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loved the visualization

There is also a solution that takes O(N) time and O(1) space

great expalantion, thanks. Would be great to see a way of arriving at this solution from first principles or from hints/thoughts, etc

🙄 "PromoSM"

Please upload more videos, your way of teaching is very awesome. Thanks

Really nice and clear explanation.

nice explanation mam !!!

Please don't jump directly to 2D matrix. Explain the intuition first. I hope this helps!

This code crashed in binSearch function for me for below input data [10, 22, 9, 33, 21, 50, 41, 60, 22, 68, 90] I took code @time 16:40 in video

Hey Suyashi, great explanation for the problem and the approach to solve that. Although I have one issue with the correctness of the Binary Search solution. It fails for test cases like [10,9,2,5,3,7,101,102,103,104,4] or [10,9,2,5,3,7,101,4]. You'd understand the malfunctioning once you try printing the lis vector. Instead of getting [2,3,7,101] for [10,9,2,5,3,7,101,4] input it'd return [2,3,4,101]. Try to work such examples by yourself and you'd get my point.

I don't get how binary search is an approach to finding a subsequence. Usually binary search is used on a sorted array. How is it possible to find the answer with binary search?

hi can you make a video on regular expression matching and it will be great if u can start with the recursive approach and then move to tabulation one. thanks

Hi

Nice and clear DP walkthrough of LCS algorithm. Many thanks!!

Awesome 🙌

😇✌️

Very well explained!! Thanks for making this amazing video👏

Brilliant!

Nice one. I found this video after solving same Leetcode question earlier today. Really liked your clear explanation in C++. Looking forward to future Leetcode videos on this channel.

Excellent walkthrough of NlogN solution for Leetcode 300. It finally makes sense. I watched other videos of NlogN solution for this Leetcode problem and none were as clear as this. It was a nice bonus comparing it with DP solution.

Very clear and lucid explanation, thanks. By the way C++ offers std::lower_bound function to conduct binary search for this type of queries. FYI.

this is the only video .... that made the nlogn approach clear to me .... thanks a lott

Great explained. I saw lots of video of this problem but your videos make understand finally 😍

Thanks a lot for the clear explanation. Once the logic is clear the code becomes simple as well.

Crystal clear explanation. Well done.

this nlogn method is WRONG! it does not work for my arr = [5,2,8,6,3,6,9,5]. just before i check the last element of "arr", i.e. 5, my "lis" so far looked like this [2,3,6,9]. but then we take the last element 5, and we find the smallest element which is greater than equal to 5, which is 6 in this case and it replaces it; making the final seq as [2,3,5,9], which is incorrect. its not a valid sequence from the original array. also we don't need to do binary search to find out if all elements in "lis" are smaller than the current element we are trying to add in. since lis is sorted we can just check the last element directly and append if it meets the condition.

The O(NlogN) algorithm is not actually maintaining a LIS array. The i-th position of the array is the smallest possible "ending" value of an increasing subsequence (IS) of length i+1. Considering nums = [0, 2, 3, 1], it'll produce [0, 1, 3] which is not a legal subsequence. [0, 1, 3] actually corresponds to IS [0] (of length 1), [0, 1] (of length 2) and [0, 2, 3] (of length 3)

Finally!! Found someone who explained nlogn solution in an understandable manner. Thanks a lot.

Great video!

easy to understand :) thanks for sharing :-))

Nice! Make a few more of these, with slightly differences, so they will be longer! TIming is good, just a bit too short! I also have an RUclips channel, go check it out?

Shorts lao

Very beautifully explained!!

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👍

Voice is awesome

💯🤟🏻

Great 🤩

That was very good, keep it up 👍

Nicely explained

that was greatly explained :-)

You made this so easy to understand !!

Nice explanation of nlogn solution

Nice

Check this out 🙃 #include<bits/stdc++.h> using namespace std; int main(){ int n; cin>>n; int a[n]; for(int i=0;i<n;i++){ cin>>a[i]; } int dp[n+1]; //length from 0-n dp[0]=INT_MIN; for(int i=1;i<=n;i++){ dp[i]=INT_MAX; } for(int i=1;i<=n;i++){ int idx=upper_bound(dp,dp+n+1,a[i]) -dp; if(dp[idx-1]<a[i] and a[i]<dp[idx]){ dp[idx]=a[i]; } } int ans=0; for(int i=1;i<=n;i++){ if(dp[i]!=INT_MAX){ ans=i; } } cout<<ans; return 0; }

Hi, great content. Thanks a lot!

Great🤟🏻

🔥🔥🔥

#classmate 😀