Відео

Is there a different preliminary round for each region in ICPC, or is there only one round for all regionals? This time, the timings for the Amritapuri and Kanpur regions are the same.

May i know interview will be in online mode or offline mode

thank you!!!

Great work uday !!! Congratulations subhankar 🎉

is there scope for full time conversion?

You know 😢

Thank you senior ❤

Recently i gave oa for amzon and i completed oa successfully like i passed all test cases and when will i get interview call like aprrox time if u have an idea?

Congratulations Subhankar

Are You shortlisted for intern ?

can i use recursion + memo?

Hey Hi., Thank You for the video From the morning I am literally searching for good explanation approaches for these 2 problem. Like I have a doubt in the problem they mentioned that "Two rooms are adjacent if they share a common wall, either horizontally or vertically." You haven't talk about this need some clarity on this. In addition to that I have a small suggestion 1. I get what your saying but it would be even more good if you took 2 more examples while explaining 2. as Dijkstra's Algorithm relate graph problems you can also mention this as well. 3. explain while writing the code. Thank You !!

Great work!! Need more of these

Great work bro ! keep doing

Great Work bro ! Keep Doing !

Detailed Process from Selected Candidates : www.linkedin.com/posts/karthikeya-t_jman-activity-7256580715784351744-opT0?

🔥

Hey amazon has again given a chance for sde 6m role .. i want to apply for this time .. help me how to clear the test

hii i got shortlisted for the 2nd round virtual interview .... can you tell how the interview will be and what questions were asked

how many days it took to get notification after interview. i have completed my interview on 16th .

how to construct the graph for 1st question in coding round2

when was his placement completed?

Anna nee branch and year cheppu I want to Meet u

Nice job guys

this was very informative !!! #cfbr

nice explaintion

Great soln :)

Can you explain the part where are mid + p[0] < p[1], still we accept it and went further with p[1]?? What if elements are like [1,5,9] and d=2, so p = 1 is not possible still according to our mathod it's true, why?? I know we need to look for bigger elements than this, but how can we apply binary search on such inconsistent data.

great explanation . thankyou

Great solution, we could reduce the search space from 10**10 to max(arr) + d for the high

Best solution, Uday! THank you for sharing! But why you didn't use DP here? Because DP came as the first solution to my mind and I could not think of anything.

you're a legend bro 🙏

Hey wanna ask like how can i reach expert from where to practice so that can come up with div2 C have been struggling a lot plz help

Thanks man, Great explaination

How did you choose high?

Nice explanation found got the intuition

Good explanation!

Nice Explanation Bro. You earned Subscriber here <3

very good explanation bro

The core logic remains the same, but eliminating constant time operations is necessary to fix the TLE. I’m sharing this code to provide better readability and clarity of my approach. Feel free to review it! #include <bits/stdc++.h> using namespace std; #define ll long long vector<ll> dijkstra_algo(ll st, vector<pair<ll, ll>>adj[], ll n) { priority_queue<pair<ll, ll>, vector<pair<ll, ll>>, greater<pair<ll, ll>> >pq; vector<ll>vis(n, LONG_MAX); pq.push({0, st}); vis[st] = 0; while(!pq.empty()) { auto it = pq.top(); pq.pop(); ll x = it.first; ll y = it.second; for(auto i: adj[y]) { ll curr = x + i.second; if(curr < vis[i.first]) { vis[i.first] = curr; pq.push({curr, i.first}); } } } return vis; } int main() { ll m, n; cin>>n>>m; vector<pair<ll, ll>>adj[n]; vector<pair<ll, ll>>radj[n]; vector<vector<ll>>edges; for(int i=0; i<m; i++) { ll x, y, z; cin>>x>>y>>z; x--, y--; adj[x].push_back({y, z}); radj[y].push_back({x, z}); edges.push_back({x, y, z}); } vector<ll>vis1 = dijkstra_algo(0, adj, n); vector<ll>vis2 = dijkstra_algo(n-1, radj, n); ll ans = 1e18; for(auto it: edges) { ll curr = vis1[it[0]] + vis2[it[1]] + it[2]/2; ans = min(ans, curr); } cout<<ans<<endl; }

Best explanation of this question on Utube

nice explanation

nice explained 👌👌

Well explained <3, but why you initialize curr = 1 instead of 0?

Thanks bro its really help full

You explained sooo beautifully❤ excellent teaching skill

Small Correction at Line 24 - (i < arr.size() ) not i < n. #include <bits/stdc++.h> using namespace std; int main() { int n; cin>>n; vector<pair<int,int>>arr; for(int i=0; i<n; i++) { int start, end; cin>>start>>end; arr.push_back({start, 1}); arr.push_back({end, -1}); } sort(arr.begin(), arr.end()); long long prefix = 0; long long maxi = 0; for(int i=0; i<arr.size(); i++) { prefix += arr[i].second; maxi = max(maxi, prefix); } cout<<maxi<<endl; }

Your explanation is super bro, thanks a lot.....

Bro, great explanation .. Kudos!!!!

either you can make curr += arr[i] or curr *= 2 . Both will give you the same because here (arr[i] == curr) both are equal. Even while you are adding a new coin, you will add the current searching coin . So it will be either (curr += curr or curr *= 2)