2 Sum Problem | 2 types of the same problem for Interviews | Brute-Better-Optimal 

Let's march ahead, and create an unmatchable DSA course! ❤ Timestamps pleaseeee Use the problem links in the description.

Striver, I cracked my coding interview by providing your optimal solution. Can’t thank you enough. You have changed lives of many. Keep doing great work ❤

I can’t thank enough because you teach so well about the concepts of DSA compared to others. None of the paid resources out there have good content. All I can say is May God bless you with long life and happiness. You are making all of us to land in our dream job.

bhaiya you're so much doing for us, even you live alone in Poland, you sacrifice your time for you and I understand every thing you teach as you are so good in explaining and making people understand. once again thanks a lot for this

We need more problem solving videos. Explanation is super. I have gone through a few channels but this is the best! Thank you so much.❤️

Bro, I am a guy from Africa you are the first person in this tech community that inspires me that I can do it.. Hats off for you bro 🔥🔥🔥

00:06 The video covers the 'two sum problem' and its two varieties. 02:04 Two sum problem can be solved using brute force method 04:05 Optimizing 2 Sum problem using a better solution 06:08 Using hashing to easily retrieve elements from a data structure. 08:13 Find the indexes of two elements in an array 10:30 Two pointer approach is a slightly better solution to solve the problem without using a map data structure. 12:38 Using the two-sum problem to find pairs that add up to a given target 14:51 The time complexity of the algorithm is O(n) 16:49 Space complexity and array manipulation explanation

Man you have changed a lot, this new version feels like a super saiyan mode.

Understood! Thanks a billion for your top-notch explaination brother!

Man after watching the first 5 videos and solving the problems on my own , I was also able to solve almost all medium and some hard problems. Watching these videos really built my logic , these lectures are a gold mine.

GREAT BHAIYA I buy multiple dsa courses but i only understand with ur lectures 🤗🤗

Raj, Thanks a lot for This Amazing Video about C++ Arrays Video - 5 Completed ✅

2 pointer approach was very beautiful

Thankss bhaiya for this consistency ❤️🙌 My placement is coming soon I'm in 6th semester!!🙂

Understood ! This is the first ques of my challenge 😃

int main() {int n=3; int array[3]={3,2,4}; int target=6; for(int i=0; i

10:36 mpp.insert({nums[i], i}); is more optimised than mpp[num] = i; (used in video)

🎯 Key Takeaways for quick navigation: 00:45 📚 *The video is part of the Striver's A to Z DSA course, covering comprehensive content with a guarantee of deep understanding in DS algo for interviews worldwide.* 01:12 🎯 *The Two Sum problem has two varieties: one asks for a yes or no if a pair with the given sum exists, and the other requires returning the indices of the pair.* 02:50 🧠 *The initial Brute Force solution involves nested loops to check every pair's sum.* 05:24 ⏩ *The improved solution uses hashing, mapping array elements to their indices, reducing time complexity to O(n).* 09:17 🚀 *The optimal approach, sorting the array and using two pointers, achieves O(n log n) time complexity without extra space.* 15:16 🔄 *The space complexity for the two-pointer approach is O(1), considering no additional space is used.* 17:50 📂 *Code for all three approaches (Brute Force, Hashing, Two-Pointer) is available in C++, Java, and Python in the video description.* Made with HARPA AI

Understood, Thank you strivers for this amazing video.

You are Great Sir! I love your explanation !!

Thanks to you for the video Sir . Had a query : In Type 2 of the problem , why do we need the extra Data structure for the indices ? Because I think "left" and "right" these should be giving us the indices of the 2 numbers whose sum equals to the target .

Brute -> Better > Optimal = BEST👌

The question is to find the indices of the elements which on adding give us the target value. In the last part of video, where we need to find the solution in O(1) space, we sort the array and hence lose the indices of input array. So how can we return the indices of elements then ?

Understood 🎉 40 lakh

Hey Striver I am from Jupiter... I love your DSA playlist

As usual you rock as you explain 🔥 Hey Raj, what if the array has duplicate elements for example, arr = [2,3,5,1,2] and target = 4 Will hashing work for this case too?.................Because hashmap has unique keys

Understood! Super awesome explanation as always, thank you very very much for your effort!!

i got an small logic striver not so optimal for(int i = 0; i < nums.length; i++) { for(int j = i + 1; j < nums.length; j++) { if(target-(nums[i]+nums[j])==0){ return new int[]{i,j}; array:[8,4,6,12[ } ex: target =14 14-(8+4)=!0x 14-(8+6)=0 true so we wil return the indexes

in the brute force approach you should have initialized j=i+1;

Hats off to you striver bro

12:36 sorting the array itself will take O(n*n) then another O(n) for the 2 pointer traverssal whick adds up to O(n*n) ruining the whole point of optimisation !

Thank u soo much Dada, u r the real inspiration. Respect++;

3:33 here you can keep j=i+1 instead of 0 so you don' need to write i==j

HE just killed it as it says🤗

Top notch 🤩. Understood

For the 2 pointer approach, if we are sorting the array it is changing the original indexex. So, it is giving a error in LeetCode.

i am having trouble in understanding that when we declare a map how can we immediately start finding elements in it without filling in the values

Awesome explanation, thanks a lot

I did not understand why you did mpp[a]=i at the last in the type-1 problem

can we do binary search and improve on approach

Thanks sir for provide us this type of content ❤❤❤

Completed 5/28!!!

