Nice idea, we will think of renaming it from next year onwards.

meooow orz

Actually, since this contest is rated for both divisions, the problems have been verified and tested by the CodeChef admins. Therefore, you can expect a decent problemset in this contest. We'll be more than glad to hear the feedbacks from you after the contest.

We hope to see you in the ranklist. :)

You can also check their last year's problem set. Coders' Legacy 2019 (Rated for all). It was combined rated round for all.
Although previous rounds don't promise that next round will be good as well. But still can be a good estimate about the upcoming round's quality.

The random contest made by an arbitrary university is never rated if CodeChef admin's feel it's not up to the mark of their regular rated (Cook-Off, Lunchtime or Long) contests.
In case of a bad quality round. They also own partial responsibility along with setters/testers.

Sure, they are ready.

The time limits have been verified by multiple testers. Try to come up with a more optimised solution.

So am I

The CodeChef dev team is looking into it.

Also getting nothing but 500-s now :-(

It's still rated. The CodeChef team is looking into it.

Kindly send an email to help@codechef.com

The format is still same as shown in sample output. There was a typo in the statement which is fixed now.

Exactly! For the first time, I sat 3 hrs straight. Thanks for such a well-balanced contest.

$$$n\log{n}$$$ sufarray, for all $$$l$$$ find $$$\min(r\,\colon\,s[l,r]\text{ is unique})$$$ and write $$$i$$$ at the point $$$(l, r)$$$, where $$$i$$$ is the position of $$$l$$$ in the sufarray, then each query $$$(l, r, k)$$$ can be reduced to looking for the minimal number in the rectangle with $$$l \leq x \leq r - k + 1$$$ and $$$y\leq r$$$. Build a segment tree for $$$x$$$'s where in $$$[x_l, x_r]$$$ we store all pairs $$$(r, i)$$$ where the number $$$i$$$ is written in $$$(l, r)$$$ and $$$x_l\leq l\leq x_r$$$, sorted by $$$r$$$, then we sort queries $$$(l, r, k)$$$ by $$$r$$$ and iterate over them in this order, lazily maintaining the minimal $$$i$$$'s in the segtree's nodes, thus making $$$O(n\log{n})$$$ single modifications in total.

DFS from n,m node and mark all nodes which can lead to it. Do a BFS kind of soln from 1,1 and go to nodes which are valid and having minimum char at particular depth. ( Basically solve diagonally )
Time complexity O(n*m)

You can see the editorial here.

Check my comment above.

$$$O(nq)$$$ lol https://mirror.codeforces.com/blog/entry/67001

just HLD -> Seg tree -> Trie

The intended solution was with Trie and DSU. Check the editorial here

You can use binary search to solve it.

Each wall is a straight line of the form x+y=a. For a point (x1, y1) to lie outside the line, x1+y1>a should satisfy. So for each point, you just have to find the last line/wall such that it is outside the line/wall. This can be done by binary searching (x1+y1) in the array of sorted walls.

sarthakmanna avijit_agarwal First I found out all the cells reachable from n,m. Then I did the queue implementation of bfs starting from 1,1 which took the coordinates of the cell and the letter in its parent cell as input. For the lth character of my answer I considered all cells with i+j-1=l and having parent cell equal to the (l-1)th character of the answer string.

I don't know exactly which test cases are failing in CodeChef. But your code fails on this test:

2
2 3
zba
bbb
2 3
zb
bb
ab

The answer for both of the cases should be same: zbab. But on first test case, your solution prints zbaz.