Love U Aryan Mutthal

16:46 This two Balls 🫡

so much effort, it so great to see someone putting that hardwork to make lives easier

perfect explanation, thanks

goated explanation. will be a hit yt channel soon

I solved this using bfs but I had to use one more variable which will keep track of different paths for a particular node. class Solution { public: vector countPairsOfConnectableServers(vector& edges, int signalSpeed) { int n=edges.size(); vectoradj[n+1]; for(auto it:edges){ adj[it[0]].push_back({it[1],it[2]}); adj[it[1]].push_back({it[0],it[2]}); } vectorresult; for(int i=0;i0){ tp.push_back(it.first); kl--; } } } int resu=0; for(int i=0;i

You are too good man !! Keep it up brother !!

Thank you

your are the best for me!

Thnx bro well explained

TQ bro❤

thnx bro , for such detailed explanation .

hi aryan - the trick that u used to solve for avoiding an n^2 loop for components I did not use that trick, but did not get a tle for my submission how is it possible? i think that the worst case is "star graph" - all these components will be multiplied with each other but, advantage is it will happen for only one node, not all n nodes in the star graph but still, how do I prove this?

You are great sir !!!

epic explanation

bro i solved this question using warshall but its giving tle......but generally on leetcode n cube is acceptable for 10^3

@umeshhbhat

4 ай бұрын

warshall is cubic. so TC will come around 10^9, anything above 10^8 will give TLE

First ✋✋

great work bro

@ARYANMITTAL

4 ай бұрын

Thank you so much 😀❤️❤️

I too did with the same approach, but I got TLE in the contest

@dumpster-jackson

4 ай бұрын

same here

aryan bhai mera to abhi bhi tle arrha hai same code par bhi and during the contest par aa rha tha

class Solution { private: int ss,ct; void dfs(int node,int par,int d, unordered_map&gr){ if(d%ss==0)ct++; for(auto [ch,dis]:gr[par]){ if(ch!=par){ dfs(ch,node,dis+d,gr); } } } public: vector countPairsOfConnectableServers(vector& edges, int signalSpeed) { int n=edges.size()+1; ss=signalSpeed; unordered_mapgr; for(auto vec:edges){ int u=vec[0],v=vec[1],d=vec[2]; gr[u].push_back({v,d}); gr[v].push_back({u,d}); } vectorfans(n,0); for(int par=0;par

floyd warshall

27:07 😂😂

I knew the intuition but didn't implement it because of - O(V^2 * (V + E)) Time complexity why it is accepted??

@thomasshelby6780

4 ай бұрын

it must be v2 only

@39_jatinjain4

4 ай бұрын

How??

Nahi aaya mujhe ye question

8:20 ping 999+ms🤣🤣🤣🤣