awoo's blog

By awoo, history, 13 months ago, translation, In English

1626A - Equidistant Letters

Idea: BledDest

Tutorial
Solution (awoo)

1626B - Minor Reduction

Idea: BledDest

Tutorial
Solution (awoo)

1626C - Monsters And Spells

Idea: BledDest

Tutorial
Solution (awoo)

1626D - Martial Arts Tournament

Idea: BledDest

Tutorial
Solution (awoo)

1626E - Black and White Tree

Idea: BledDest

Tutorial
Solution (BledDest)

1626F - A Random Code Problem

Idea: BledDest

Tutorial
Solution (BledDest)
 
 
 
 
  • Vote: I like it
  • +121
  • Vote: I do not like it

| Write comment?
»
13 months ago, # |
  Vote: I like it +16 Vote: I do not like it

it was a really cool round! it's a pity that he was unrated :(

»
13 months ago, # |
  Vote: I like it +4 Vote: I do not like it

For problem C, you don't have to worry about half-intervals. Here is my (badly written) code for C with segments that can be of length 0.

143049989

»
13 months ago, # |
  Vote: I like it -6 Vote: I do not like it

Can anybody please tell me how this code is working for problem D : 143147171.

Is this logic correct or are the test cases weak?

»
13 months ago, # |
  Vote: I like it +17 Vote: I do not like it

Same solution of B, but with regex (with code pattern): Perl — 143030768, (13+).

»
13 months ago, # |
  Vote: I like it +7 Vote: I do not like it

Can somebody please explain me how Binary search is applied on problem D????

»
13 months ago, # |
  Vote: I like it +26 Vote: I do not like it

C can be solved in $$$O(n)$$$, walking though array backwards or using stack.

  • »
    »
    13 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Also using two pointers can achieve %O(n)%

»
13 months ago, # |
  Vote: I like it +30 Vote: I do not like it

D can be solved in $$$O(n)$$$. the submission.

The steps of the algorithm is:

  1. Calculate if we want to choose at most $$$i$$$ smallest numbers, how many numbers can we choose.

  2. Calculate if we want to choose at most $$$i$$$ largest numbers, how many numbers can we choose.

  3. Go through $$$i,j$$$ in the powers of two. $$$i$$$ means the number of the first division after inviting extra participants. $$$j$$$ means the number in the third division. Then we can know how many original participants will be in the second division. So it's easy to find the number of participants in the second division. For every $$$i,j$$$, we choose the smallest sum of three divivions, and minus it with $$$n$$$. It will be the answer.

  • »
    »
    13 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    what about the case when i smallest numbers and j largest numbers are not disjoint there is some overlap ?

  • »
    »
    13 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    For example: n=2 a={1,2}. Let i=1 and j=1, which means there is one person in the left segment while one person in the right segment. However there is no way to add another person to the middle segment since there is no integer between 1 and 2. In short, your code gives the right answer in a wrong way.

    • »
      »
      »
      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Did't it say x < y? there is always a number between.

»
13 months ago, # |
  Vote: I like it +3 Vote: I do not like it

In Problem D,

I am not getting why the length of the middle segment should be power of 2 for the optimal solution.

Can someone help?

  • »
    »
    13 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    I also don't understand this

  • »
    »
    13 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    It's not the middle segment that should be a power of $$$2$$$. We iterate on the size of the middleweight category, which has to be a power of $$$2$$$.

»
13 months ago, # |
  Vote: I like it +10 Vote: I do not like it

Can anyone please explain C in further depth, I'm unable to understand

  • »
    »
    13 months ago, # ^ |
      Vote: I like it +4 Vote: I do not like it
    Example
»
13 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

I have a little problem about problem F: Why should we add $$$k\cdot n^{k-1}\cdot x\cdot L$$$ to the answer instead of $$$n^k\cdot x\cdot L$$$?

  • »
    »
    13 months ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    The multiplier $$$k \cdot n^{k-1}$$$ is the total number of times an element is chosen over all ways to choose elements on the iterations. It can be treated as the number of occurrences of some integer $$$i$$$ in all $$$k$$$-element vectors, where each element is in $$$[1, n]$$$. It is $$$k \cdot n^{k-1}$$$ since the total number of integers in these vectors is $$$k \cdot n^k$$$, and each of $$$n$$$ integers occurs the same number of times.

