Please Someone Explain Why i am Getting TLE in Test case#5 of Question D

My Submission is : 85376735

For problem D-Zero Remainder Array, why does my code get a TLE on test 8 despite using a similar approach? My code: https://codeforces.net/contest/1374/submission/85382181

Finally Editorial Published. Thanks vovuh and team.

After Long Time

Please organize another div3 round. This was a div3 round after a long time. Newbies like me need div3 rounds continuously.

Am I the only one who think that this was the best round ever ?!

Can E1 be done using DP?

https://codeforces.net/contest/1374/submission/85385019 Solution for B

Ahsant. problem E2 is really nice (and hard). But the proof is not clear to me. I mean the analysis of number of times the operation is done.

Upd: Yeah got it now. The editorial was also clean. Well that's why I always take part in Div3s.

I solved E2 by applying ternary search on number of common books, but i don't have any proof that it is unimodal. Can someone explain me how it is unimodal or my solution is hackable?

Can anyone explain other solutions of E2, as I saw many solution using bit tree ,treaps, ternary search.

Gr8 set of questions.....All of them seem doable.... although problem E(s) was a bit heavy on maintenance side....

Could someone please help me with the problem E1 easy version. I implemented the idea which is iterating over the books of the three vectors and finding the cheapest price, but it always gives a wrong answer on test case 13.My answer is cheaper than the actual answer given by the system. Thanks in advance

This is my submission to the problem

For Problem D: Find the ai % k for each element in a. Then find the one with the highest frequency (mode).

Let x be the number with highest frequency and f be the number of times it occurred .

If the new formed array (with ai % k) is empty or full of zeros the answer will be 0. Else the answer will be x + (k * (f — 1)) + 1.

Solution Python3 -- > 85419324

In problem B, I have another simple solution

int query(ll n)
{
    int cnt = 0;
    while (n % 6 == 0) n /= 6, res += 1;
    while (n % 3 == 0) n /= 3, res += 2;
    cout << (n == 1 ? res : -1) << '\n';
    return 0;
}

E1 to E2 so big gap.E2 tooo hard for div3..... 
E1-1600 dificulty & E2-2500 dificulty  :(

In E1 we can sort all books and pick books greedily. We need to keep track of number books liked by Alice and Bob, and a stack that stores books that are liked by single person only. Code

Can somebody pls find the error in my code for E2 https://codeforces.net/contest/1374/submission/85464212. My procedure: Fist solve the problem like E1 and find out the number of items.If it is greater than M then erase 10 and 01 items and add 11 items.If it is less than M then we can add one 01,00,11,10 or we can remove one 11 item and add 01 and 10 items.

problem E1..wrong ans test case 5...i can't find wrong my code...please help me to finding wrong or give me some small test http://codeforces.net/contest/1374/submission/85463861

Codeforces Round #653 (Div. 3) D. Zero Remainder Array

In this Problem , if we consider a input — 1 2 5 4 4

then as per the solution provided,the output is 7 but if we conside like —

1 .when x = 1 , we will server one of the 4 (4+1=5 and 5%5=0) then x will become 2 2. when x = 2 , we will add it to the remaining 4 and it will become 6 and x will become 3 3. At this step , x is incremented and becomes 4 4. Now if we add 4 to the 6 , it will become 10 and 10%5==0 and then x becomes 5

So the answer should be 5 . . Is it ? ? ?

My Video solution for the contest

Problem A,Problem B, Problem C The video is time stamped for your convenience (check the description).

Problem D
Problem E

Can someone please help me, why my submission of problem A is getting TLE if I'm using a single while loop for a test case.

My submission: https://codeforces.net/contest/1374/submission/85356771

Thank you!

Can someone please tell me what is wrong with this greedy approach for E2 :

Let's find the maximum amount of books of Type 1 or Type 2 that can be picked. They would simply be (m-k). Let's denote it with allowed

Greedily pick those books that either Alex or Bob likes in the following way :
Let t1 denote least time taken for an unpicked book of type 1. 
Similarly t2 for type 2 and t3 for type 3 following above criteria.

We can fill in the rest of the remaining books with those unpicked books taking least time.

This approach fetched me a WA on test 12. I looked in the status and a lot of people got the same result on same test case. Can someone pls help. Thanks in advance

"and move it to the beginning (so the negative balance becomes zero again and the answer increases by one). This move is much more optimally than moving the current closing bracket to the right."

"Moving a closing bracket to end" or "an opening bracket to beginning",doing either of these will result in the same answer ,so why the second one is optimal ?

Can someone explain me why I am getting Wrong Answer in E2. https://codeforces.net/contest/1374/submission/85472085

My logic: lets call the 1 1 books as both. if both+Alice < k: then not possible also if both+Bob <k then not possible

Otherwise I sort the books. Then fulfill the condition of k by taking the cheapest book. Finally I take the remaining elements and sort them and take the cheapest books to complete the set

Edit: I am new to competitive programming so can you also help me to make my code efficient.

