In my opinion, I think that the positions of E and D should have been swapped.

In the problem E-Replace the Number, alternate solution mentions the small to large method. Can anyone explain how it has complexity of O(n log n)? Any resource added would also be helpful.

Mike, please please remove cheaters, I am at 1899

I upsolved E using map of list as merging list is O(1) operation.

Can someone please explain why this DSU based solution for E is failing TC 4?
I am using a hashmap and a pointer array to map the representatives to the values and vice-versa; and everything seems to be working fine.
My Submission.

In case you are interested in video solutions, Here they are(for A-E)

For problem E, we can have map of number vs linked list of indexes. Then for type 2 query, we just need to append linked list for x behind linked list for y which can be done in o(1). and for type 1 also be handled in o(1). overall o(n).

Can somebody please help me figure out what is wrong with my submission for E

Edit: Silly mistake... I reversed the digits in my answer

someone, please explain to me the solution of Problem C in easy language

In C , are there any constraints on the total number of strings possible ? Will it be always in the INT range ?

Can anyone check my E solution. I have stored the values and a vector of pointers along with it and and when changing value i just move the pointers from one to another https://codeforces.com/contest/1620/submission/139817827

Getting TLE in Problem E, with this solution. Can somebody tells why because it seems more efficient than one in the solution 2 of tutorials.

https://codeforces.com/contest/1620/submission/139879128

https://codeforces.com/contest/1620/submission/139877833 Can someone explain why this submission gives TL on E? It should be O(n) since double-linked list concatenation is O(1) operation... Thanks.

While going through the solutions of other participants for problem E I noticed everyone is using a large vector to store the indices for each value. So I am wondering, is there anything wrong with using a map? I used an unordered_map<int, list<int>> and it did work fine for this problem. Is there something I have to worry about when using maps for such problems (other than the hashtable overshooting the memory limit)?

In problem E why my solution is wrong this one.any suggestion or testcase

https://codeforces.com/contest/1620/submission/139826532 I have a question about E. in my opinion i have O(n) solution, but it takes TL at 5th test, can someone tell me what's wrong? Is it solution or it's just Python can't make it

Can somebody explain how we are converting x-1 to the mixed base?

Can someone tell me some similiar problem as E, or some algorithm tag?

In the editorial of D,

no need to take more than $$$3$$$ coins $$$1$$$ and more than $$$3$$$ coins $$$2$$$

I think you meant,

no need to take more than $$$2$$$ coins $$$1$$$ and more than $$$2$$$ coins $$$2$$$

Because, in the given solution code, you are iterating from $$$i = 0$$$ to $$$i \lt 3$$$. And, I have also done the same :)

cur = (j — i + 1) * k + 1 what is the use of plus 1 in the (j-i+1)

C problem is really amazing. And the solution is lot more amazing.

I have a different solution to F

following the editorial, there can't exist $$$i<j<k$$$ such that $$$a_{i} > a_{j} > a_{k}$$$. if an array satisfies this condition then the array can be decomposed entirely into two or less increasing subsequences.

Proof: consider doing a pass through the array and greedily picking out an increasing subsequence. Do two such greedy passes through the array let $$$S$$$ be the sequence chosen by the first pass and $$$T$$$ the sequence chosen by second pass. Assume there exist an index $$$k$$$ that isn't chosen by either sequences pick such smallest $$$k$$$, indexes $$$1$$$ to $$$k-1$$$ can't be all in $$$S$$$ otherwise $$$k$$$ would be in $$$T$$$, this implies there exist an index $$$i$$$ with $$$i+1$$$ < $$$k$$$ such that $$$i$$$ is in $$$S$$$ and $$$i+1$$$ is in $$$T$$$. Pick $$$j$$$ such that all indexes from $$$i$$$ to $$$j$$$ are in $$$T$$$ but $$$j+1$$$ is not. then $$$a_{i} > a_{j}$$$ otherwise some index between $$$i+1$$$ and $$$j$$$ would had been chosen by $$$S$$$ and $$$a_{k} > a_{j}$$$ otherwise $$$k$$$ would have been chosen by $$$T$$$ so we get $$$a_{i} > a_{j} > a_{k}$$$ which is impossible. For the other direction if there exist a length 3 decreasing subsequence then all three must be in different increasing subsequences.

Great!now we can construct a dp that tries to builds two or less increasing subsequences. Let $$$dp_{0 / 1,i}$$$ be the minimum possible value of the end of the increasing subsequence that $$$i$$$ is not part of(consider the end of the subsequence to be $$$-\infty$$$ if its empty) , $$$0/1$$$ denotes whether we choose to flip the sign of $$$p_{i}$$$, the transitions are deciding whether or not to stick $$$p_{i}$$$ into the same subsequence as $$$p_{i-1}$$$ ,the answer is yes if either $$$dp_{0,n}$$$ or $$$dp_{1,n}$$$ $$$\neq\infty$$$

my submission :140109080 (in hindsight the dp I described should be equivalent to the editorial dp, but I think this interpretation is cleaner and easier to understand)

For E I think you can also use dsu.

