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😉
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.
@@karthikeyan.s2565 This is a long process bro, keep in mind the techniques you have already employed and then try to come up with ideas for how we may shorten the loop as much as possible. Actually, after successfully answering an easy issue, I usually attempt to answer it in three or four ways (if at all feasible). Occasionally, after solving the problem, I peruse other people's answers on Leetcode to discover new strategies, and the Striker video assisted me in approaching the problem.
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--; } } } }
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. 👍
#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.....
we can keep the indexes of the elements in the hashmap as well, in one loop; then while taking i and j, we can check whether the index is same as that of i or j or not. That would require less time complexity than the better solution.
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.
When sum is less than 0 or greater than 0, then also we should skip for duplicates right? As sum will be same for the next duplicate value. EG: -4 -2 0 0 0 2 2 2. when i is at index 2, j is at index 3, and k is at the last index, and the sum is greater than 0, you want to skip duplicates for k as long as the value at k is the same as the previous one.
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.
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
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.
in better solution ; time complexicity :-> O(n^2 * log(no. of unique triplets)) but for seraching to hashset using find and insert operation of hashset should we count ? why ?
dont forget to add if nums[i]>0: break at starting line in the for loop, since it is a sorted one, if your 1st element itself >0 then you cant find the sum that is ==0 so add this, which improves runtime drastically
I found the optimal initially. And I thought it might not be the optimal. so,i had been trying to find the better optimal 😂.but I couldn't And came to your video.guess what I successfully wasted a day..😅😢
If we directly use hashset and use pointer approach then we dont need to check many different conditions, which is easy for as a beginner class Solution { public List threeSum(int[] nums) { Arrays.sort(nums); HashSet res = new HashSet(); for (int i = 0; i < nums.length; i++) { int lo = i+1, hi = nums.length-1; while (lo < hi) { int sum = nums[i] + nums[lo] + nums[hi]; if (sum == 0) { res.add(Arrays.asList(nums[i], nums[lo], nums[hi])); lo++; hi--; } else if (sum < 0) lo++; else hi--; } } List resultList = new ArrayList(res); return resultList; } }
13:07 we can actually use a hashmap, by sorting and storing all the elements of given array as key and indexes as values. If 3rd element of triplet i.e. -(arr[i]+arr[j]) lies in the map and you don't want it to clash with value at current 1st and 2nd elements then simple check if index of that 3rd triplet ele in map is greater than index of 2nd triplet j. Arrays.sort(arr);//o(n*logn) HashMap map = new HashMap(); for(int i=0;i > set = new LinkedHashSet(); for(int i=0;i > list = new ArrayList(set);
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.
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
I've 2 queries please resolve them 1) Set datastructure stores sorted and unique elements only then why are we storing in temp and then sorting it. 2) 2D vector "ans" is called only once at end so it will only have 1 list with all elements rather than list of lists
@@aviralmishra181 class Solution { public: vector threeSum(vector& nums) { //first sort the array. sort(nums.begin(), nums.end()); int n = nums.size(); vector res; for(int i = 0; i < n-2; i++) { //to skip duplicate values in nums[i]. if(i == 0 || i > 0 && nums[i] != nums[i-1]) { int low = i + 1; int high = n - 1; int sum = 0 - nums[i]; //reminder a+b+c = 0, so b+c = -a; nums[low] = b //nums[high] = c so if b + c = -a (can be found that will be the triplet.) while(low < high) { if(nums[low] + nums[high] == sum) { //means triplet found. vector temp; temp.push_back(nums[low]); temp.push_back(nums[high]); temp.push_back(nums[i]); res.push_back(temp); //to ignore duplicate values. while(low < high && nums[low] == nums[low + 1]) low++; while(low < high && nums[high] == nums[high - 1]) high--; //[....1 1 0....] after above loop low is in first 1 so do low++ low++; high--; } else if(nums[low] + nums[high] < sum) { while(low < high && nums[low] == nums[low + 1]) low++; low++; //low++; } else { while(low < high && nums[high] == nums[high - 1]) high--; high--; //high--; } } } } return res; } }; This code passes all the test cases try it.
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
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
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
I also solved with two pointer approach on my own after 3 loops got time exceeded 😊 he built my logical thinking
@@karthikeyan.s2565
This is a long process bro, keep in mind the techniques you have already employed and then try to come up with ideas for how we may shorten the loop as much as possible. Actually, after successfully answering an easy issue, I usually attempt to answer it in three or four ways (if at all feasible). Occasionally, after solving the problem, I peruse other people's answers on Leetcode to discover new strategies, and the Striker video assisted me in approaching the problem.
