4 Sum | Brute - Better - Optimal with Codes

[@takeUforward]

Please watch our new video on the same topic: kzbin.info/www/bejne/m3WcZoqIm81rZ8U

[@uniqueyuvrajgaming3392]

You said it is O(n3) but sorting will take some time to what about that?

[@janardhan2jordan]

​@@uniqueyuvrajgaming3392nlog n is basically negligible in comparision of n3 thats why is it not included

[@TaimoorKhan-h6l]

Sir,I have one confusion about the time complexity. Inside the innermost loop we have the insert operation in set which take O(logN) in the worst case considering we have n elements inside the side. If we are taking the worst case means that the if(sum==target) is true for each condition.Then Why we are not considering it's complexity. According to my calculation,Time Complexity is O(N^3(logN)). Please sir reply. Thank you in Advance.

[@TaimoorKhan-h6l]

@@uniqueyuvrajgaming3392 Sorting will have always 4 elements which will take constant time.

[@mohitshahu4601]

​@@TaimoorKhan-h6l do you want to use ordered set?

[@shailesh_rajpurohit]

Note: 0:42 Make sure you watch the 3Sum before doing the 4Sum . 🙂

[@TrashAccount-ye3ul]

What if I have seen since a long ago, will it work ?🙃 But why he was laughing while saying this ?

[@nikhilKumar-xv4mt]

😂😂

[@kanhiyasharma3648]

naughty audience😂😂😂😂

[@MandeepSingh-rr8kk]

ok 🙂

[@ashishashish1403]

Whole video is about 4 sum but 0:40 to 0:50 is wholesome

[@yashsharma6396]

Excited to do 4sum after completing 3sum yesterday

[@gareer2436]

Damn bro include me next time

[@yashsharma6396]

​@@gareer2436 you know how to bend?😂😂

[@gregorirasputin659]

Yaar tum log bohat tharki ho 😂

[@gareer2436]

@@yashsharma6396 no but I know what to do when someone bends...😏 Lay them down and.. Massage their backs of course!

[@gareer2436]

@@gregorirasputin659 ma to sath me problem solve krne ki baat kr raha tha bro... tum kya soch liye?😏

[@Akash-yr2if]

0:44 Striver's Smile Says it all

[@brijmohan_f_s]

you got my point bro😁

[@AkOp-bf9vm]

bhai bhai🫡🫡😂. what a detailing bro

[@guneeshvats46]

Damn I was about to comment this

[@SIVAKUMAR-bj2vl]

Hhahahahahahaha

[@jatinsareen7771]

3 sum was awesome, 4 sum was fantastic!!!!

[@suyashrahatekar4964]

legends like 5 sum

[@Akash-yr2if]

Ultra Legends like 6 sum

[@takeUforward]

Do give us a like and subscribe, it won't cost you anything, but it will highly motivate us to create more such type of videos :)

[@rishav144]

striver ..king of DSA

[@shra1]

The if condition inside while to check if(k

[@codemania2878]

sir which whiteboard pen app you use for explain the code

[@itsmrcrazy7670]

hey striver you forgot to add time complexity required for --sorting array --optimal approach so it will be O(NlogN) + O(N^2). plz update

[@BhaveshKumar-z1h]

@@codemania2878 He is using an iPad and its default Notes app​

[@shubhamagarwal1434]

#Free Education For All.. # Bhishma Pitamah of DSA...You could have earned in lacs by putting it as paid couses on udamey or any other elaerning portals, but you decided to make it free...it requires a greate sacrifice and a feeling of giving back to community, there might be very few peope in world who does this...."विद्या का दान ही सर्वोत्तम दान होता है" Hats Off to you man, Salute from 10+ yrs exp guy from BLR, India.

[@arujgarg7267]

Finally did 4 sum, amazing experience

[@gareer2436]

Can you elaborate your experience? Asking for a friend

[@catalyst8654]

can you explain about your experience...🙂🙂

[@leisuretime4198]

@@catalyst8654 🤣🤣

[@Akash-yr2if]

Would you take a moment to share your experience with us.

[@cinvest2411]

You should not take I upto

[@SasukeUchiha-pl8le]

it doesn't create a big difference

[@selvaarumugam370]

the way you explain the two pointer approaches and the use of them makes two pointer approach the easiest way to solve a problem. I fell in love the way you are teaching these concepts.

[@mehulthuletiya497]

00:54 Problem Statement 02:24 Brute force approach 03:08 Code 05:38 Complexity 06:14 Better approach 11:31 Code 13:22 Complexity 14:54 Optimal approach 21:59 Code 27:04 Complexity

[@takeUforward]

You watch all the videos? I see you daily updating me with timestamps, massive thanks for that.

