LeetCode Biweekly Contest 19 (2nd) + Explanations

I started 3 minutes late but still got 2nd! Anyways, explanations start at 10:11. If they are a bit unclear, feel free to ask questions in the comments below so that I can improve!
Bonus: Solve the 2nd problem in O(nlogn), with the modification that you need to count the number of subarrays of any length! (atcoder.jp/con...)
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[@zqz9112]

pretty much helpful. Thanks for the clear explanations!

[@josephwong2832]

love these weekly LC contest explanations!

[@himanshusingh3056]

Bonus question is indeed interesting! Will try..

[@akashgupta13]

Good work buddy! 🙂

[@rohithgupta9672]

Thanks Lin

[@iitnakanpur..]

Cool stuff

[@yojahuang]

tmworz

[@shashanktiwari4442]

You can now get some tshirts from leetcode with ur leetcoins..😀

[@kasthurishravankumarhpc]

Thank you, sir, an explanation is nice.

[@alirezaghey8763]

Your explanations are great. The whiteboard you use for visualization purposes is also great. Would love if you could solve randomly some leetcode problems and also explain. Have checked some other competitive programmers and they lack your ability to clearly explain your intuition and thought process. Thanks!

[@saikat93ify]

Can you share the reason we are only relaxing the edges only if it's weight is oo ? We aren't checking if steps[v] > steps[u] + 1 at every step

[@tmwilliamlin168]

steps[v] > steps[u]+1 is the same as steps[v] >= 1e9 in my bfs

[@saikat93ify]

@@tmwilliamlin168 What I'm trying to ask is that we reduce the edges from oo to a finite value the first time, but we never relax the edges after that. Is there any reason for this ? That the first time we can visit an edge is the lowest distance for that edge ?

[@saikat93ify]

@@tmwilliamlin168 Can you please explain the purpose of the clear statement ? I am getting TLE without it ?

[@tmwilliamlin168]

In this graph, all edge weights are the same, so the first time we visit an edge to v, say u1->v, will always give the lowest distance. If we visit another edge u2->v later on, d[u2] must be >= d[u1], so d[u2]+1 >= d[u1]+1, and we don't need to relax the edge.

[@tmwilliamlin168]

Consider the array [1, 1, ..., 1]. Without the clear statement, the algorithm becomes O(n^2).

[@sumantopal558]

Liked, subscribed

[@naveenverma2390]

can you explain why am i getting error if put 2 instead of 2.0 in clock angle question for this Your input 3 15 Output 7.00000 Expected 7.50000