4 Sum | Brute - Better - Optimal with Codes
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- čas přidán 9. 07. 2024
- Problem Link: bit.ly/3It5SyP
Notes/C++/Java/Python codes: takeuforward.org/data-structu...
We have solved the problem, and we have gone from brute force and ended with the most optimal solution. Every approach's code has been written in the video itself. Also, we have covered the algorithm with intuition.
Full Course: bit.ly/tufA2ZYt
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0:00 Introduction of Course
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
Please watch our new video on the same topic: czcams.com/video/eD95WRfh81c/video.html
You said it is O(n3) but sorting will take some time to what about that?
@@uniqueyuvrajgaming3392nlog n is basically negligible in comparision of n3 thats why is it not included
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.
@@uniqueyuvrajgaming3392 Sorting will have always 4 elements which will take constant time.
Note: 0:42 Make sure you watch the 3Sum before doing the 4Sum .
🙂
What if I have seen since a long ago, will it work ?🙃 But why he was laughing while saying this ?
😂😂
naughty audience😂😂😂😂
ok 🙂
Excited to do 4sum after completing 3sum yesterday
Damn bro include me next time
@@gareer2436 you know how to bend?😂😂
Yaar tum log bohat tharki ho 😂
@@yashsharma6396 no but I know what to do when someone bends...😏
Lay them down and..
Massage their backs of course!
@@gregorirasputin659 ma to sath me problem solve krne ki baat kr raha tha bro... tum kya soch liye?😏
3 sum was awesome, 4 sum was fantastic!!!!
legends like 5 sum
0:44 Striver's Smile Says it all
you got my point bro😁
bhai bhai🫡🫡😂. what a detailing bro
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 :)
striver ..king of DSA
The if condition inside while to check if(k
sir which whiteboard pen app you use for explain the code
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
@@codemania2878 He is using an iPad and its default Notes app
Whole video is about 4 sum but 0:40 to 0:50 is wholesome
Finally did 4 sum, amazing experience
Can you elaborate your experience? Asking for a friend
can you explain about your experience...🙂🙂
@@catalyst8654 🤣🤣
Would you take a moment to share your experience with us.
You should not take I upto
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.
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
You watch all the videos? I see you daily updating me with timestamps, massive thanks for that.
@@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.
@@takeUforward striver please update timestamp in disucssion so that ppl can access it easily
Understood beautifully!! 🔥❤
Understood,Thanks striver for this amazing video.
Thanks a ton for explaining in simplest way 🙏🙏🙏🙏🙏🙏
Thankyou for the wonderful series ❤️🔥
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
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 :)
@@_Arunvfx glad you finally understood it. Keep pushing forward 😃
Understood!! Thanks for the nice explanation
wow,....wow....wah..awesome expaination. undestood bro..hatsoff to you.
I have done this question using subsequence technique thanks u raj sir 😊 u teach really simple
uffffffff.... understood , bahut bhala se bujhi hela bhai
The consistency level of striver is really the source of motivation 🔥.
Amazing, thankyou!
Amazing Explanation
Great bhaiya ❤️🇮🇳 , ready for next one
understood,thnx for the amazing video ❤❤❤❤👌👌💕💕
Awesome Lecture Sir............
best explanation on you tube
understood , thank you!
Striver..... you are amazing.................................... Thanks so much bro
excellent question....
Understood! Super wonderful explanation as always, thank you so so much for your effort!!
Thank you bro. Understood
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
Thank you!
you are the best 🧿🧿
Understood ❤
Understood 💯💯💯
Understood ❤️
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.
understood !
Understood🤙
Nice to do 4 sum.
thank you
Understood!
Thank You : )
understood ❤
This has to be the horniest coding problem 🌚
Thxu striver bhai 😊😊
Keep going 👍
Understood 🎉
Thanks Lord
Understood Srtiver!
Sir app apne resume k project me konsa tech stack use kya haii. ❤ love your videos
understood👍
People who are aware of 2 pointer approach, watch from 14:56, saves a lot of time.
Understood.
UNDERSTOOD
easy peasy✨✨
We should consider the time complexity for sorting the given array right ?
For optimal solution time complexity, we have to add sort() time too ig
Awesome
Nice explanation. But the title is a bit too sus don't you think?
When striver submits on Code Studio , it also showed him time complexity as O(n4 log n) but it doesn't show in mine. Why?
After doing 4 Sum.... I think Sky is the limit 🌝
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 .