I had a really simple solution for E1:

I claim that Alice and Bob will both read exactly $$$k$$$ books that they each like. If we had a selection where both of them read more than $$$k$$$ books that they liked, then we can subtract out books (which decreases the total time) of $$$11, 10, 01$$$ until one of them has read exactly $$$k$$$ books that they like and the other has read $$$K \geq k$$$ books that they like. However, if one of them has liked $$$K-k$$$ more books, then they must have read at least $$$K-k$$$ books where they were the only one to like it. Thus, we can take out these books as well, which gives us a selection of books where they read exactly $$$k$$$ books that they each like, which is better than the original selection (where at least one person liked more than $$$k$$$ books).

Now, based on the claim, if they read a $$$10$$$ book, they must also read a corresponding $$$01$$$ book in the optimal solution. Thus, the problem now just comes down to pairing these $$$10$$$ and $$$01$$$ books. We can do this greedily (I'll leave the proof for you). We sort the $$$10$$$ and $$$01$$$ books independently, and match them from lowest time to highest. Now, we essentially have created $$$11$$$ books from $$$10$$$ and $$$01$$$ books, where the time of this double-book is the sum of times of its two parts. We can just treat this double-book the same as a normal $$$11$$$ book, so we can sort the $$$11$$$ and $$$10+01$$$ books all together and take the first $$$k$$$ smallest times to find our answer.

Code: https://codeforces.net/contest/1374/submission/85336049

Has anyone solved problem D using sorting and 2 pointer approach?

Can anyone please help why am i getting wrong answer on test case 12 of problem E2. My code — https://codeforces.net/contest/1374/submission/85427982 Initially I have taken all books which is interesting for both.Then if it is less than k then i have taken from remaining books which are in sorted order until the count of interesting books for both is equal to k.After that i have replaced the books interesting for both with books interesting to 1 only(2 books added for each 1 book interesting to both removed).Afterwards I have just added more books to make the total count m.I am not being able to figure out what is wrong in this approach. Plz. help.

I had different approach to E1, solved it greedily.

We can separate books into 3 groups : the ones that Alice, Bob and both of them like respectively.

Sort them, and if the sum of the minimum element in "Alice" group and "Bob" group is less than the minimum element in "Both of them like" group, then i add up this sum to the answer, otherwise i add to the answer the minimum number from the "Both of them like" group.

Proof:

There are finite number of possibilities so the answer exsistes. If we replace books from the answer, that only Alice likes, by the same number of the smallest elements from the "Alice" group, answer will stay the answer, i mean that value of the answer won't get bigger. The same for the books from the answer, that only Bob likes.

If the sum of the smallest(unused) book from the asnwer, that only Alice likes and the one only Bob likes is smaller than biggest book that both Alice and Bob like, then we replace that one book by these two books, value of the answer won't get bigger, so it will remain an answer. Do it untill can't be done.

If the smallest book(unused) that both ALice and Bob like, has value smaller or equal to, the sum of the biggest by value book from the answer, that only Alice likes, and from the ones, that only Bob likes, then we can repalce these two books by that one book, and answer will remain answer, i mean answer value cannot get bigger. We do these untill it can be done.

And if we replace all the books from the answer, that both Alice and Bob like, by the same amount of smallest by value books from the group "Both ALice and Bob like", answer cannot get bigger, so it will remain an answer.

It is obvious, that there is only one such set of books to which any replacements from the above cannot be done and as shown above it is the answer.

The result we get from our greedy approach cannot have any of the replacement above, so it is the answer.

My submission

Why i am getting wrong answer for this submission(Problem E1): https://codeforces.net/contest/1374/submission/85476850 Please help!!

https://codeforces.net/contest/1374/submission/85367241 problem D.. idk why i am getting WA on tc 2...i had the same approach as in editorial...can someon help please!!

Imho this approach for E2 is easier: first, let's take k elements from 11 group (if this group is smaller than k — take all books from 11 group and first k - size(11) books from each 01 and 10). If number of books that we took is greater than m — the answer is -1. Now, we have some set ans of books that we took (size(ans) <= m), and we have 5 possible transitions: first 4 is to simply add the cheapest book from one of the sets, and 5th is to delete the most expensive book from 11 set and add 2 books from 01 and 10 sets (after this transition we will add cost(added) - cost(deleted) to the answer. After each of this transitions number of books in ans increases by 1, thus we have to make m - size(ans) iterations, (in each iteration we will try to add as less as possible to the answer). This approach can be easily implemented by storing pointers for each group. My code — 85363378.

Can somebody explain E2

Another solution for E1: I sorted 01, 10, 11 books independently. Now you need to reach k books for both Alice and Bob. So we either have to pick 11 or (a 10 and a 01). We can greedily pick whichever is cheapest. Create a new v10[i] + v01[i] books. Combine v11 books with this new tuple of books, sort them and pick the k smallest values. 85376690

please someone explain me this..

Then we need to find two consecutive pairs with the same first values and swap these two elements in the permutation. Because in fact these two numbers are equal in the array and have consecutive values in the permutation, we guaranteed change the parity of number of inversions.

