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I suggest you should do some code to first make a problem statement let if you want to get 10 problm that he has solved how can you get it and then code may be that will be better

❤❤

please make for E

finally u r back, please don't stop making videos please, I love ur explanations,please make for every contest

bro make more videos ,luv ur content

I am getting 403 error

thanks

Consider the case n=5 and k=12, so in the first iteration we are looking in the set for the value 6+1=7, which should return st.end() and dereferencing it should give a garbage value unless we handle this case. I don't understand how did it get accepted.

Thank you for the explanation.

keep it up bhai, nice explanation

could you explain why this code is not getting accepted though I went with a similar approach #include <bits/stdc++.h> using namespace std; #define ll long long int main() { int t; cin >> t; while (t--) { long long n, k; cin >> n >> k; // Check if k is odd if (k % 2 != 0) { cout << "NO" << endl; continue; } // Calculate the maximum possible sum long long sum = 0; long long m = n / 2; for (int i = 1; i <= m; i++) { sum += (n - 1 + (i - 1) * 2); } sum = 2 * sum; // Check if k is larger than the sum if (k > sum) { cout << "NO" << endl; continue; } else if (k == sum) { // If k is equal to the sum, print the array in descending order cout<<"YES"<<endl; for (int i = n; i >= 1; i--) { cout << i << " "; } cout << endl; continue; } else if (k == 0) { // If k is zero, print the array in ascending order cout<<"YES"<<endl; for (int i = 1; i <= n; i++) { cout << i << " "; } cout << endl; continue; } // Initialize the array x vector<ll int> x(n + 1); for (int i = 1; i <= n; i++) { x[i] = i; } // Adjust the array based on the value of k ll int low = 1, high = n; while (k != 0 && low < high) { ll int diff = high - low; ll int adjustment = 2 * diff; if (adjustment <= k) { swap(x[low], x[high]); k -= adjustment; low++; high--; } else { high--; } } // Print the resulting array cout<<"YES"<<endl; for (int i = 1; i <= n; i++) { cout << x[i] << " "; } cout << endl; } return 0; }

Can you give the mathematical proof to say that this one pass algorithm does not give false negatives?

why are u using remain + i + 1 why not just remain

bro crystal clear explanation

Thanks for this explanation. Do you have any recommendations for improving observation skills?

Please keep posting 🙏

edaa monee

Believe it or not, this is too close to my graduation work, I will present it on thursday 20 june, exactily one week from now "UTC+1", Moxico, Angola...

Just love your video...

Can't we do it like we find sum of all rows or columns of both given Matrix and compare them if they have same elements print yes eles no ??

Show the code as well

Amazing explanation 🔥

op .

super amazing, I understood very very good keep going

sorry but do you have some speaking problems? like sneezing, speaking slow and unclear, talking in self doubt. Improve this bro, doesnt sounds decent.

Is that code got accepted?

really worthy explanation. keep it up

Good

Can you please put the link to the code ?

Very few people has explained this you explained great nice job

Amazing brdr!!

nice explanation 👍

is good thing

thx bro!!

didnt get what is the use of prefix array?

Genius!!!....really liked the way u explained it was amazing

bhai tu bhagwaan hai

nice logic

awesome brother!!

finally understood the solution bcz of u man!! Thanks man!!!

Can you please comment whether this video was useful or not. This will help me in making much better videos in the future

I was half sleepy when i was making this video, if you guys have any doubts, please ask. I will post the solution of problem B in another 10 minutes

If you can please comment what do you expect from my future videos