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Can anyone summarize?

Live ho

3rd one is codeforces marked 1900 question

Bro which tablet are u using?

I hope you guys are doing well🎉

Company tags on telegram, Discord & Website :) Structured DSA Practice Problems (700 Problems Structured) - codewitharyan.com/cwa-sheet/practice-problems

some feedback - make it a screen share instead of a vidcast a lot of the screen gets hidden - share the code

instead by changing condition only if(dist[i]+myweight> node[i]){then update the nodes[i] weights works?

what about calculating the level for each component individually and adding it to get final answer?

For BFS, why can't we start from node with minimum degree and instead have to try BFS from every node?

Excellent observations and Very well explained by breaking down the problems and justifications.

Great Job Bro ❤ great explanation for a good problem ..completely understood thanx

best explaination bhaiya

How we would have the changed the code if the question also demanded to list the maximum grouping? Thanks for the explaination! Simply great as usual! :)

Thanks for the intuition, bro This really helped me much to understand

Great explanation brother

very intuitive Explanation

There are multiple answers to the same graph like in the same example you can remove (1, 4) edge or (2, 3) or (3, 4) or (1, 2) to remove cycle but for the answer we need to remove the edge which appear at the last that's why answer is (1, 4). in case of dfs this wont work as we are just detecting cycle and remove the edge which causes cycle, in the same graph start dfs with 2 -> 2,1,5,4,3,2 so here 2 is again visited to (2, 3) would be answer but it is not correct. In case of dsu it will give correct answer as we are going in order

Bipartite (DFS) - kzbin.info/www/bejne/jWWvmZaVqbZ0qqc Bipartite (BFS) - kzbin.info/www/bejne/nHjJnX-cgJqHfNE DSU - kzbin.info/www/bejne/d6DWqGuJh9VqsJY

thanks for sharing your explanation

thanks for the video :)

DSU dsu(N); for (auto edge : edges) { // If union returns false, we know the nodes are already connected // and hence we can return this edge. if (!dsu.doUnion(edge[0] - 1, edge[1] - 1)) { return edge; } } // Why above code works ? // according to question this should be the implementation right vector<int> lastCycleEdge; for (const auto& edge : edges) { if (!dsu.doUnion(edge[0]-1, edge[1]-1)) { lastCycleEdge = edge; // Store the cycle-forming edge } } return lastCycleEdge;

Good solution. To keep it much simpler, this can be solved using post order traversal as well.

SYSTEM DESIGN WHEN?????

This guy gives the simplest and efficient solution there ever is

DSU - kzbin.info/www/bejne/d6DWqGuJh9VqsJY My Uber Interview Experience - kzbin.info/www/bejne/jHi8e3yOat6fbZY My Coinbase Interview Experience - kzbin.info/www/bejne/f5uydGRul8SXedk My American Express Inteview Experience - kzbin.info/www/bejne/mWS4mYybm82hp80 My JP Morgan & Chase Interview Experience - kzbin.info/www/bejne/Y5vElIemjppqpNE ..... more coming soon (along with LLD course on Second Channel)

In the dfs approach, we can avoid using the extra space of a visiting array by changing the value of a grid to zero after adding.

Nice Explanation, Eagerly waiting for 3435. Frequencies of Shortest Supersequences.

Bad explanation 😢

god level explanation.........

is it live or recorded video

I'M UNABLE TO SEE YOUR CODE . CAN ANYONE PLEASE HELP ME HOW I CAN I SEE SOLUTION AND SUBMISSIONS SUBMITTED BY HIM?

this can be solved in O(n + qlogq) by doing lazy updates on all and here and prefix sums + binary search for eliminating overcounting.

Hi @aryan why cant we solve this by first forming a map of the toposort array (element,level in the toposort) so if for (a,b) if map.get(a) is less this means this comes first in toposort but 1 things is left how will this ensure that a,b are connected or not so can use DSU for that so wont then time complexity by just (n^2) i.e. of toposort and dsu just takes ~ o(1) Please suggest am I correct ?

very well explained, thank you Aryan

THIS IS THE BEST EXPLANATION OF THIS QUESTION ON THE WHOLE INTERNET

This guy is genius🎉

Lots of love ,just loved your content watching day and night

brother where is 3435. Frequencies of Shortest Supersequences

brother where is 3435. Frequencies of Shortest Supersequences

brother where is 3435. Frequencies of Shortest Supersequences

code quality

Wow. Best explanation for this problem.

Awesome Explanation ❤❤❤

Topo sort is fine but dfs also works right given the bound is only 100 ?

can anyone provide the C++ code with the same logic?