Idea: Vladosiya, MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
void solve(){
int n;
cin >> n;
string s;
cin >> s;
transform(s.begin(), s.end(), s.begin(), [] (char c) {
return tolower(c);
});
s.erase(unique(s.begin(), s.end()), s.end());
cout << (s == "meow" ? "YES" : "NO") << "\n";
}
int main(){
int t;
cin >> t;
while(t--) solve();
}
```

1800B - Count the Number of Pairs

Idea: myav

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int N = 26;
void solve(){
int n, k;
cin >> n >> k;
string s;
cin >> s;
vector<int>big(N, 0), small(N, 0);
for(auto &i : s){
if('A' <= i && 'Z' >= i) big[i - 'A']++;
else small[i - 'a']++;
}
int answer = 0;
for(int i = 0; i < N; i++){
int pairs = min(small[i], big[i]);
answer += pairs;
small[i] -=pairs; big[i] -= pairs;
int add = min(k, max(small[i], big[i]) / 2);
k -= add; answer += add;
}
cout << answer << endl;
}
int main(){
int t;
cin >> t;
while(t--) solve();
return 0;
}
```

1800C1 - Powering the Hero (easy version)

Idea: Vladosiya

**Tutorial**

Tutorial is loading...

**Solution**

```
def solve():
n = int(input())
s = [int(x) for x in input().split()]
ans = 0
buffs = [0] * n
for e in s:
if e > 0:
buffs += [e]
j = len(buffs) - 1
while buffs[j] < buffs[j - 1]:
buffs[j], buffs[j - 1] = buffs[j - 1], buffs[j]
j -= 1
else:
ans += buffs.pop()
print(ans)
t = int(input())
for _ in range(t):
solve()
```

1800C2 - Powering the Hero (hard version)

Idea: Vladosiya

**Tutorial**

Tutorial is loading...

**Solution**

```
from queue import PriorityQueue
def solve():
n = int(input())
s = [int(x) for x in input().split()]
ans = 0
buffs = PriorityQueue()
for e in s:
if e > 0:
buffs.put(-e)
elif not buffs.empty():
ans -= buffs.get()
print(ans)
t = int(input())
for _ in range(t):
solve()
```

Idea: MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <iostream>
#include <vector>
#include <queue>
#include <map>
#include <set>
using namespace std;
void solve() {
int n;
cin >> n;
string s;
cin >> s;
int res = n - 1;
for (int i = 1; i + 1 < n; ++i) {
if (s[i - 1] == s[i + 1]) {
res--;
}
}
cout << res << '\n';
}
int main(int argc, char* argv[]) {
int t;
cin >> t;
for (int i = 0; i < t; ++i) {
solve();
}
}
```

1800E1 - Unforgivable Curse (easy version)

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
#define sz(v) (int)v.size()
#define all(v) v.begin(),v.end()
#define eb emplace_back
void slow_solve(int n, int k, string s, string t) {
set<string> was;
queue<string> q;
q.push(s);
was.insert(s);
auto add = [&](string& s, int i, int j) {
if (i >= 0 && i < j && j < n) {
swap(s[i], s[j]);
if (!was.count(s)) {
was.insert(s);
q.push(s);
}
swap(s[i], s[j]);
}
};
while (!q.empty()) {
s = q.front(); q.pop();
for (int i = 0; i < n; ++i) {
add(s, i, i+k);
add(s, i, i+k+1);
add(s, i-k, i);
add(s, i-k-1, i);
}
}
cout << (was.count(t) ? "Yes" : "No") << '\n';
}
void solve() {
int n,k; cin >> n >> k;
string s; cin >> s;
string t; cin >> t;
if (n <= 5) {
slow_solve(n, k, s, t);
return;
}
map<char, int> cnt;
for (char c : s) {
cnt[c]++;
}
for (char c : t) {
cnt[c]--;
}
bool ok = true;
for (auto [c, x] : cnt) {
ok &= x == 0;
}
cout << (ok ? "Yes" : "No") << '\n';
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
}
```

1800E2 - Unforgivable Curse (hard version)

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
#define sz(v) (int)v.size()
#define all(v) v.begin(),v.end()
#define eb emplace_back
void solve() {
int n, k; cin >> n >> k;
string s; cin >> s;
string t; cin >> t;
vector<int> cnt(26, 0);
bool ok = true;
for (int i = 0; i < n; ++i) {
if (i >= k || i+k < n){
cnt[s[i] - 'a']++;
cnt[t[i] - 'a']--;
} else {
ok &= s[i] == t[i];
}
}
cout << (ok && count(all(cnt), 0) == 26 ? "YES" : "NO") << '\n';
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
}
```

Idea: Gornak40

**Tutorial**

Tutorial is loading...

**Solution**

