(vector& arr, int target) { vector ans; int n = arr.size(); if(n

excellent :)

@techdose4u

3 жыл бұрын

Thanks :)

sec one was good

love from USA

@techdose4u

2 жыл бұрын

Thanks 😊

Thank you so much for such wonderful videos, these videos have helped me learn so much in so little time!

Sir there is 2 playlist of hash... One is hashing with 5 videos and one is with more than 10-15 videos... I have done this 5 video playlist so is the other playlist continuation of this playlist

@techdose4u

3 жыл бұрын

No it's not. Other playlist has only problems. It's not a continuation, other videos are pending

I am getting signed integer overflow bro for the sum value

tq anna

@techdose4u

3 жыл бұрын

Welcome :)

👌👌

@techdose4u

3 жыл бұрын

👍🏼

why are you using n -4, n-3, n-2, n-1 where does the -4 come from

please solve Water Connection Problem on gfg

Nice

@techdose4u

3 жыл бұрын

:)

Plz make this video for python as well

@ismail8973

3 жыл бұрын

DSA Concepts are same whatever be the language doesn't matter.

i am getting runtime error: signed integer overflow: in this line sum=nums[i]+nums[j]+nums[left]+nums[right];

@MantuKumar-xu3nf

2 жыл бұрын

Define long long int sum,sum1,sum2; instead of int sum and sum1=nums[i]+nums[j]; sum2=nums[left]+nums[right]; sum=sum1+sum2; It will not give you any error

This an amazing explanation. How can i contract with you brother?

I love you! I love you!

@techdose4u

2 жыл бұрын

😊

var sum = target - (arr[i] + arr[j]);

Code translated to Java: public class Sum4 { public static List fourSum(int[] nums, int target) { List resultado = new ArrayList(); int n = nums.length; if(n return resultado; Arrays.sort(nums); Map mapa = new HashMap(); for(int i = 0; i for(int j = i + 1; j List lista = new ArrayList(); lista.add(new Pair(i, j)); mapa.put(nums[i] + nums[j], lista); } } for(int i = 0; i if(i > 0 && nums[i] == nums[i-1]) continue; for(int j = i + 1; j if(j > i + 1 && nums[j] == nums[j-1]) continue; int sum = target - nums[i] - nums[j]; if(mapa.get(sum) != null) { for (Pair pair : mapa.get(sum)) { int k = pair.getIndex1(); int l = pair.getIndex2(); if(k > j) { if(!resultado.isEmpty() && resultadoContainsCurrentValues(resultado, nums[i], nums[j], nums[k], nums[l])) { continue; } List temp = new ArrayList(); temp.add(nums[i]); temp.add(nums[j]); temp.add(nums[k]); temp.add(nums[l]); resultado.add(temp); } } } } } return resultado; } private static boolean resultadoContainsCurrentValues(List resultado, int iValue, int jValue, int kValue, int lValue) { for (List list : resultado) { if(list.get(0) == iValue && list.get(1) == jValue && list.get(2) == kValue && list.get(3) == lValue) { return true; } } return false; } public static void main(String[] args) { System.out.println(fourSum(new int[]{2,2,2,2,2}, 8)); } static class Pair { int index1; int index2; public Pair(int index1, int index2) { super(); this.index1 = index1; this.index2 = index2; } public int getIndex1() { return index1; } public void setIndex1(int index1) { this.index1 = index1; } public int getIndex2() { return index2; } public void setIndex2(int index2) { this.index2 = index2; } } }

Please explain some hard questions in leetcode

@techdose4u

3 жыл бұрын

I am explaining topicwise. Whether hard or medium. It shall be done.

Please explain how the 3 approach takes n^4, I think it takes n^3

@techdose4u

3 жыл бұрын

If chain length is N^2 and (Target-2sum) = 2sum then you will get N^4.

sir this can be even done through backtracking ryt ❔

@samprasdsouza6993

3 жыл бұрын

will give tle

Coool

@techdose4u

3 жыл бұрын

:)

Second approach 5:27

Pls read the leetcode problem carefully. Unique quadruplets doesn't mean unique i,j,k, and l. It means unique n[i], n[j], n[k], and n[l]. Your code is correct but you should have mentioned this way in explanation. Thanks for the answer though.

@ranirathore4176

2 жыл бұрын

yes

some questions about the complexities, for set insertion, hashset insertion should be constant time, isn't it? same constant time for get. The 3rd approach - generating all the two-sum takes n^2, plus looping through the sums, it's 2* n^2, may not be (n^2 * n ^2)

@techdose4u

3 жыл бұрын

O(1) is amortized time, meaning average time. But worst case is not the same. For the 3rd approach, your chain length can go to N^2 and you might have to traverse the entire chain for each 2sum. Possible 2sums are N^2. So time is N^4.

@sachinthakan8211

3 жыл бұрын

@@techdose4u I think xian is right, even if we consider the worst-case we will find all two sums for a single pair (i, j), and the rest of the pairs will do nothing. So we will traverse all N2 pairs and in the worst case, we will end up including all two sums from the map to our answer. So-net operations are the sum of all operations needed to fetch two sums from the map, which is still O(N^2) EDIT: though making map initially can take O(N^4) in worst case

I would like to join your live course, could you please help me to register?

@techdose4u

3 жыл бұрын

Yep. You can register when the forms are released.

Your first code is not running

I guess, Approach 3 will fail in case all elements are same?

@ismail8973

3 жыл бұрын

No it will work even if all elements are same. But this is worst case situation and the time complexity will become O(N^4)

link to buy the mic ?

BTW MOST OPTIMAL SOLUTION USES NO DATA STRUCTURE ONLY 2 POINTERS

I have done using the second algorithm.But my set is containing duplicate vectors.How is this possible xD

@Mandeepsingh-jo5cf

3 жыл бұрын

did you sort.

@sameerswankar

Жыл бұрын

same

I change speed from 2x to normal I was not able to identify his voice

@techdose4u

2 жыл бұрын

😂