@@ArjunS-hi3vl Yes
have you completed the course ? im also quite late in starting the course and moving at about the same pace as you
@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?
understood ;)
understood
Understood
The complexity of better solution is n^4 *logn according to codestudio
thanks
bro couldn't control his laugh during 3sum , 4sum intro and had to cut the clip 🤣
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?
Yes, the complexity will keep on increasing tho
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
kuch bhi, same solution hai upar wala time complexity n3 hai, apki nlogn kaise aa gayi
@@akashchoudhary6876 the overall time complexity of the fourSum function is O(nlogn+n^3)which simplifies to O(n^3)in Big O notation, assuming the input is sufficiently large.
@@akashchoudhary6876 sorting in cpp will take O(nlogn)
Just wanted to add the Time Complexity will also contain log(n) since we are sorting the array, right?
US , THANK U
Aaj maza aa gya 3 sum karke🌚🌚🌚 ab 4 sum krunga
Bhaiya is everything fine any health issues video is not coming ??
Under stood
Tapped into pleasure back to back with 4Sum followed by 3Sum
Hi, thanks for the video explanation. I had one doubt. When we get the sum==target, we move the k and l pointers front and back respectively [not just once but until we get a different value for k and l pointers so as to not repeat the same ans]. Why are we not doing the same when {sumtarget}. [i.e. in these 2 cases we just do k+=1 or l-=1; shouldn't we here also move the k and l pointers ahead until we get a different value for them.] Please let me know if possible. Thanks!
It goes the same as well. If we move the pointer and again we get the same set , the sum will be the same right ? therefor if the sum is less than zero , it remains the same and vice versa. Then the iteration completes and we will hit the same condition. This goes until we get another pair.
I think first i should learn 3SUM then will do 4SUM as you said
Please learn 2 sum, then 3 sum, then only 4 sum should be implemented ! Yes in terms of problem.
@@takeUforward 😂😂
@@takeUforward 🤣🤣🤣
@@takeUforward sir but partner is not available how to do 2sum
I am still with 1sum
while i'm taking sum==target case in else it is not working like we did in three sum and if i put if(sum==target) at initial position it is working . why it is so ?
understood.
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!.
Imagine someone getting your video, on Searching 3sum video or 4sum video on google 🌚😂
Optimal Orgasm 🤣
😂😂
After Understanding this concept ,
I can Say that Sky is the limit..
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?
23:16 shouldnt j= i+1 will be second element? I cant understand
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.
did you found how does it works ? same doubt here
Bhaiya , Can uh please post an article about k sum similiar to this type , which is applicable for all k sums . It can help to build intution!!!!
Hi sir
We request you
Can you please make a video on Alex goes Shopping problem. We cannot understand the question.
04:20 - what does exceeding the integer limit mean? (when we are storing it in a long long)
if you store in int it might exceed
@@takeUforward but in 3 sum we sum up like sum = nums[i]+nums[j]+nums[k] but in this 4 sum you sum up all the indices separately which is correct but I am getting error in a test case when I am doing sum = nums[i]+ nums[j]+ nums[k]+ nums[l]; why?
@@SANDEEPKUMAR-in6li the compiler has an algorithm for adding these numbers . while adding four numbers(big numbers) the algorithm needs a lot of space , but while adding two or three the space is under limit. hence adding them one by one is recomended.
why we are putting condition while(k
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
is this the end..or more videos will come....I'm about to start this course.
Gr8
In the better approach why the code didn't gave runtime error for j and k when i = n - 1, because j = i + 1 and k = j + 1, making j and k going out of bounds of array so it should've thrown the error right?
while starting for loop we also stated j and k should be less than n if it goes out of bound whole loop will end.. so no chance
@@statuscreation9493 ohh got it thankyou bro
Tree ka v new playlist aayega kya...
Bhaiya s.c in optimal approach return ans ki hai O(no of quad) ??
jitne quads hongey answer me utna hi space lagega bro
@@takeUforward space answer return ma use hora ha na solve ma nahi hora na
Why removing duplicates I'm not getting the concept in 3 sum problem too in optimal solution
In the optimal code solution, why are we checking for k against k + 1, It seems to work for :
arr[k] == arr[k - 1]
this will work but at end. k will have value of last element of (common element sub array) e.g. [1,1,1,0,0,0,0,0,0]// after loop k will be 3 .
its cleaver trick to break loop and when access k. it will give new value (different from copy elements).
if we first decrese k(usual procedure of algo. nothing extra done); and then check arr[k] == arr[k + 1] in loop,
k- - // k=7
e.g. [1,1,1,0,0,0,0,0,0]
// after loop breaks, k will be 2