```
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx2,avx,fma,bmi2")
#include <bits/stdc++.h>
#include <immintrin.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
#define endl '\n'
//#define int long long
#define all(arr) arr.begin(), arr.end()
#define multitest() int _gorilla_silverback; cin >> _gorilla_silverback; while (_gorilla_silverback --> 0)
#define pikachu push_back
#define ls(id) (id << 1 | 1)
#define rs(id) ((id << 1) + 2)
#define sqr(x) ((x) * (x))
#define dlg(x) (31 - __builtin_clz(x))
#define ulg(x) (32 - __builtin_clz(x))
typedef pair<int, int> ipair;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> treap;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MAXN = 200200;
const int L = 26;
int n;
string srr[MAXN];
int arr[MAXN], brr[MAXN], crr[MAXN];
void build() {
for (int i = 0; i < n; ++i) {
for (char c: srr[i]) {
arr[i] ^= (1 << (c - 'a'));
brr[i] |= (1 << (c - 'a'));
}
}
}
long long calc(int c) {
int k = 0;
for (int i = 0; i < n; ++i)
if (brr[i] >> c & 1 ^ 1) crr[k++] = arr[i];
sort(crr, crr + k);
int mask = -1 & ((1 << L) - 1) ^ (1 << c);
long long ans = 0;
for (int i = 0; i < k; ++i) {
auto itl = lower_bound(crr, crr + k, crr[i] ^ mask);
auto itr = upper_bound(crr, crr + k, crr[i] ^ mask);
ans += itr - itl;
}
return ans >> 1LL;
}
long long solve() {
long long ans = 0;
for (int c = 0; c < L; ++c)
ans += calc(c);
return ans;
}
signed main() {
ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
cin >> n;
for (int i = 0; i < n; ++i)
cin >> srr[i];
build();
cout << solve() << endl;
}
```

Idea: Vladosiya

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
#define int long long
#define pb emplace_back
#define mp make_pair
#define x first
#define y second
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
typedef long double ld;
typedef long long ll;
using namespace std;
mt19937 rnd(time(nullptr));
const int inf = 2e18;
const ll M = 1e9;
const ld pi = atan2(0, -1);
const ld eps = 1e-6;
int last;
map<vector<int>, int> eq;
map<int, bool> symmetrical;
int dfs(int v, int p, vector<vector<int>> &sl){
vector<int> children;
for(int u: sl[v]){
if(u == p) continue;
children.emplace_back(dfs(u, v, sl));
}
sort(all(children));
if(!eq.count(children)){
map<int, int> cnt;
for(int e: children){
cnt[e]++;
}
int odd = 0, bad = 0;
for(auto e: cnt){
if(e.y & 1){
odd++;
bad += !symmetrical[e.x];
}
}
eq[children] = last;
symmetrical[last] = odd < 2 && bad == 0;
last++;
}
return eq[children];
}
void solve(int tc){
int n;
cin >> n;
eq.clear();
symmetrical.clear();
eq[vector<int>(0)] = 0;
symmetrical[0] = true;
last = 1;
vector<vector<int>> sl(n);
for(int i = 1; i < n; ++i){
int u, v;
cin >> u >> v;
sl[--u].emplace_back(--v);
sl[v].emplace_back(u);
}
cout << (symmetrical[dfs(0, 0, sl)]? "YES" : "NO");
}
bool multi = true;
signed main() {
int t = 1;
if (multi)cin >> t;
for (int i = 1; i <= t; ++i) {
solve(i);
cout << "\n";
}
return 0;
}
```

P.S. I haven't found good articles about hashing root trees so I'll post one soon.

UPD: Here it is

Isn't that blog enough for understanding hashing root trees? http://rng-58.blogspot.com/2017/02/hashing-and-probability-of-collision.html

I just hope to make a bit better topic with showing approaches as main point.

could u check my solution for e1 it gives a wrong answer when a char array is used but accepted when a string is used

C1 was easier than A and B.

no it wasnt, A is easier

why?? i spent apporximately 2 hour solving C1 by using dp 2-D.however, i can not complete it :)) maybe i thought too much

StringForces

i dont understand BFS solution of task E can anyone explain please?

There is an observation that if you turn the string to a graph where the edges are like this: $$$(i,i+k)$$$ when $$$i+k \le n$$$ , $$$(i,i+k+1)$$$ when $$$i+k+1 \le n$$$. So you can swap characters in indices in the same component however you like. You can try a few examples if you want.

