3 Sum | Brute - Better - Optimal with Codes
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- čas přidán 5. 07. 2024
- Problem Link: bit.ly/3X34JSI
Notes/C++/Java/Python codes:
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
00:41 Problem Statement
02:56 Brute force approach (Using 3-pointer)
04:55 Pseudocode
07:36 Code
09:27 Complexity
10:26 Better approach (Using Hashing)
12:23 Dry run
18:09 Code
20:15 Complexity
22:20 Optimal approach (Using 2-pointer)
23:20 2-Pointer Technique
31:50 Code
36:41 Complexity
Please do give us a like and subscribe, it won't cost you anything, but it will motivate me to make such kind of content more and more.
Legendary stuff Raj bhai. Your explanation clearly shows you actually have a very strong depth on these fundamentals. And of course you do, you work for Google😉
Awesome stuff. Liked and subscribed. Keep going. 👍
great content love u bhai
I start by like for all the video
Bhaiya code nhi chal raha hai 3 sum ki u give the condition sum is greater than 0 , less than 0 but didn't give the condition sum is equal to 0 why?
Tried 2 sum, 3 sum and 4 sum problems together as a beginner. It was so frustrating after a point before I understood the optimal approach 😂
true bro
u r strong bro u have tried all as a beginner 😅
You have a lot more experience than the whole comment section combined.
The optimal approach for 3sum is just the extension of optimal approach of 2 sum when the array given is sorted
@@user-ye4xi4py2k yup !! you are right
It's interesting-I initially tackled this problem with three nested loops, but when the time exceeded, I decided to find a way to eliminate one loop and ended up developing a two-pointer solution. Although I found the solution, I still enjoy watching Striver's videos to refresh my mind, spark creativity, and discover new approaches to problem-solving.
Bro 3 loops was the best I could think of, I can't able to optimize it
How do you develop this logical thinking ?
Could you help me with this ?
Me too pls
00:41 Problem Statement
02:56 Brute force approach (Using 3-pointer)
04:55 Pseudocode
07:36 Code
09:27 Complexity
10:26 Better approach (Using Hashing)
12:23 Dry run
18:09 Code
20:15 Complexity
22:20 Optimal approach (Using 2-pointer)
23:20 2-Pointer Technique
31:50 Code
36:41 Complexity
That's what a explanation beginner require for these type of problems
Just cameback for a quick revision, and now it's indeed got into my head, thanks for your crystal clear intuition!
Yes! I was onto this optimal approach but my implementation failed because I wasn't thinking it through. Simply lovely explanation!
Completely Understood your explanation! Thank you for what you are doing, and please continue the good work. You are an amazing teacher. Have watched 3 videos of yours and I was able to understand all 3 with out any confusions. Big thumbs up for the video. 👍
Thanks brother for helping and providing us amazing solutions of the most important questions that asked in MNC's. Thanks a lot brother🙏
I love the way you teach bhayiya.❤❤ I don't have seen the teacher like you....you are God of DSA.
how can one explain so smoothly man...Hats of STRIVER bhaiya
We will never get such a detailed explanation of 3 Sum problem. You are a Legend for reason....Striver.....!!!!
Are u a genius how do u know what doubts a newbie would have . U r just superb in explaining the Algo
Explained all 3 approaches very clearly. Thank you so much!!!
Understood! Super amazing explanation as always, thank you very much for your effort!!
My man is doing God's work, thanks for this amazing playlist!
Thanks for the in-depth explaination with in-depth time and space complexity
Great examples, which helps understand the algorithm very clearly even for non CSE folks!!
I was looking for brute force approach tried myself but couldn't remember I have to used set DS but now , I can understand where I was wrong. Thanks for the tutorial.
Bhai maine aapki dsa sheet aaj first time dekhi hai
kya banayi hai bhai , sach main mja aa gya ....
Thanku dil se striver bhai ❤
I found your explanation the best among all available... gd job
Amazing content learn a lot every day from your course.Thanks for creating such an amazing course.
THANK YOU FOR EXPLANING IN SIMPLE WAY
Hi, you are doing extremely good work DSA topics. You making concepts very clear. Glad that I got your channel reference. But un luckily I am from JavaScript background , i am finding a resources like anything for DSA, I dint get any . Your help will be appreciated on this.
Too good man, more and more kudos to you for such explanation. now im getting grip on building logic...finally.
nice all the three approaches, helped a lot.
awsm video striver ❤❤ the free education you are providing is helping a us alot.
no need of condtion j
Awesome explaination. Thank u for such a great content.
Great job! your code is so clean.
what a solution. MINDBLOWING!!!!
understood and came up with the optimal solution myself almost same. just used an extra set to store triplets 😅
Great explanation Striver ❤
Now i can finally answer someone if someone ask me have you done 3Sum 😆. Thanks Striver 😉
Beautiful dry run!! Understood😄
gained a subscriber with your amazing explanation
Hi Raj, there are 2 slight mistakes in your optimal solution.
1: The for loop will run till n-2 instead of n, because when i=n-1, j becomes n (j=i+1=n-1+1) and num[j] throws out of bound exception.
2: We could also insert another check in the for loop(nums[i] = 1 there's no way any 3 elements sum would be 0 since the array is now sorted.
3: Here's the more readable code(JAVA):
class Solution {
public List threeSum(int[] nums) {
int n = nums.length;
List result = new ArrayList();
Arrays.sort(nums);
for(int i=0; i low && nums[high] == nums[high+1])
high--;
}
else {
result.add(Arrays.asList(nums[i], nums[low], nums[high]));
low++;
high--;
while(low < high && nums[low] == nums[low-1])
low++;
while(high > low && nums[high] == nums[high+1])
high--;
}
}
}
}
mind blown, dopamine released, love u striver
Thank you so much bhaiya....you are the best teacher ❤❤❤
amazing exlanation , loved this video
Another Awesome Lecture................
what a fantastic explantion!!!!