[@mehulthuletiya497]

​​@@takeUforward Yes, I watch the all the videos. After the watching video i comment the Timestamp. Recursion and Backtracking series is the one of the best explanation and understanding with your teaching style. I completed 2 days ago.

[@salihedneer8975]

@@takeUforward striver please update timestamp in disucssion so that ppl can access it easily

[@dishankbaid05]

The consistency level of striver is really the source of motivation 🔥.

[@impalash_ag]

Hi Raj, there are 2 slight mistakes in your optimal solution. 1: The 1st for loop will run till n-3 and 2nd for loop will run till n-2 instead of n, because we need to allocate last 2 elements to our two pointers and as per your code when i=n-2, j becomes n-1 (j=i+1=n-2+1) and num[k] throws out of bound exception, because k becomes n(k=j+1) 2: We could also increment low and decrement high when sumtarget respectively until there are duplicates for code to be more optimised. 3: Here's the more readable code(JAVA): class Solution { public List fourSum(int[] nums, int target) { int n = nums.length; List result = new ArrayList(); Arrays.sort(nums); for(int i=0; i

[@_Arunvfx]

hatsoff bro i was confused for 4 hours for optimal solution that passes all test cases.after seeing your comment it helped me to understand it.thankyou :)

[@impalash_ag]

@@_Arunvfx glad you finally understood it. Keep pushing forward 😃

[@rockpaperscissors1781]

Afaik, it won't throw out of bound exception, because there is a check being done (k < l). In ur case, k takes n and l takes n - 1.

[@shubhrajyotipoddar1684]

bro couldn't control his laugh during 3sum , 4sum intro and had to cut the clip 🤣

[@omkarsawant9267]

Time Complexity: O(n log n) Space Complexity: O(n) #include #include #include using namespace std; vector fourSum(vector &nums, int target) { // sort the array sort(nums.begin(), nums.end()); // initialize result vector vector result; int n = nums.size(); // iterate through array with two pointers for (int i = 0; i < n; i++) { // skip duplicate elements if (i > 0 && nums[i] == nums[i - 1]) { continue; } for (int j = i + 1; j < n - 2; j++) { // skip duplicate elements if (j > i + 1 && nums[j] == nums[j - 1]) { continue; } // use two pointers for remaining elements int left = j + 1; int right = n - 1; while (left < right) { int currentSum = nums[i] + nums[j] + nums[left] + nums[right]; if (currentSum == target) { result.push_back({nums[i], nums[j], nums[left], nums[right]}); // skip duplicate elements while (left < right && nums[left] == nums[left + 1]) { left++; } while (left < right && nums[right] == nums[right - 1]) { right--; } // Move pointers left++; right--; } else if (currentSum < target) { // If the sum is less than the target, move the left pointer to the right left++; } else { // If the sum is greater than the target, move the right pointer to the left right--; } } } } return result; } int main() { // Example usage vector nums = {1, 0, -1, 0, -2, 2}; int target = 0; // Call the fourSum function vector result = fourSum(nums, target); // Print the unique quadruplets for (const auto &quadruplet : result) { cout

[@akashchoudhary6876]

kuch bhi, same solution hai upar wala time complexity n3 hai, apki nlogn kaise aa gayi

[@omkarsawant9267]

@@akashchoudhary6876 the overall time complexity of the fourSum function is O(nlog⁡n+n^3)which simplifies to O(n^3)in Big O notation, assuming the input is sufficiently large.

[@omkarsawant9267]

@@akashchoudhary6876 sorting in cpp will take O(nlogn)

[@THEReFleX]

@@akashchoudhary6876 hahahahah

[@THUNDER-kw3wq]

Striverr bhai ke liye toh 10some bhi asaan hai...Jokes apart hes the best teacher on youtube

[@AbhimanyuKumar-pw3je]

if u are stuck at test case 281 in leet code then :- class Solution { public: vector fourSum(vector& nums, int target) { int n =nums.size(); vector ans; sort(nums.begin(),nums.end()); for(auto it:nums) cout

[@hareshnayak7302]

Understood,Thanks striver for this amazing video.

[@shubha_jagadeesh]

fell in love with your way of explanation❤

[@asgarh4589]

People who are aware of 2 pointer approach, watch from 14:56, saves a lot of time.

[@ankitadas5833]

Though I am little bit late, in this A to Z DSA Playlist But I covered All previous Videos within a week & now just completed 4 Sum. Striver Sir You are great! You will going to create a history in DSA course very soon around all over the World … I don't have the proper words to express my gratitude to you. All the Best Sir God bless you .

[@ankitadas5833]

@@ArjunS-hi3vl Yes

[@DR-mq1le]

have you completed the course ? im also quite late in starting the course and moving at about the same pace as you

[@ankitsharda1131]

@ankitadas5833 hey I would like to know if you have completed entire striver a2z dsa course? If yes, how much time did you take to complete entire course?

