BledDest's blog

By BledDest, history, 12 months ago,

1765A - Access Levels

Idea: BledDest, preparation: awoo

Tutorial
Solution (awoo)

1765B - Broken Keyboard

Idea: vovuh, preparation: vovuh

Tutorial
Solution (vovuh)

1765C - Card Guessing

Idea: DStepanenko, preparation: BledDest

Tutorial
Solution (awoo)

1765D - Watch the Videos

Idea: BledDest, preparation: DmitryKlenov

Tutorial
Solution (DmitryKlenov)

1765E - Exchange

Idea: BledDest, preparation: BledDest

Tutorial
Solution (BledDest)

1765F - Chemistry Lab

Idea: awoo, preparation: awoo

Tutorial
Solution (awoo)

1765G - Guess the String

Idea: BledDest, preparation: BledDest

Tutorial
Solution (BledDest)

1765H - Hospital Queue

Idea: Neon, preparation: Neon

Tutorial
Solution (Neon)

1765I - Infinite Chess

Idea: DmitryKlenov, preparation: dmitryme

Tutorial
Solution (awoo)

1765J - Hero to Zero

Idea: BledDest, preparation: BledDest

Tutorial
Solution (BledDest)

1765K - Torus Path

Tutorial

1765L - Project Manager

Idea: BledDest, preparation: awoo

Tutorial
Solution (awoo)

1765M - Minimum LCM

Idea: BledDest, preparation: Neon

Tutorial
Solution (Neon)

1765N - Number Reduction

Idea: Neon, preparation: Neon

Tutorial
Solution (Neon)
• +50

| Write comment?
 » 12 months ago, # |   +5 Unfortunately, the editorials for two problems are not ready yet. They will appear here as soon as they're written.
 » 12 months ago, # |   +3 In problem Torus Path wording "Note that you can't visit all vertices on the antidiagonal (vertices .... at the same time." — is quite confusing.
 » 12 months ago, # |   0 can problem N can be solved by using stack data structures ?
•  » » 12 months ago, # ^ |   0 Yes indeed! You need to check for all 1 <= i <= n-1, if s[i] > s[i+1] then we can delete s[i] to obtain a smaller number. You will continue doing this till k is 0. One special case, when stack's first element is greater than 0 and all the remaining values are 0, when placing a value x > 0 and x < stack.top, if k >= stack size, delete all the elements from the stack and push the current one . P.S — You need vector to do this stack operation. Check my submission
•  » » » 7 months ago, # ^ |   0 Thanks :)
•  » » » 3 months ago, # ^ |   0 nice solution!
•  » » » 5 weeks ago, # ^ |   +3 I liked your implementation! This code indeed requires a good sense of implementation! How did you come up with this?Have you done similar problems before?
•  » » 2 weeks ago, # ^ |   0
 » 12 months ago, # |   0 The problem K is easily solvable using DP. But the greedy method is quite tricky, at least for me.
•  » » 11 months ago, # ^ |   0 Can you please explain your solution.. And how do you store which cells you have already visited? Thanks in advance
•  » » 10 months ago, # ^ |   0 I was wondering if DP was possible, but I'm not so experienced with DP, and then I the greedy anyways.
•  » » 4 months ago, # ^ |   0 In your code,using 'for loop', you have have handled the case of going from last to first within a row.But you have not done anything in case of going from last column to first column,whenever lvl==n if prev!=n-1 , you are just returning INT_MIN.why have you ignored that case??
 » 12 months ago, # |   0 Having a problem not understanding the solution to F.Please, can someone explain it more?
 » 11 months ago, # |   0 The problem D can be done in O(N) using just two pointers.
•  » » 11 months ago, # ^ |   0 Sorting does require O(NlogN) operations, you probably meant O(N) operations after sorting.
•  » » 11 months ago, # ^ |   0 Your solution seems ease to understand. Thank you for sharing!. But TC is O(Nlog(N)) because of sorting.
 » 11 months ago, # |   0 My implementation for N 187904357 using priorityqueue
 » 11 months ago, # |   0 can we do N. Neon by using monotonic stack ??
•  » » 10 months ago, # ^ |   0 Yes we can. Check out my solution 190315580. I am using string itself to simulate stack.
 » 10 months ago, # |   0 Neon Can u plss explain why the limit of for loop in problem M is g * g <= n??
•  » » 6 months ago, # ^ |   0 It is for finding divisor of the number which you will get within root n
 » 10 months ago, # |   0 Is there anyone to fail my code ? imo my logic is true and working every test cases that i used. Any help appreciated. 191443868
•  » » 10 months ago, # ^ |   0 Try: 1 906 2 
•  » » » 10 months ago, # ^ |   0 thanks ^^
 » 10 months ago, # |   0 How to approach problem K that at max n-1 anti-diagonal elements can be visited editorial proof is understandable but how to approach such a mathematical problem???
•  » » 8 months ago, # ^ |   0 can you explain how at max n-1 anti diagonal elements can be visited. I was unable to understand this part of editorial
 » 4 months ago, # |   0 Thank you BledDest for one of the most beautiful problems I've recently seen (J).
 » 4 months ago, # |   0 In problem G, using the exact same idea from the editorial, and going backwards from the end instead of forwards, you end up with $1000/1.5 \approx 667$ steps on average. Even better, when you get a reply $x > 2$, you can find $x$ values in one move.