That's what a explanation beginner require for these type of problems
Yes! I was onto this optimal approach but my implementation failed because I wasn't thinking it through. Simply lovely explanation!
how can one explain so smoothly man...Hats of STRIVER bhaiya
Just cameback for a quick revision, and now it's indeed got into my head, thanks for your crystal clear intuition!
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--;
}
}
}
}
no need of condtion j
I love the way you teach bhayiya.❤❤ I don't have seen the teacher like you....you are God of DSA.
Are u a genius how do u know what doubts a newbie would have . U r just superb in explaining the Algo
Thanks brother for helping and providing us amazing solutions of the most important questions that asked in MNC's. Thanks a lot brother🙏
Great examples, which helps understand the algorithm very clearly even for non CSE folks!!
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. 👍
We will never get such a detailed explanation of 3 Sum problem. You are a Legend for reason....Striver.....!!!!
Explained all 3 approaches very clearly. Thank you so much!!!
God bless you for all the help you do
we can do a small improvement int the optimal code
if nums[i]>0 then we can break the loop and return answer directly
My man is doing God's work, thanks for this amazing playlist!
#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.....
Was asked in Adobe interview for DEI hiring for specially abled candidates.
Woah, thanks !
we can keep the indexes of the elements in the hashmap as well, in one loop; then while taking i and j, we can check whether the index is same as that of i or j or not. That would require less time complexity than the better solution.
never got an opportunity to do 3sum 😢 koi na LC pe hi krleta hu 🙂 seh lenge
agreed🤣
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.
Amazing content learn a lot every day from your course.Thanks for creating such an amazing course.
I found your explanation the best among all available... gd job
Too good man, more and more kudos to you for such explanation. now im getting grip on building logic...finally.
Bro that hash map solution is so genius.
understood and came up with the optimal solution myself almost same. just used an extra set to store triplets 😅
Noice title 😉
;)
When sum is less than 0 or greater than 0, then also we should skip for duplicates right? As sum will be same for the next duplicate value. EG: -4 -2 0 0 0 2 2 2. when i is at index 2, j is at index 3, and k is at the last index, and the sum is greater than 0, you want to skip duplicates for k as long as the value at k is the same as the previous one.
Understood!
I went somewhere near the optimal approach but wasnt able to come up with the concrete solution
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.
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
In the brute force approach discussed couldn't we have just sorted the array in the starting and then use 3 pointer technique instead of using sets
Now i can finally answer someone if someone ask me have you done 3Sum 😆. Thanks Striver 😉
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! Super amazing explanation as always, thank you very much for your effort!!
Thanks for the in-depth explaination with in-depth time and space complexity
Understood 👍
in better solution ;
time complexicity :-> O(n^2 * log(no. of unique triplets))
but for seraching to hashset using find and insert operation of hashset should we count ? why ?
God of DSA❤
THANK YOU FOR EXPLANING IN SIMPLE WAY
2 sum , 3 sum , 4 sum...what is this Google...Name it GANGBANG...KATHAM TATA BYE BYE SEE YOU 😂😂😂
Java Solution that is accepted on Leetcode:
public List threeSum(int[] nums) {
int n = nums.length;
HashSet set = new HashSet();
for (int i = 0; i < n - 2; i++) {
HashSet s = new HashSet();
for (int j = i + 1; j < n; j++) {
int third = -(nums[i] + nums[j]);
if (s.contains(third)) {
List temp = new ArrayList();
temp.add(nums[i]);
temp.add(nums[j]);
temp.add(third);
Collections.sort(temp);
set.add(temp);
}
s.add(nums[j]);
}
}
List ans = new ArrayList(set);
return ans;
}
Bro can please explain me new ArrayList(set); this line where u passed set how is this fetching List in hashset to arraylist of ans.
Understood
Understood🔥
dont forget to add if nums[i]>0: break at starting line in the for loop, since it is a sorted one, if your 1st element itself >0 then you cant find the sum that is ==0 so add this, which improves runtime drastically
Understood!
Great explanation but what about the time complexity of those 2 while loops from the optimal solution? Can you elaborate a bit here please?
what a solution. MINDBLOWING!!!!
Consdering the third solution cant we just use Set of List and return as List of List, then we avoid the two while loops :
while(j
But it will increase your space complexity, those two while loops are light loops so Strivers 3rd solution I think is best as its saves space.
I found the optimal initially. And I thought it might not be the optimal. so,i had been trying to find the better optimal 😂.but I couldn't
And came to your video.guess what I successfully wasted a day..😅😢
Great job! your code is so clean.
mind blown, dopamine released, love u striver
its java solution even if i copy paste from your sheet is not working, only 3 or 4 test cases are getting passed for brute and better solution.😫😫😫😫
Line 29: Char 10: error: type 'vector' does not provide a call operator
ans(st.begin(), st.end());
^~~
1 error generated. brute force code
As Smooth as Butter 😃
nice all the three approaches, helped a lot.
awsm video striver ❤❤ the free education you are providing is helping a us alot.
