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It's so cool !!

Hi can you please help me debug my code..I have a logic and implementation style very similar to yours but my answer seems to be incorrect

Mind blowing explanation 🫡

Thanks .

nice

u made it easy :)

thanks so much :D

beautiful problem and solution ... loved it😇

very beautiful explanation! i was unable to understand from many other places, but this video made it in one go! You are putting so much effort in explaining little things, that is appreciable! Thank you 😊

Nicely explained :)

Thanks a lot brthr

dhang se to samjha jhat ke baal , google se solution to mai tep du

Great job.. the way you explain the application is too good. just one request if you use marker or pointer then it will be super great. Thanks.

Keeping the code out here for future reference ;) ********************* Recursive Memo ************************* /* कर्मयोग */ void TanTanmayMay() { ll n, m; cin >> n >> m; unordered_map<ll, vector<ll>> adj; for(ll i=0; i<m; i++) { ll a, b; cin >> a >> b; adj[a].push_back(b); } vector<ll> dp(n + 1, -1); auto dfs = [&] (auto f, ll cur) -> ll { if(dp[cur] != -1) { return dp[cur]; } ll x = 0; for(auto node : adj[cur]) { x = max(x, (1 + f(f, node))); } return dp[cur] = x; }; ll ans = 0; for(ll i=1; i<=n; i++) { if(dp[i] != -1) { ans = max(ans, dp[i]); continue; } ans = max(ans, dfs(dfs, i)); } cout << ans << ' '; return; } ********************** Topological ************************* /* कर्मयोग */ void TanTanmayMay() { ll n, m; cin >> n >> m; unordered_map<ll, vector<ll>> adj; for(ll i=0; i<m; i++) { ll a, b; cin >> a >> b; adj[a].push_back(b); } vector<ll> inDegree(n + 1, 0), topo ; for(ll i=1; i<=n; i++) { for(auto neigh : adj[i]) { inDegree[neigh]++; } } queue<ll> q; for(ll i=1; i<=n; i++) { if(!inDegree[i]) { q.push(i); } } while(!q.empty()) { ll curNode = q.front(); q.pop(); topo.push_back(curNode); for(auto neigh : adj[curNode]) { inDegree[neigh]--; if(!inDegree[neigh]) { q.push(neigh); } } } vector<ll> dp(n + 1, 0); for(ll i=(n - 1); i>=0; i--) { for(auto neigh : adj[topo[i]]) { dp[topo[i]] = max(dp[topo[i]], (1 + dp[neigh])); } } ll ans = 0; for(ll i=1; i<=n; i++) { ans = max(ans, dp[i]); } cout << ans << ' '; return; }

damn, i thought it could be solved by greedy; that we simply add two smallest elements each time. Can you find me any counterexample which cause this idea fail?

Bhai please add more video on. Advance dp.

perfect explanation to be honest

No comments! Underrated channel...insightful explanation!

Can be do it by recursive dp approach sir.

sir you write very nicely all the cases, but you stop while explaining very frequently. Please try to be more fluent. Thank you for the tutorials. And also please have a good mike.

Why you do plus 1 on 2nd and 3rd conditions in recursive function

Nice video

Nice Nice Nice

Just Wow

explanation is very good.. keep it up

bhai you didnt memoize the ans in solve function

You are the worst explainer I've ever encountered 🤢

🔥

nice work, thanks!

Which software you are using for note taking ?

Can't believe there's no likes you're literally the best person describing out here, thank you.

Nice Nice

thanks for this, it explains it well

Nice

THANK YOYOYOOOU little goat

Was wondering how it can be done using iterative DP

thanks for clear explanation

Nice

// हर हर महादेव // जय महाकाल जय महाकाल जय महाकाल जय महाकाल /* Author :- Himanshu Shekhar , IIIT Bhagalpur */ #include <bits/stdc++.h> using namespace std; const int MOD = 1e9 + 7; const long long INF = 1e15; const int MAX_N = 200 * 1000 + 13; # define ll long long vector<vector<int>> graph; ll a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z,sum,maxi; bool flag; string str; // IN-OUT DP or UP-DOWN DP ll in[MAX_N]; // // number of ways to colour the nodes outside the subtree of i ll out[MAX_N]; // number of ways to colour the nodes outside the subtree of i void dfs(int source, int parent){ in[source]=1; // choice 1: for painting black for(auto it: graph[source]){ if(it!=parent){ dfs(it,source); in[source]=(in[source]*in[it]); } } in[source]++; // for painting all whites } void dfs2(int source, int parent){ vector<ll> pre,suff; for(auto it: graph[source]){ if(it!=parent) pre.push_back(in[it]); suff.push_back(in[it]); } int sz = pre.size(); for(int i = 1; i<sz; i++){ pre[i]=(pre[i]*pre[i-1]); } for(int i = sz-2; i>=0; i--){ suff[i]=(suff[i]*suff[i+1]); } int cnt = 0; for(auto it: graph[source]){ if(it==parent) continue; ll t2 = in[it]; // re-rooting from node to ch ll l=1, r=1; if(i-1>=0) l=pre[i-1]; if(i+1<sz) r=suff[i+1]; out[it]=(l*r*out[source]); out[it]++; dfs2(it,source); cnt++; } } void solve(){ cin>> n >> m; graph.resize(n+1); // constructing the graph for(int i = 1; i<n; i++){ cin >> a >> b; a--; b--; graph[a].push_back(b); graph[b].push_back(a); } fill(in,in+MAX_N,1); fill(out,out+MAX_N,1); dfs(0,-1); // calculation in[i] for each node i dfs2(0,-1); // calculation out[i] for each node i for(int i = 0; i<n; i++){ cout << ((in[i]-1)*out[i])%m << endl; // "-1" to remove case of for all whites we did earlier for "i" while calcualting in[i] } cout << endl; fflush(stdout); cout << flush; } int main() { #ifndef ONLINE_JUDGE //for getting input from input.txt freopen("input1.txt", "r", stdin); //for writing output to output.txt freopen("output.txt", "w", stdout); #endif ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(NULL); solve(); return 0; } please check this approach why AC on 8 TC while rest are showing WA ???

Nice

Just keep uploading…

very helpful...

Nice editorial )):

Hatasi

Can you add some questions on this cyclic dp state

You are doing a great job man. Your content is on an upper level. On top of that you are consistent despite the low views. All the best Bhai..your all work and hard work will pay you off one day for sure.

Nice~~

GG