Відео

bro please upload hallowen array solution

Thanks Akash, Its helpful:D

code

2nd solution please

Temporary Temperature plese make this solustion

does this not take more than n^2 time

did this work in time?

Explanation - ua-cam.com/video/r0CKXtVofrI/v-deo.html

Explanation - ua-cam.com/video/456BT4IIKIk/v-deo.html

Great explanation

class Solution { public: bool isZeroArray(vector<int>& nums, vector<vector<int>>& queries) { int n = queries.size(); for (int i = 0; i < n; i++) { int l = queries[i][0]; int r = queries[i][1]; for (int p = l; p <= r; p++) { if(nums[p]==0) { continue; } nums[p]--; } } for (int i = 0; i < nums.size(); i++) { if (nums[i] != 0) { return false; } } return true; } }; why showing tle in last two cases meanwhile same code this is that you write

Encountered Such type of Problem for the First time. Thank You for The Great Explaination. I did through Brute Force O(m*n) and Encountered TLE. A small doubt, How did you come up with this solution? Like I mean did you solve similar concept problems prior?

Can you please explain how the diff[si] - - and diff[ei+1] + + helped here? The logic behind doing this is still unclear😅

Explanation- ua-cam.com/video/eE_-c6YnfVQ/v-deo.html

Great explanation , can you suggest some sources to read more about cumulative sum?

code

Helpful, Thanks for the explanation.

Nice Explanation

code

wrong

bro explaination pls

Thanks, It's helpful

Great explanation

It's helpful, Thanks:)

Tiwari ji, well explained

Well Explained⚡

4th question ans is available?

Nice and Clear Solution Understand it within 3 min THx

Thanks

i solve 2nd using priority queue class Solution { public: #define p pair<ll, pair<ll, ll>> #define ll long long long long minNumberOfSeconds(int mountainHeight, vector<int>& workerTimes) { ll ans=0; priority_queue<p, vector<p>, greater<p>>pq; for(auto it : workerTimes){ pq.push({it, {1, it}}); } while(!pq.empty() && mountainHeight){ p top = pq.top(); pq.pop(); // Manually extract the pair ll val = top.first; ll level = top.second.first; ll mul = top.second.second; ans = max(ans, (ll)val); pq.push({val + (level + 1) * mul, {level + 1, mul}}); mountainHeight--; } return ans; } };

bhai 2nd and 3rd ki explaination details me batana step wise

send me code bro

Thnx man keep uploading contest answers helps a lot! Also clear voice mei bolo bina hesitation ke impact acha aata h❤

what is the intuition for this particular problem

Have you solved similar problem before, I was not able to think of this solution.

mai to bhai isme Dynamic programming lgga rha tha

bhai iss question ki maine 3-4 lagah explaination dekhi abb jaa kr samajh me aaya thanks bhaiya

thanks very easy explanation

First viewer🫣🙋

I think you not understand question , I tell you , If i am wrong correct me, In question at any point of time we reaches to alice , alice can move N steps let suppose alice wants to move R R R D D but when he travelled R R R and bob catch in this position than it is also correct. as per my understanding if he is able to catch on any step<=N Bob win and I think you are talking about after completing N steps , bob can reach or not.

Bro if you make a full contest solution video it would be really grt help🙏

I really appreciate the effort you put into breaking down the information in such an understandable way.✨

Nice explanation ✨

q3 and a4 plz brother

Nice explanation:))

It's helpful, thanks Ak👍🏻

voice is not louder

bro first two are very easy. Pls make solution videos for next two...

good

bro kahase sikha apne please guide us