F. i guess it should be bj = bi XOR (1<<26 -1) XOR (1<<k)

Except C1, C2 and F, all other problems are about strings

F is also string

ah sorry, I meant except C1, C2 and G

So many FST's on A, such a tricky easy question !

Amazing, I have never seen so many FST on a problem A

So many FST's on F too, jiangly and SSRS got hacked, I don't feel too bad now after getting hacked, XD

what is FST . sorry i m a newbie

Failed System Test

What does

`count(all(cnt), 0) == 26`

mean in solution of E2? What is the logic behind this? Can anyone tell me, pleaseHe is simply checking if count of all letters that are not in the interval [n-k; k] is equal.

`Count(all(cnt), 0)`

returns number of zero elements in the array.So, with this line:

`cnt[s[i] - 'a']++;`

he is counting letters in string s, and with this:`cnt[t[i] - 'a']--;`

he is deleting letters from cnt, and when all letters were processed there will be only zeros in cnt if counts of letters were equalThanks!)

D&G two hashing problems!

E can be solved using disjoint set union. Just check whether each connected component are the same.

can u plz elaborate more on this?

195970968

Can you please explain it, its really hard to understand the code.

Thank you for the contest, loved all the questions. Kudos to authors, coordinators. I may finally become expert after being a specialist for 6 months now, :')

Good tutorial. Thanks for explaining the solutions properly.

G is too obsivious for people who know hash of trees.

I thought my solution of G could fail on system test by hash-collision, but unexpectedly my solution of A got FST. There should be something like "eow" in the pretest of A.

Looks like this would have been my expert round had I not made a stupid error and got a penalty for overflow on problem C. Still, I won't cry about +100 delta :)

Can not understand why O(26⋅n⋅log n) for F got TL. Can anyone suggest what I am doing wrong? https://codeforces.com/contest/1800/submission/195673951

all you need is to lower your constant. change bitset to long long or using unordered_map instead of map. it finally runs 2000ms

same my solution of O(26.n.log n) is also gave TLE on test 26 195813129 Can somebody please explain why is this happening? (also unordered map is also not working)

In your case I think that problem is that it is O(26 * 26 * n * log n) instead of O(26 * n * log n). Upd, actually it is O(26 * 26 * n + 26 * sum_of_len + 26 n * log n), but probably it also should be fine.

You are right it is because of the constant. If I use find from std::map instead of operator[] it pass tests. https://codeforces.com/contest/1800/submission/195806091 But anyway too close to TL.

yes, I also tried find instead of [] and it passed, can you please tell why is this happening shouldn't both take same O(log n) time?

Vladosiya Could you consider changing the time limit for F? The map based solution should also pass tests. It is div3 you should not care about constant too much.

did not understand the "abudance" example in E1. Can anyone explain it to me?

D is a nice problem. I worried about the cases where 2 removals are not overlapping. Did it worry you too? Consider: R1+R2+S+R3+R4 where S is a string, and removals of R1+R2 and R3+R4 gives the same results: S+R3+R4 = R1+R2+S then S[even index]=R1 and S[odd index]=R2. If S has even length, R3=R1 and R4=R2; if S has odd length, R3=R2 and R4=R1. Thankfully, this shows checking overlapping removals will cover the non-overlapping ones too.

A,B,C1,C2,D were leaked on YT. This isn't fair. people think and try to solve and may get negative delta and others just copy and paste and get positive delta. I think Codeforses must prevent the new accounts and that solved less than N of problems in problem set to participate in Div 4 or 3

like that one : 195674461 which is for Mohamed_712 and got +44 and this answer is 100% simmilar that was leaked !

[deleted] Oh, my method is the same with editorial.

Could someone explain why we need to do "1&" and "-1&" in calc() in the solution of problem F? I thought that they wouldn't take any effect since 1 & 1 = 1 and 1 & 0 = 0, but removing them indeed resulted in WA. Here's the WA submission: 195819749

Oh I think I know why. The "-1&" is unnecessary but the "1&" is not. Sorry for the stupid question

in the -1& case, i guess it just trying to find the mask where only the position of letter c is unset. So in my view, -1& is completely not required.

How to proof C? I am not convinced even the solution magically works.

As we know all the cards before we take action, we can just discard those that aren't good enough. That ensures the greedy solution using priority queue.

I solved G without using hashing, but instead I "sorted" the subtrees in a certain order. Did I do a fakesolve? or is it a valid method?

I also solve G without hashing, it seems like we have similar idea, I scan nodes in depth ascending order to maintain some equivalent groups among them.

Really good contest!

https://codeforces.com/contest/1800/submission/195826604, isn't the tc of this code same as the tc suggested for problem F? Why did it TLE? Can anyone help?