For each number $$$x$$$ in the array, all of the indices $$$i$$$ such that $$$a[i] = x$$$ will be in the same group. Maintain an array $$$root$$$ that keeps track for each number in the array what the current representative of that number's group is. It will be $$$0$$$ if the number doesn't exist in the array. Also maintain an array $$$val$$$ that tracks the leaders' value. So the number at index $$$i$$$ will be $$$val[Find(i)].$$$

For the first type of query, we check if $$$x$$$ is in the array. If it is, we merge the index of the added element with the pre-existing group for $$$x.$$$ To find the correct group we just look at $$$root[x].$$$ (It will be $$$0$$$ if $$$x$$$ does not in the current array.) If $$$x$$$ is not in the array we set $$$root[x]$$$ to the last index of the array of the newly added element.

For the second type of query, we again check to make sure that both $$$x$$$ and $$$y$$$ are in the array. If they are both in the array since we're doing quick union we first merge, and then set the $$$val$$$ of the root of the merged group to be $$$y.$$$ If $$$y$$$ doesn't exist in the array then it's easier, we just set $$$val[root[x]]$$$ to be $$$y.$$$ And of course if $$$x$$$ doesn't exist then there's nothing to do. Finally we update $$$root$$$ if needed, ie. set $$$root[x] = 0$$$ because $$$x$$$ no longer exists in the array.

please can someone tell me why is this code failing for C ?? https://codeforces.com/contest/1620/submission/140169385 I have lost my mind at debugging this for 3 hours at this point,also is there a way to see the full testcase in this website ?

Nice problems

In problem C, for this test case

TC:
                1
                8 3 18
                a***a*

The output of the editorial code is abbabbb i.e; (2+1) * (3+1) => 12th lexicographically smaller but I asked for 18th right. How is this correct?

My code for problem C gives TLE on test case 5. Can someone please help me out? Here's the link to my solution. https://codeforces.com/contest/1620/submission/140430241 Thanks!

I really dont understand the problem D. Why there is no need to take more than 3 coins 1 and more than 3 coins 2? In my opinion, there is no need to take more than one coin 1 (i.e. one coin 1 at most) because if any ai required two coins 1, two coin 1 could be replaced by one coin 2; and ,similarly, there is no need to take more than two coins 2. This really confuses me. Could anyone explain it? Thanks in advance!

in problem A if the string is (NNNE) in this case the answer should be NO but the editorial says something different..... can anyone make me understand how it is possible pls????

I can't understand why my code is wrong. It can't pass the example.

#include<iostream>
#include<cstdio>
#include<cmath>
#define ll long long
#define rint register int
#define INF 0x7fffffff
using namespace std;
int t,n,a[101],m;
int maxElem(){
	int maxn=0;
	for(int i=1;i<=n;i++){
		maxn=max(maxn,a[i]);
	}
	return maxn;
}
bool single_judge(int val, int c1, int c2, int c3) {
	for(int cur1=0;cur1<=c1;cur1++){
		for(int cur2=0;cur2<=c2;cur2++){
			if(cur1+2*cur2>val){
				continue;
			}
  		    if((val-cur1-2*cur2)%3!=0){
				continue;
			}
  		    if((val-cur1-2*cur2)/3<=c3){
				  return true;
			}
		}
	}
	return false;
}

bool judge(int c1,int c2,int c3){
	for(int i=1;i<=n;i++){
		if(!single_judge(a[i],c1,c2,c3)){
			return 0;
		}
	}
	return 1;
}
void solve(){
	int ans=INF;
	for(rint c1=0;c1<3;c1++){
		for(rint c2=0;c2<3;c2++){
			if(m-c1-c2-c2<0){
				continue;
			}
			int c3=max(0,m-c1-c2-c2);
			if(judge(c1,c2,c3)){
				ans=min(ans,c1+c2+c3);
			}
		}
	}
	cout<<ans<<endl;
}
int main(){
	cin>>t;
	while(t--){
		cin>>n;
		for(int i=1;i<=n;i++){
			cin>>a[i];
		}
		m=maxElem();
		solve();
	}
	return 0;
}

Could not understand the solution for A. Can anyone explain in a simpler way ?

https://codeforces.com/contest/1620/submission/140748031

My subission for C is giving wrong ans at tc 1 in codeforces compiler , while it works properly in my system, can any one check

Counterexample to B.

"Since there are points on all sides, the points on the opposite side are the furthest"

https://ibb.co/G3FnRmW

I solved E using Linked list in O(n) https://codeforces.com/contest/1620/submission/141979731

Here is the casework solution for D https://codeforces.com/contest/1620/submission/142005976

Does anyone know why this solution to E is getting an MLE on testcase 4? 143353188

Edit: I see why because of this link: https://codeforces.com/blog/entry/64625 check it. Problem G: Some contestant solved this problem using inclusion-exclusion, let cnt be the number of strings that are subsequence of the strings given by the mask, they do +cnt when the amount of bits is odd and -cnt when is even. Then, they do dp[mask] = cnt and apply SOS DP at the end. SOS Dp is clear, but what is the argument of that inclusion-exclusion when adding and decreasing depending on the bits?

Sorry for my engish, thanks in advance.