If we directly use hashset and use pointer approach then we dont need to check many different conditions, which is easy for as a beginner
class Solution {
public List threeSum(int[] nums) {
Arrays.sort(nums);
HashSet res = new HashSet();
for (int i = 0; i < nums.length; i++) {
int lo = i+1, hi = nums.length-1;
while (lo < hi) {
int sum = nums[i] + nums[lo] + nums[hi];
if (sum == 0) {
res.add(Arrays.asList(nums[i], nums[lo], nums[hi]));
lo++; hi--;
}
else if (sum < 0) lo++;
else hi--;
}
}
List resultList = new ArrayList(res);
return resultList;
}
}
Awesome explaination. Thank u for such a great content.
Another Awesome Lecture................
understoood
understood!!! please came up with string series...please
13:07 we can actually use a hashmap, by sorting and storing all the elements of given array as key and indexes as values. If 3rd element of triplet i.e. -(arr[i]+arr[j]) lies in the map and you don't want it to clash with value at current 1st and 2nd elements then simple check if index of that 3rd triplet ele in map is greater than index of 2nd triplet j.
Arrays.sort(arr);//o(n*logn)
HashMap map = new HashMap();
for(int i=0;i > set = new LinkedHashSet();
for(int i=0;i > list = new ArrayList(set);
return list;
I have a doubt... Why are we running the i loop till less than n-2 and j till less than n-1
@@misty6129 according to my intuition which I mentioned above , i always lies before j and k , j always lies before k.
amazing exlanation , loved this video
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.
Great explanation Striver ❤
Thank you bro, love from Tamil Nadu ❤
best teacher in the world...........
also, we can stop when nums[i] reaches some positive value in the most optimal solution, because after then zero cannot occur as our sum
Understood 🙏🏻
Beautiful dry run!! Understood😄
gained a subscriber with your amazing explanation
Great Explanation 💯💯
Gajab BHAI ❤❤❤❤❤❤
Amazing explanation
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
what a fantastic explantion!!!!
Thank you so much bhaiya....you are the best teacher ❤❤❤
Awesome explanation....
u r motivating all the students of tire 3 colleges of West Bengal and India . You are like Virat Kohli in coding.
Dry run 🔥
Thank you ❤
Mza aagya
Understood!!
understood
I've 2 queries please resolve them
1) Set datastructure stores sorted and unique elements only then why are we storing in temp and then sorting it.
2) 2D vector "ans" is called only once at end so it will only have 1 list with all elements rather than list of lists
Understood bro. Thank you
understood raj sir
HashSet in C# still stores duplicate List in C# . Is there any alternatives in C# to store unique List of Integers?
why unordered set of vector is giving error in leetcode
Understood, thank you.
Sir,
Please continue your solution videos for A-Z DSA course
Perfect explanation
Keep going brother ❤
Great Job again !!
Understood
I think it should work but it gives wrong answer on leetcode it does not passes all the testcases.
If you find solution please comment back.
@@aviralmishra181 class Solution {
public:
vector threeSum(vector& nums) {
//first sort the array.
sort(nums.begin(), nums.end());
int n = nums.size();
vector res;
for(int i = 0; i < n-2; i++) {
//to skip duplicate values in nums[i].
if(i == 0 || i > 0 && nums[i] != nums[i-1]) {
int low = i + 1;
int high = n - 1;
int sum = 0 - nums[i]; //reminder a+b+c = 0, so b+c = -a; nums[low] = b
//nums[high] = c so if b + c = -a (can be found that will be the triplet.)
while(low < high) {
if(nums[low] + nums[high] == sum) {
//means triplet found.
vector temp;
temp.push_back(nums[low]);
temp.push_back(nums[high]);
temp.push_back(nums[i]);
res.push_back(temp);
//to ignore duplicate values.
while(low < high && nums[low] == nums[low + 1]) low++;
while(low < high && nums[high] == nums[high - 1]) high--;
//[....1 1 0....] after above loop low is in first 1 so do low++
low++;
high--;
}
else if(nums[low] + nums[high] < sum) {
while(low < high && nums[low] == nums[low + 1]) low++;
low++;
//low++;
}
else {
while(low < high && nums[high] == nums[high - 1]) high--;
high--;
//high--;
}
}
}
}
return res;
}
};
This code passes all the test cases try it.
Thank you so much sir for such a nice explanation your are super sir❤