Welcome to my channel! Hope you love it ❤️
This channel is for anyone who is interested in coding (beginner or advanced). It has different and interesting coding videos and tutorials.
You can also find many Mock Interviews with me as both interviewee and as an interviewer with tons of knowledge and tips, not just from me but from many inspiring people. I am sure these videos will help you prepare for interviews better!
I love making learning process easier, faster and better for everyone and so, I have started my LIVE courses. We already have 2200+ participants in HLD, LLD, DSA, C++ courses. You can checkout all the details with week-wise curriculum and FAQ sections are here - www.educosys.com
Everything on the channel is arranged properly in playlists, so it should be easy to navigate.
Don't forget to subscribe! 👻
Thank you for all the support!💙
Пікірлер
I think there is an edge case in 7 - 8 - 9 - 10 - 7 - 9. We can remove 7-8 and then 8-9 then we have 1 more component = 8. So we don't really need to remove 2 edges to get one more component. Instead I would find the bridges. If a + bridges >= k thats a return. Else keep track of the sizes of cycles like in this case 3, 3. Then individually check for those cyclic components
What is the difference between builder and abstract factory design patterns?
i think you are mistaken on the stocks , atlassian stocks also gets vested continuously right after you join , thanks .
Thanks!! I am preparing for an interview at Epic Games
Notes folder is not accessible for free, can anyone share the doc link
Beautiful
where is the 3rd and 4th video?? please tell me
where is the next video??
Not sure about Cassandra, I feel when we have a very complex search query structure like - filter on age, location, interests, gender, hobbies etc a search engine that supports text based search (basically quick indexing on props) like ES may be an option. But Cassandra looks like a horrible choice for a read-heavy system like Tinder :( Surprised to see Gaurav was okay with this lol
Hey, this mock is for which exam exactly. Someone please tell me
like exchange connector, orderManager and OrderValidator will also be singleton, right ?
orderValidator is a function related to order, we can have it in orderManager/orderService only, it should not be considered as violation of SRP. or am i getting it wrong.
Really good info .What software did you use?
Quality of this content is extremely high thank you so much maam
Thanks a lot for clear and nice explanation
beautifully done
Didii woww such an informative video!! Flipkart Grid ke upar ek video banao na please, 1 lakh ka stipend and 32 LPA salary mil raha hai.
The completeUpload controller is not fully visible
Notes section is not free !!!
Thanks nice content. Do we have separate video for strategy design pattern?
Yes we need a OOPs video
Suppose there are 2 seats available 2 users want to book 1 seat each at 1 sec gap lets say so when first user starts booking, u will lock all the seats?? NO right?? so i think we will need multiple enteries for seats rather than one +there are different kind of berths etc + quotas (senior citizen, ladies, gn, foreign, tq) so it is not that easy in real life🤯
This can also be one solution 1. Sort the array 2. Find min as a[0] 3. Subtract min-1 to all numbers 4. Now we have min as 1 and our max can he n. 5. Make another array as visited. 6. Ittrate in the array and for each value in the array mark visited[value] =1. 7. At last count number of zeros in visited array. Is it correct?? 😊
How much civil engineer watched this video 😂😂 This video clearly massage it is promoting scaler ....😂😂
can't we do using priority queue
BFS is much easier to implement for me but a bit slow in finding valid path :/
ple provide me whole code kindly post here only .
u explained really well..!
Aap java use nhi karte ho kya 😅
ok
total waste of time,
Intuit is building 8 and my office is in building 9
Yes mam
Problem very similar to travelling salesman problem and he is grandmaster it would be kpop for him 😂 but the way he came up with soln and just code it with no problems .. i mean thats incredible
Nicely explained. Understood it completely
we can use sliding window to find maximum elements that is inside (maxelement - minelement) <= n-1 and the answer is n -maxi
for the first question i think two pointer approach would work pushing all index of thiefs in a vector and separate vector for police then try to give a thief who is as left as possible and if you cant then go for next atmost k difference of distance from the current police is that it works or not mam
this is a problem which is similar to Polise Chase in cses
Great how-to-get-80-lpa-package-in-atlassian video!
mam apki salary kitni hai 80 LPA,1.5 CR, 2.4 CR and which position you are now SDE1, SDE2, SDE3
i made my application responsive so no need for open closed principle? 😂
thankyou mam..
Kuch error ho rha hai mere me, koi help kar skta hai??
take mock coding interview with kunal kushwaha
take mock coding interview with love babbar
Idk but while watching this, I felt that the interviewer is more confused than the interviewee
Hare Krishna Hare Krishna Krishna Krishna Hare Hare Hare Rama Hare Rama Rama Rama Hare Hare ❤❤ Raadhe Raadhe ❤❤ Jai Shree Ram ❤❤❤
awesome content, servers & storage estimation is something which can be included after functional & non functional requirement gathering.
ll n;cin>>n; ll m;cin>>m; ll k;cin>>k; vll adj[n+1]; vll indegre(n+1,0); while(m--){ ll u,v;cin>>u>>v; adj[u].pb(v); adj[v].pb(u); indegre[u]++; indegre[v]++; } k-=connectedcomponet; ll ans=0; set<pll>s; fl(i,1,n+1){ s.insert({indegre[i],i}); } vll vis(n+1,0); while(!s.empty()){ ll node=(*s.begin()).ss; ll x=(*s.begin()).ff; vis[node]=1; s.erase(s.begin()); k--; ans+=x; if(k<=1){ break; } for(auto it:adj[node]){ s.erase({indegre[it],it}); indegre[it]--; if(!vis[it]&& indegre[it]>0) s.insert({indegre[it],it}); } } cout<<ans<<endl; } Can anyone confirm this solution is correct or not? I have checked on some test cases and it giving correct answer.