[@bkmeher9005]

uffffffff.... understood , bahut bhala se bujhi hela bhai

[@praveennemagoudar3442]

wow,....wow....wah..awesome expaination. undestood bro..hatsoff to you.

[@gauravsoni5424]

best explanation on you tube

[@MdKaif-bf3nz]

Thank you so much Striver brother, you taught the 3sum and 4sum very clearly. 💀💀 ok jokes apart lol, this series is just too good, i can literally find every concept DSA related a lot of questions of each and every concept.

[@Mythri333]

Ahaaa... excellent explanation 😊

[@kaichang8186]

understood, thanks for the perfect explanation

[@helloworld2054]

This has to be the horniest coding problem 🌚

[@jatinukey4062]

😂😂😂😂

[@adebisisheriff159]

Striver..... you are amazing.................................... Thanks so much bro

[@sarangkumarsingh7901]

Awesome Lecture Sir............

[@ra-onegaming1383]

I have done this question using subsequence technique thanks u raj sir 😊 u teach really simple

[@footballcreativeeverywhere260]

Great bhaiya ❤️🇮🇳 , ready for next one

[@VishalGupta-xw2rp]

After doing 4 Sum.... I think Sky is the limit 🌝

[@kanyapandey]

Thanks a ton for explaining in simplest way 🙏🙏🙏🙏🙏🙏

[@akankshajain3997]

helpful to understand integer overflow

[@gregorirasputin659]

Nice explanation. But the title is a bit too sus don't you think?

[@isheep9025]

instead of taking elemens between j and k we can also take hashset elements to be between i and j it will also give coreect ans #include vector fourSum(vector& nums, int target) { int n = nums.size(); //size of the array set st; //checking all possible quadruplets: for (int i = 0; i < n; i++) { set hashset; for (int j = i + 1; j < n; j++) { for (int k = j + 1; k < n; k++) { // taking bigger data type // to avoid integer overflow: long long sum = nums[i] + nums[j]; sum += nums[k]; long long fourth = target - sum; if (hashset.find(fourth) != hashset.end()) { vector temp = {nums[i], nums[j], nums[k], (int)(fourth)}; sort(temp.begin(), temp.end()); st.insert(temp); } // put the kth element into the hashset: } hashset.insert(nums[j]); } } vector ans(st.begin(), st.end()); return ans; }

[@arunmahajan1028]

After Understanding this concept , I can Say that Sky is the limit..

[@tejaspatel2212]

Imagine someone getting your video, on Searching 3sum video or 4sum video on google 🌚😂

[@shis10]

Optimal Orgasm 🤣

[@himanshukaushik9223]

😂😂

[@D3vansh]

I think first i should learn 3SUM then will do 4SUM as you said

[@takeUforward]

Please learn 2 sum, then 3 sum, then only 4 sum should be implemented ! Yes in terms of problem.

[@gandhamphotos]

@@takeUforward 😂😂

[@shis10]

@@takeUforward 🤣🤣🤣

[@godson200]

@@takeUforward sir but partner is not available how to do 2sum

[@mohsin7343]

I am still with 1sum

[@vagabondfoodie5788]

Aaj maza aa gya 3 sum karke🌚🌚🌚 ab 4 sum krunga

[@Algorithmswithsubham]

Sir app apne resume k project me konsa tech stack use kya haii. ❤ love your videos

[@nquanta1548]

Thankyou for the wonderful series ❤️🔥

[@shuvbhowmickbestin]

Code is not running for this test case: nums = [-1, -1, -1] target = -2 if you're writing in a language like C# or Java where lists are considered reference types and the equality are based on references rather than actual values. The workaround in C# is to sort the list like usual and then use a *ValueTuple* to extract the values from the list and then adding this to the set as ValueTuples equality checks are value-based rather than reference which means two different valuetuples with the same numbers in the same order will still be considered as *EQUAL* whereas the same in a list would be considered as *UNEQUAL* due to a difference in references.

[@ayanakash9900]

For optimal solution time complexity, we have to add sort() time too ig

[@AniketKumar-hf2bo]

understood,thnx for the amazing video ❤❤❤❤👌👌💕💕

[@cinime]

Understood! Super wonderful explanation as always, thank you so so much for your effort!!

[@joeljacob4685]

Understood!! Thanks for the nice explanation

[@AbhishekKumar-cv1dh]

Understood beautifully!! 🔥❤

[@rishabgupta2733]

The complexity of better solution is n^4 *logn according to codestudio

[@PrinceTripathi-j7u]

Nice to do 4 sum.

[@satheeshbhukya9970]

excellent question....

[@thomasshelby6780]

Tapped into pleasure back to back with 4Sum followed by 3Sum

[@Lakshya-f4l]

Mark for revision

[@shivanichaudhary1427]

Amazing, thankyou!