»
13 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Can someone please explain for Problem E, I tried this test case - for the solution mentioned in editorial

input —

13
1 0 0 0 0 0 0 0 1 0 0 0 1
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9

the output is —

1 1 1 1 1 1 1 1 1 0 0 0 1 

As per my understanding, according to the problem st. shouldn't it be, only black nodes and the ones adjacent to it, i.e. 1, 13, 9, 2, 12, 8

1 1 0 0 0 0 0 1 1 0 0 1 1

Can someone explain this?

Upd: Sorry I gave wrong input.

  • »
    »
    13 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It doesn't have to be only black nodes and the ones adjacent to it. Consider an edge which links two white nodes. If the number of black nodes on the left side of this edge (in the entire tree) are 2 or more, then we can go from right to left along this edge, and similarly for left to right. The adjacent-black-node condition is only needed when the number of black nodes on one side of an edge are less than two.

    Example: If your tree is 1-2-3-4-5 and 1,2 are black nodes. Then from 5 you can select 1 and move the chip to 4. Then you can select 2 and move the chip to 3, and so on. So, whenever you have 2 or more black nodes on one side of the edge, you won't get forced back as you move along the edge in that direction.

    • »
      »
      »
      13 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Got it, thanks for the explanation!

»
13 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Hi , Can Anyone please point out what is wrong in my Approach for problem C . I have used a greedy approach while mantaining a stack that holds the index of all the monsters that i defeat by starting a spell somewhere between a[i] and a[i-1] . So if for the ith monster , lets say that it can be defeated by continuing to add 1 to the spell casted at the i-1th index until we reach the ith monster , so i directly do that , if i can defeat the current monster by starting from 1 again in between the time of the previous monster and the current , while pushing the index of the current element in the stack. So only 1 case is left where , i am not able to defeat my current monster even by adding 1 to the (i-1)th spell through the time k[i]-k[i-1] , in that case i start popping my element from the stack until i reach an index x where instead of starting my spell from 1 before x , if i had continued adding 1 from the x-1th spell , our current monster too could have been defeated . This also gives the minimum mana that we will need to defeat the current monster . I would really appreciate the help , here's my code : 143039544

»
13 months ago, # |
Rev. 2   Vote: I like it +1 Vote: I do not like it

Problem C is a lot simpler using stack, when we introduce a new segment, we pop the stack for the segments that were overlapping, then in the end answer is simply derived from the segments left in stack as they are all non overlapping.

Here is my code for the same 143004113

»
13 months ago, # |
  Vote: I like it +11 Vote: I do not like it

In Problem D, I ran 3 nested loops i, j, k for the power of 2 the first second and third segment needs to be, and checked if it was possible to divide the array into these groups, if it was possible, simple store the minimum for every iteration of i, j, k. answer in any iteration will be 2^i+2^j+2^k-n.

Here is my code for the same 143024306

  • »
    »
    13 months ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    got the same idea as you!

    it's cool to solve it in $$$O(n + \log^4 n)$$$ instead of $$$O(n \log n)$$$

    can even solve in $$$O(n + \log^3 n)$$$ after some optimization

    143025919

»
13 months ago, # |
  Vote: I like it -19 Vote: I do not like it

question statement of problem B says "You are given a decimal representation of an integer x". Initially, I didn't get that the input in problem B is a string or an integer....?

firstly, I have tried my solution with an integer and got runtime error in test case 3 and later have tried with a string as input and my solution got accepted.

Moral: a decimal representation of an integer x is a String.

  • »
    »
    13 months ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    Correct Moral: read constraints carefully next time.

    • »
      »
      »
      6 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      can you explain how with help of constraints we came to know that we have to input a string . Sorry for the silly question though!

»
13 months ago, # |
  Vote: I like it 0 Vote: I do not like it

C has a tag of binary search, can someone please share/explain a binary search based solution if they have implemented it?

»
13 months ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it

E.

Another solution and explanation of E.
»
13 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Problem E was not that tough, I have solved it with in out dp; Solution link

»
13 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Problem C can be solved with $$$\mathcal O(n)$$$.

»
12 months ago, # |
Rev. 6   Vote: I like it 0 Vote: I do not like it

Hi everyone, I want to share my solution for C which is $$$O(n)$$$.

Here is my submission:144945360.

Spoiler
»
8 months ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it

It’s interesting to notice that in D’s solution the while loop has the condition of “mid <= n” instead of “len1 + mid <= n”. Actually I can prove “len1 + mid / 2 <= n” will not miss cases but cannot prove the other two condition’s correctness.

»
2 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

D can be solved with dp + segment tree in $$$O(nlog^2 n)$$$.

Here