Understood. Very clear explanation

Job ke saath saath Padhana koi inse sikhe..🙌🙌

00:06 The video covers the 'two sum problem' and its two varieties. 02:04 Two sum problem can be solved using brute force method 04:05 Optimizing 2 Sum problem using a better solution 06:08 Using hashing to easily retrieve elements from a data structure. 08:13 Find the indexes of two elements in an array 10:30 Two pointer approach is a slightly better solution to solve the problem without using a map data structure. 12:38 Using the two-sum problem to find pairs that add up to a given target 14:51 The time complexity of the algorithm is O(n) 16:49 Space complexity and array manipulation explanation Crafted by Merlin AI.

bhaiya Third approach will not be apply if we return index of both element because if we do sort the element then automatically element indexes will be changed...

Understood Thank You :-)

understood sir

Can we do this using recursion like picking up the index and not picking....please share the code for it.

Your way of teaching is amazing but I struggle sometimes to convert your code into Python only then when I m facing complexities otherwise its manageable. I would request if you can add Python code of the solutions as well. Thank you !

for optimal solution why cant we return the pointer values as indexs ?? for variety 2

thank you anna

The Last Optimal Solution of Sorting an array and using two pointer technique will be difficult to use if Array is not Sorted and I have been told to return index of the pairs.

Understood;

Understood

Understood !

you jus make things easier..

while (left

understood brother💞

Average and worst csse TC of unordered Hashmap is O(1) and O(N), so total TC can be O(N) or O(N^2). So how did you arrive at total TC of O(NlogN) for the better approach?

Where can I find the python code for the same? In the notes its C++ and Java

Understood. Thanks a lot. Make More videos please......

Understood! great explanation

it does not work on tc=-3,-2,2,3,3 any please help me out in this

understood everything .... thanks striver

What to do if for the optimal solution there are duplicate elements? The next occurrence of the element overwrite the previous index and it leads to wrong answers if we were to return all the solutions. However the current code works if there are no duplicate elements and if only one unique solutions exists. Can someone tell me if they found a good approach to deal with duplicate elements (when we have to return index)

Understood 🔥

in the optimal approach why cannot we return left and right pointer for indexes

Can anyone pls explain why it is checked whether remaining is not equal to mp.end() 👀👀

def two(): nums = [3, 3] target = 6 dic = {} for i, v in enumerate(nums): dic[v] = i nums.sort() left = 0 right = len(nums) - 1 iteration = len(nums) - 1 for _ in range(iteration): if nums[left] + nums[right] == target: if dic[nums[left]] == dic[nums[right]]: a = dic[nums[left]] dic.pop(nums[left]) b = dic[nums[right]] return a, b else: return left, right elif nums[left] + nums[right] > target: right -= 1 elif nums[left] + nums[right] < target: left += 1 print(two()) Hello! Bro I am getting problem in line 22 while popping the key[3] with value (0, 1), Because its a test case of num = [3, 3] and target = 6. So, while in a dictionary when we remove a key it removes all the matching keys. Try to give us the solution with this test case --> nums = [3, 3] and Target = 6 👈🏻👈🏻 👆🏻👆🏻👆🏻👆🏻 This is my only request, Don't get frustate with my code just plz resolve this test case with your Opinion and Logic.🙏🏻🙏🏻 Please..

Understood !!

I am starting leetcode from 1st January 2024 1st problem two sums

I have a doubt at 10:12 at line number 10. Why that mpp[a] = i is used there.... If any one knows please explain

Understood bhaiya 🙏 ❤️

1.Understood

Understood💌

Understood : )

understood 👍

how the time complexity is O(n), sorting also has a time complexity, please explain

// Two Sum in Java using HashMap TC --> O(N) import java.util.*; class HelloWorld { static int[] twoSum(int arr[] , int target) { HashMap map = new HashMap(); for(int i=0;i

understood bhaiya. u r best

Understand…….👍🏻

Koi muja batayega hum return parenthesis ma kyu kar rahe ha curly braces ma kyu nahi ??

Understood bro.. thank you

striver bhaiya, thank you so much!!

understood bhaiya

thank you bhaiya

mpp.find(more) != mpp.end() can anyone explain why its like that why comparing with end?

Understood

As usual lovely.

Can anyone help me When solving with 2 pointer approach Raj is mentioning greedy What does he mean Greedy here

Samaj aa gaya!!

I am still looking for hash function, which can work for similar values. example : input array contains : [ 1 1 1 1 1 1 4 1 1 1 1 1 7 1 1 1 1 ] where some constant number of value are other integers and mostly all values are same. in other ways: BigOh (n - c) values are similar and number of different values are "c" constant # of values as we can see above input array as example. If anyone have understood the problem, please share your thought, I will highly appreciated. I know the problem with Hashing is collisions, and find will take BigOh(n) for above example. Can anyone help me to find the solution??

What if duplicate elements exist ?In that case two pointer approach will fail …!!

I feel confident till step 3.1, shall i jump to next topic (like linkedlist, trees) or continue the medium and hard sections? Please someone reply.

how to handle duplicates here brother ?

Understood! Sir

Great job. Thanks.

Thanks

I have to do clap for your optimal solution