Відео

You're doing an amazing job. Keep up the great work

That solution for E was great!

leetcode or codeforces

Everything is really good explaination but I can not understand the H question .. is there anyone who can help me out from that

In last question, why did u initialize l for binary search=-1 instead of 0 as median>=0. I wasn't getting right answer with l=0. Thanks for the explanation btw !!

thanks alot bro I get really good approach from your channel

1:05:27 this mint (atcoder library) allowed in ICPC or not (also on codechef and leetcode as well)

Bro I started attempting codeforces, after watching your videos.thank you for your such a nice explanation.

I still don't understand the 3rd case in problem F. how did expected value of 1 2 3 4 5 became 500000012 ?

Sir if possible can u pls explain java code too..

Your stream is mostly watched by 1200 below rated coders, it would really help if you start your solution by first explaining a test case and then doing a dry run over your solution that would really be helpful and please go a bit slow for questions after d(e,f,g...) for div3 and after b(c,d...) in div2, you just need to understand why is your stream not watched by enough even after so many newbies in the contest something needs to change in your style otherwise blatantly giving out solution won't help and your channel would remain stagnant.

why gcd==0? can't we write? ke whole array GCF ==1 then it's always possible ke array ko ham 0 to n-1 bana sakte

For problem E I didn't attempted because I thought O(t x 26 x n) will timeout as t is too high

to be honest jis level pe app ho , apka minimum 1m subs hone chahiye , but ye duniya , kya karein bhaiya , acche log ka kadar hi nhi hai , faltu content me million , billion view aate hai

where is error in this code void func(){ int a, b; cin>>a>>b; if(a&1){ no; return; } a/=2; if(b<=a){ yes; }else if(b>a){ b = abs(b-a); if((b&1)==0) yes; else no; }else no; }

Can the last problem be also solved using small to large merging? Thanks for wonderful explanation !

That thumbnail picture looks really nice👏👏

thanks for the valuable content !

congrats\\\\\\\\\\\\\\\\\

Aryan level on cf when?

your new subscriber

Leetcode contest winner💥

can you do abc367 also?

Can you please make a video on selecting problems for practicing and which resources do you use learn for learning topics?

Worst explanation

In problem D1, it says you start with some element X but what exactly is that X? We don't really take anything X as input so how do we exactly guess what x to take?

please explain what you have written in contest do not code a new code while explaining it creates problem in understanding the code

Need a video on parallel binary search

Thanks for codeforces. Can you also do this for atcoder?

sir D1 wala bilkul smjh mai nhi aya

In question C, we can store the characters along with their frequency without sorting and then keep popping the front element for one time and then the other element. This continues while size of q is greater than 1 and at last we can pop the last character. #include <bits/stdc++.h> using namespace std; #define ll long long int #define pp pair<int,char> int main() { ll t; cin>>t; while(t--) { ll n; cin>>n; string s; cin>>s; unordered_map<char,int> mp; for(auto &x : s) mp[x]++; queue<pp> pq; for(auto &x: mp) pq.push({x.second, x.first}); string st=""; while(pq.size()>1) { auto tp= pq.front(); pq.pop(); st.push_back(tp.second); if(tp.first-1>0) { auto tp1 = pq.front(); pq.pop(); st.push_back(tp1.second); pq.push({tp.first-1,tp.second}); if(tp1.first>1) pq.push({tp1.first-1,tp1.second}); } } if(pq.size()) { auto tp = pq.front(); pq.pop(); for(int i=0;i<tp.first;i++) st.push_back(tp.second); } cout<<st<<endl; } } This also shows accepted.

bhai yeh aapke baal pure white kaise ho gaye ek dum se ?

Sir can you please continue doing this for upcoming codeforces contest as well?

In google OA there was a question in which given a tree we needed to find sum of medians of all odd length path of the tree.. i tried solving it using ordered set but got TLE.. tips on how to solve it

Can someone suggest me like it took me almost 1 hour to hit the approach for the second one.

Number of question solve by you in today contest can send someone to depression 😢 but motivating at same time

Thanks a lot for this .

congratulations! nice performance today

Will u Atcoder regular contest solution?

didnt know pedro pascal started cp

Amazing thank you so much

C: 8:30

plz add subtitle for better understanding.

don't stop bro. Love from Bangladesh!

useful video,hope u do more vid like this

Sir , can you explain distance colour problem , why we divide grid into k*k grids in n*m grid problem of codeforces

I was trying G1 using binary search. But for checking if the division is possible. I was doing like consider that length is mid. So first mid character should be same. Now to count the number of subarrays which can have the common prefix of mid length I was doing something like int j = 0; int cnt = 0; for(int i = 0; i < n;i++){ if(s[i] == s[j]){ j++; if(j == mid){ cnt++; j = 0; } }else{ if(s[i] == s[0]){ j = 1; }else{ j = 0; } } } I have a pointer j which is telling which character to compare.. and when it reaches mid it means we have got one substring with the prefix and increase cnt. But I don't know why Is it giving wrong answer on test 3. Can you please help me out on this ? codeforces.com/contest/1968/submission/277009975

Waiting for codechef staters 147 discussion video

Problem E - > Plz compile this code and told me error this give me TLE ON 2ND TEST CASE #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<int> answr; for (int i = 0; i < n; i++) { int a = 0; int b = 0; cin >> a; cin >> b; vector<int> dp(200001, -1); vector<int> prefixsum(200001); for (int i = a; i <= b; i++) { int count = 0; int p = i; while (p) { count++; p /= 3; if (dp[p] != -1) { prefixsum[i] = dp[p]; continue; } } prefixsum[i] = count; dp[i] = count; } for (int i = a + 1; i <= b; i++) { prefixsum[i] += prefixsum[i - 1]; } int ans = prefixsum[a] * 2 + (prefixsum[b] - prefixsum[a]); answr.push_back(ans); } for (int i = 0; i < n; i++) { cout << answr[i]; cout << endl; } return 0; }