Agree.... feeling sad after making wrong submissions blindly :(

And that surprise "Case 1:" twist was the ultimate test :)

No, leetcode contests are harder.

Enough with the direspect bro, this can’t be accepted,

leetcode contests are much better than this

full shit sherlock

They added extra tests and rejudged your submission. After rejuding, execution time has not been updated.

just write it iteratively, and/or use booleans rather than max

mathura laxmi chit fund

You can't rely on 2d-vectors where your complexity is $$$\mathcal{O}(N^2 \cdot T)$$$ which is probably $$$8 \cdot 10^8$$$ iterations, with time limit of 3s.

Use 2d-arrays instead.

More like 1200 into 1400 but yeah, that was so annoying (:

What when s[i] is equal to value at both pointers!

I used the same approch but sill not able to understand why equal value would effect

do you have the same for Amritapuri regionals?

For me it didn't

Centroid decomposition ig

Let sz(x) mean the size of the subtree rooted at x. Formally $$$sz(x) = 1 + \sum sz(v)$$$ for all children $$$v$$$ of $$$x$$$.

Choose an arbitrary root and walk through the tree normally with a dfs calculating values of $$$sz$$$. For any node where $$$sz(x) \geq \sqrt{n}$$$, add $$$x$$$ to the subset and consider $$$sz(x) = 0$$$ for further calculations.

Why does this work?

• For any child $$$v$$$ of $$$x$$$, $$$sz(v) \lt \sqrt{n}$$$, so the distance between any node in this subtree from $$$v$$$ must also be $$$\lt \sqrt{n}$$$. Since $$$v$$$ is a child of $$$x$$$, the distance of any node in the subtree from $$$x$$$ is $$$\lt \sqrt{n} + 1$$$ or in other words $$$\leq \sqrt{n}$$$.

• Since $$$sz(x) \geq \sqrt{n}$$$ whenever we perform this, we add at most $$$\frac{n}{\sqrt{n}} = \sqrt{n}$$$ nodes to the subset.

1) Pick any spanning tree and root at node $$$1$$$.

2) Pick the node having the max depth, go up $$$\sqrt{n}$$$ for that node following parents, push this one in the answer, and remove that subtree.

3) Do step 2 until the tree is not empty.

Yeah Amritapuri was relatively better, getting a lower rank didn't hurt in Amritapuri but it did in Mathura ;(

Ya I guess my worst case would be n^(3/2), you can refer here

is that common?

Nope, full tests are run during contest time.

For each L you can find the smallest R by binary search and if L,R is good then L,(R+1) is also good.

What are the numbers next to the college name? The number of teams qualifying?

Our team has rank 45 from Kanpur-Mathura and we are the 2nd team from our college and the 1st team has rank 10 so do we stand a chance for regionals

Results for Amritapuri have been announced: link

Depends upon the no. of slots :)

today or tomorrow, there are 150 seats

Source — codedrills telegram channel

(someone claimed to got reply from icpc mathura-kanpur via email )