Thank you bro, love from Tamil Nadu ❤
Great Explanation 💯💯
While solving this problem, the very first approach which comes to my mind was optimal. Although the way I was handling duplicates was giving time limit exceeded error so I have to took help from gpt but rest of the logic was correct. Feeling extremely happy.
Bro that hash map solution is so genius.
Thank you so much sir for such a nice explanation your are super sir❤
superb explanation brother👏
understood!!! please came up with string series...please
Understood bro. Thank you
Great Job again !!
Awesome explanation....
Keep going brother ❤
Understood bhaiya 🙏 ❤️
Great explanation 💯
Amazing explanation 👌💯
HashSet in C# still stores duplicate List in C# . Is there any alternatives in C# to store unique List of Integers?
best teacher in the world...........
never got an opportunity to do 3sum 😢 koi na LC pe hi krleta hu 🙂 seh lenge
agreed🤣
Best explanation 🔥
Perfect explanation
3-sum code using 2 for loops and 1 while loop. Using for loops helps in more readability in this problem.
// Sort the vector and create ans array.
sort(arr.begin(), arr.end());
vector ans;
// Iterate the sorted array.
for (int i = 0; i < n; i++) {
// Make sure first element of triplet is unique.
// Now whole array on right of 'i' is a 2-sum problem.
if (i > 0 && arr[i] == arr[i-1]) continue;
// Take -ith element as target.
int twosum = -arr[i];
int right = n-1;
// Look for unique 2nd element in the search space. Notice that for
// each unique i, the pairs that create 3sum with it are unique.
// Similarily, for each unique j, the 3rd element will be unique.
for (int j = i+1; j < right; j++) {
// Looking only for unique 2nd element.
if (j > i+1 && arr[j] == arr[j-1]) continue;
// Look for its partner.
while (j < right && arr[j] + arr[right] > twosum) right--;
// No need to skip. All the 2nd elements are unique!
if (arr[j] + arr[right] == twosum && j != right)
ans.push_back({arr[i], arr[j], arr[right]});
}
}
return ans;
Similarily, 4sum can be broken down as: first unique element + 3sum in remaining search space.
understood brilliantly 😘
Great explanation but what about the time complexity of those 2 while loops from the optimal solution? Can you elaborate a bit here please?
Understood, thank you.
Thank you bhaiya, Love from bangladesh
Awesome explanation
Understood 👍
Mza aagya
Understood!!
Understood🔥
Line 29: Char 10: error: type 'vector' does not provide a call operator
ans(st.begin(), st.end());
^~~
1 error generated. brute force code
understood ,thnx for explanation ❤❤❤❤❤❤👌👌💕💕💕💕
Thanks a lot, sir !!
Understood!
Java Solution that is accepted on Leetcode:
class Solution {
public List threeSum(int[] nums) {
Set res = new HashSet();
Arrays.sort(nums);
for(int i=0;i
Sir,
Please continue your solution videos for A-Z DSA course
got it nicely explained.
we can do a small improvement int the optimal code
if nums[i]>0 then we can break the loop and return answer directly
I have watched many lectures of your but there are some lecture whose solution link is not in the description , As in this video itself please provide the link.
best explaination !!!
Understood sir thank you soo much sir
Understood!! ❤❤
Understood ❤
Understood 💯💯💯
love it, thanks
Hey @takeUforward
While I was going through the optimal solution, ans.push_back(temp) will not eliminate duplicity. You will have to add in a set and then later use the set to derive ans as list.
nope, as we are using unique elements whenever sum gets to 0 through 2 while loops.
@@ujjwalkumarsingh5216 Yeah noticed that.. its working fine..
Understood✅🔥🔥
Excellent 👌
As Smooth as Butter 😃
thank you so much bhaiya❤
understood
thanks striver
God of DSA❤
Understood!
Thank you ❤
when implementing the second approach instead of using set, we can use unordered_map(with the second key as its index), and we will start i from 0 and j from i+1 and for the third value we can ensure by checking that the index of the last target element from the map should be greater than j :)
hey, im not able to digest this thing in the 2nd approach
arr is -1 0 1 2 -1 4
and the i & j are pointing at -1 and -1 so arr[k] = -(-2) = 2 so for this we need to search in the entire array right
and he stored [0 1 2] elements between -1 and -1
but what if the arr was like this
1 0 1 2 3 -4 and i is pointing to 1 and j pointing to 3 then arr[k] = -(4) = -4 then if we store [ 0 1 2 ] we will not find in the set as -4 is after the j pointer
could u pls make me understand this as im bit confused
@@lakshsinghaniayou are right but there is a catch here ,
When i is pointing to 1 and j is pointing to 3 we need to search for -4 in the hashset which is not available. However after that we will add the arr[j] into the hash set i.e. we will add the element 3 into hash set ..
So in the next step j will point to 4 now we have to search for -(1-4) = 3 and this element 3 is present in the hashset thereby we will get the required unique triplet
thank us sir again
Understood 🎉
can someone explain me ,, why we are redeclaring hashmap in every i ,,can not we make a one traversal and store all elements in the map at once .In last when we are inserting elements in a set after sorting ,can it will not avoid duplicacies?
Understood.🙂
Hello sir in two pointer approach when we got sum=0 why we are moving k to k-- since we can get different results with old i + new j + old k? 29:43
Understood😊