Problem F

Pls explain why I'm getting WA in testcase2 of problem D. https://codeforces.net/contest/1374/submission/85355292

For anyone having trouble understanding the 1374E2 - Reading Books (hard version) editorial: I think they meant 'but elements of 11-group are not free' instead of 'but elements of 00-group are not free'.

For problem D, consider the test case n=3, k=7 and a={1, 1, 1}. With this approach the answer would be 21. But we can do it in 7 steps only. The required remainders are {6, 6, 6}. Which can be obtained by 1+5, 2+4 and 6. So array {7, 7, 7} can be obtained in one full cycle only.

E2 is shit!

Hey guys, if any of you having difficulty in understanding the editorial of problem E1. Here's a video to help you out.
Thank you!

IS THERE ANY PROBLEM WITH THIS CODE FOR PROB B https://codeforces.net/contest/1374/submission/85385019

Can't understand cyclic shifts sorting. Can someone explain why odd parity fails to give answer ?

Also why in the end two elements wont be sorted?

Problem F

Please Someone Explain Why i am Getting TLE in Test case#5

my submission is : 85516199

Can someone help me i am getting TLE in D part Here is my submission https://codeforces.net/contest/1374/submission/85529755 Thanks

please help me with my solution for E1. Am getting WA. link

Please help.I am Getting WA in Problem E2. My code:(https://codeforces.net/contest/1374/submission/85530060)

sorry,the code of the solution of E2 have some errors, it can't run in my computer,such as

for (auto [value, position] : st) res.push_back(position);

in the line 119 and

st.erase(prev(st.end()));

in the line 26

could any one help me? thank

Can anybody help me? for E2 (reading books hard version). I m getting wrong answer in test case 12 and my answer only differs by 1 from real answer.

my code if https://codeforces.net/contest/1374/submission/85595106

Can anyone help, what I'm doing wrong? I'm getting WA in E2.

My logic for solving this —

• Make min heap for all 00, 01, 10, 11.
• Get element from 11 or (01 and 10) whoever gives me lowest sum for contributing to k likes.
• After repeating step 2 for k times, get element from 00, 01, 10, 11 whichever contributes minimum to sum and repeat this till total books read goes to m.

my submission — 85968003

Thanks in advance!

My approach:
First step:
I am greedily allocating the common taste books, followed by remaining books of alice and bob. With this, I am ensuring at-least k-books are included in the list and the time is minimized.

With this, easy version got accepted.

Second step:
If booksRead > m, I am reducing the numbers by replacing the individual taste with the common taste books. If booksRead < m, I am greedily allocating the least time taken book from the remaining entire books available.

It seems that I am missing something in my code. Due to that, the hard version got failed in the test case 12. Here's the submission link: https://codeforces.net/contest/1374/submission/86028907

Can someone please explain what I am missing in my code?

Please help me i am getting TLE in question no. D

My submission : https://codeforces.net/contest/1374/submission/86383860

I am not getting why I am getting wrong answer for problem E. Here is my solution: https://codeforces.net/contest/1374/submission/86808426

MikeMirzayanov,vovuh
In the problem 1374F - Cyclic Shifts Sorting, is the order of indices for performing the operations fixed?

While performing the greedy algorithm, I chose the maximum element instead and moved it to the last position. Wouldn't that give me a different order of indices?

The solution 86861838 having taken the maximum element isn't being accepted. Please clarify.

Could someone help my E1 code out? I got TLE at case4. CODE

Problem D

Why max count always giving the maximum final value of x? (x from problem description).

Because :

say, min possible value with maxCount < max possible value with less than maxCount

which is => (maxCount-1)*k + 1 + 1 < (maxCount-2)*k + (k-1) + 1

=> 2<0 ; Impossible .

        for(int i=0; i<n; i++){
            cin>>a[i];
            if(a[i]%k == 0) continue;
            cnt[k-a[i]%k]++;
        }
        lli ans = 0;
        for(auto [key, value] : cnt){
            ans = max(ans, (value-1)*1LL*k + key + 1);
        }

This works too for problem B

if(n==1) return 0;
		int c{0};
		while(n>1)
		{
			if(n%6==0)
			{
				n=n/6;
			}
			else if(n%2!=0)
			{
				n=n*2;
			}
			else
			{
				return -1;
			}
			c++;
		}
		return c;

Please someone, explain why am I getting Memory Limit in Test Case #5 for Question D

This is my submission: 142601616

for problem B why was it (cnt3−cnt2)+cnt3 please anyone explain me i am just a beginner

I really really hated E2 because its a really easy question, but the implementation required is nightmarish.

Anyways, alternate O(n*log^2(n)) solution using two fenwick trees which can be quite easily optimised to O(n*log(n)) using a segment tree (but I'm too lazy to implement it).

Anways, my solution is similar to the editorial in the sense that we iterate over the number of (11's) used and then we use k — (11's) used (01's and 10's). But I use the fenwick trees to simulate this entire process.

This question can also be implemented with indexed multiset but I'm too lazy for that.

Why are some people crazy? Why are they showing up the solution without hiding? There is an editorial. You can make an important contribution by writing a comment that will help someone. Don't ruin anyone's importance. thank you.