[@Manishgupta200]

It's amazing from brute to better to optimal similarly like 2Sum, 3Sum .. and my doubts in what's about 5-Sum, 6-Sum.. is any universal solution exist?

[@harshshukla7675]

Yes, the complexity will keep on increasing tho

[@myaccount09011975]

What about Time complexity for sorting the arrray? I know it will be lesser than N3 in this case. But still good to mention that!. Thanks for putting all the hard word!.

[@himaniupadhyay8201]

Thanks Lord

[@keshavgarg8407]

it is crazy how I am doing both 3sum and 4sum in a single day

[@saurabhtamta7295]

why is it when i use:- long long sum = nums[i]+nums[j]+nums[k]+nums[l] -----it says runtime error but when i use :- long long sum = nums[i]; sum+=nums[j]; sum+=nums[k]; sum+=nums[l]; ----it works fine. Why is that?

[@konankikeerthi]

Thank you bro. Understood

[@harshilpatel3205]

Understood 🙏🏻

[@gautamsaxena4647]

understood bhaiya

[@lwnflash2123]

for both brute and better ,you have missed the time complexity of inserting unique elements in the set.totally for better approach :O(n^3 *log(no of unique triplets* log m)...

[@statuscreation9493]

while inserting elements in set we are inserting 3 to 4 elements onnly so it nearly O(1)

[@ictfan-ly8gh]

Bro u are wrong . If N is large there could be more elements. So it will be added​@@statuscreation9493

[@RituSingh-ne1mk]

Understood!

[@bikramprajapati5358]

Amazing Explanation

[@hemantpatel1413]

Understood.

[@Kj89865]

thanks for threesum

[@UECAshutoshKumar]

Thank you!

[@raghavkansal3765]

you are the best 🧿🧿

[@Shivi32590]

thank you

[@senseiAree]

Understood ❤

[@hiteshpanchal5772]

Thxu striver bhai 😊😊

[@mdazizul-y7b]

understood sir

[@RohitKumar-dy2gc]

easy peasy✨✨

[@the_avii_7]

Understood !!!!!

[@himanshukaushik9223]

Bhaiya is everything fine any health issues video is not coming ??

[@per.seus._]

understood ❤

[@ArunsinghParihar-j3j]

Understood🤙

[@VishalPanwar-df5ek]

Understood Srtiver!

[@ravitejachandana6199]

understood !

[@dewanandkumar8589]

Understood

[@culeforever5408]

understood ;)

[@DR-mq1le]

understood , thank you!

[@Ubaiish]

i cant even watch dsa video without headphones now

[@AbjSir]

understood

[@satyamsrivastava3423]

Just wanted to add the Time Complexity will also contain log(n) since we are sorting the array, right?

[@203-aasritha2]

We should consider the time complexity for sorting the given array right ?

[@lakshmiprasanna7058]

Understood 💯💯💯

[@rishabpurba3268]

hey, does someone relate with this, i have trouble coming around the solution that i haven't come up with, i overthink and find it hard believing the solution?

[@her_soulmate]

Understood 🎉

[@mranonymous1982]

Did 4sum by myself (brute, better, and optimal), after understanding 3sum. Ps: who's coming up with these names for the questions bruh..

[@ShivamSingh-g2x6l]

in the brute force approach when we first added nums[i]+nums[k]+nums[k]+nums[l] and then checked if its == target then it gives integer overflow error regardless of sum being stored in long long variable but if we do it one by one like sum = nums[i]+nums[j] and then sum+=nums[k] and sum+=nums[l] then the error is resolved can anyone please explain to me how its working.

[@EC20022ELAKKIYAC]

did you found how does it works ? same doubt here

[@AbhishekBhattacharjee-j2m]

UNDERSTOOD

[@sanayaagarwal1907]

understood👍

[@VivekKumar-kx2zf]

why we are putting condition while(k

[@brp3522]

If k crosses l then we have gone through all the possibilities. Imagine this if they crosses each other they will act like each other. After crossing K will give the elements L has already given and the same goes for L

[@satyamjha-codeindwala6666]

satyam jha

[@iamnoob7593]

US , THANK U

[@inspiredomkar1239]

I didnt understand why are we calculating the sums separately for all the 4 variable : Why we need to do this :👇 ong sum = nums[i]; sum += nums[j]; sum += nums[left]; sum += nums[right]; when we can do simply this in one line 👇 long sum = nums[i] + nums[j] + nums[left] + nums[right];

[@udayprakashharsh2805]

This is because the nums[i] is int so if we assume all the 4 quad to be 1e9 then the result would be 4*1e9 which will lead to overflow int error to avoid this we are separately calculating the sum for all the 4 array elements.

[@SagarKumar-li4um]

​​@@udayprakashharsh2805but sum is already declared as long long then there should not be any problem of overflow either we do the sum in one step or in 4 steps