When viewing your code, I saw two variables: ev and od. They eat a lot of time and memory, because this is a hashmap array, try to pull them apart a little and reformat the code. If it's not clear what I'm saying, then here's an idea: do not cycle(1...n) cycle (0...26) and vice versa cycle(0...26) cycle(1...n) then you can remove huge arrays and put the use of hashmap into the cycle itself, made the code faster. Sorry for grammatical errors, I do not know English.

Literally StringFORCES!

Someone pls explain this conclusion in the last paragraph of tutorial of problem F:

To count the number of pairs that include our word, we need to count the number of words with the characteristic $$$b_j=b_i\oplus(2^{26} − 1)$$$.

Letter with odd count will be represented as digit 1, and that with even count will be digit 0. To find pair (i, j) which meets the condition, we should check if they miss a certain letter and all the remaining 25 letters have odd count. Note that $$$odd = odd + even$$$ is similar to $$$1 = 1 \oplus 0$$$. So the pairs we are finding are (i,j) which have $$$b_i \oplus b_j = 2^{26}-1$$$ (the right side is 26 1's). (In the implementation, we actually count words meeting $$$b_j = b_i \oplus ((2^{26}-1) \oplus 2^{c})$$$ to handle the missing letter $$$c$$$.)

Great tutorial!!!!

What is the meaning of this line in solution F if (brr[i] >> c & 1 ^ 1)

It aims to pick words that miss the certain letter

`c`

I wrote this java solution to D, I got out of memory exception on test case 5, this could be optimized, reading insights about my solution would be fun. Please comment.

this would have given mle in cpp as well

Why so?

You are trying to store 200 000 strings of size 200 000 in the memory. You would need 4 * 10^10 bytes of memory to do that, which is about 40 GB. Not to mention that

`substring`

has linear time complexity, so your overall solution is quadratic and would TLE even with enough memory.What does the

`-1 &`

do inside the F solution? For me it looks like no-op because -1 has`1111...111`

binary representation and bitwise-and with such number shouldn't change the second numberIt's just my understanding so that someone can point out what is wrong with it — I don't know much about bitwise operations

It does nothing and removing it won't have any problem. Here's a such AC code: 195838110

Can you explain why do we need to keep distinct maps based on parity of lengths? As per my understanding, if we can find two strings such (b_i ^ b_j) = 25, won't the odd length of the s_i.s_j concatenation hold already? In that case, do we really need to count masks of odd length strings and even length strings separately?

Can someone please tell me why my submission 195888152 with unodered_map doesn't work while the solution with map 195888396 works.

Actually, it has nothing to do with hash collisions. When you iterate over your map you add new elements by using

`operator[]`

. This leads to visiting some characters multiple times in the case of unordered maps. By iterating over a copy of the map your code gets accepted: 196522048F is easy but I didn't solve it ontime, otherwise I would be Expert now :V

1800F - 31 - Даша и кошмары I tried solving it using bitsets in C++ but it ended up giving WA on TC3.Can anyone help me with my approach. 195952648

Is the argument given in the tutorial for D sufficient? i.e. what about potential over counting (the argument only considers consecutive positions)? Or is this obvious?

In problem 1800E2 — Unforgivable Curse (hard version) it is enough to check if index of mismatching character in string s and string t is within the limit k <= index <= n — (k + 1). accepted Code

can you prove it?

Can someone explain why am I getting Wrong Answer in Problem F by using unordered_map and Accepted using map.

MAP Solution

UNORDERED MAP Solution

https://en.wikipedia.org/wiki/Hash_collision

Actually, it has nothing to do with hash collisions. When you iterate over your map you add new elements by using

`operator[]`

. This leads to visiting some masks multiple times in the case of unordered maps. By iterating over a copy of the map your code gets accepted: 196526230Thanks. But I'm still wondering that order of elements in unordered map changes while traversing as I was not updating the map, just traversing it.

If you call

`operator[]`

on an element that didn't exist before in the map, it gets created. As the map is unordered, you don't know where it gets inserted in the traversing order.Here is my solution for problem F. Can anybody tell me where am I getting wrong? My submission link

So many problems based on strings

In Promblem D,

Lets X(i,i+1) = string obtained after chars at index i,i+1 removedThen Why X(i,i+1) and Y(j,j+1) are guarenteed to be distinct ? Why cant they be same ?

Have the same doubt

There is a much easier solution to E. You don't need to use anything except one map container. Refer my submission for the same:- 244026758

I dont understand for problem E when i >= k or i + k < n it doesn't guarantee the possibility of moving the letter in any direction by 1 because if i = k we aren't guaranteed to be able to move i to the right if i also equals n.

Can someone prove it to me? Thank you in advance