Відео

Good job sir

Harris Kenneth Thomas Angela Taylor Sharon

Worst explanation ever

Well explained

Sir I am bengali medium student...Which language would be better to prepare for TCS, bengali&english?please tell me

Video title is wrong for this video.

Please Bring back Yash Dwivedi sir or Sid Sir :)

worst

good explanation!

Thank you very much! It was extremely helpful in understanding the paths in the grid problem.

NYC explaination 👍

Is there more optimized approach for the same or any algorithm to reduce LOC

The explanation is ok but I love those videos in which the approach is explained in the way that iit is live video. This video seems to be more scripted then an explanation.

Good Explanation!!! class Solution { public: Node *reverse_linked_list(Node *head) { Node *curr = head; Node *prev = NULL; Node *next = NULL; while (curr != NULL) { next = curr->next; curr->next = prev; prev = curr; curr = next; } return prev; } Node *merge2Sortedlist(Node *ptr1, Node *ptr2) { Node *Dummy_node = new Node(-1); Node *temp = Dummy_node; Node *t1 = ptr1; Node *t2 = ptr2; while (t1 != NULL && t2 != NULL) { if (t1->data <= t2->data) { temp->next = t1; temp = t1; t1 = t1->next; } else { temp->next = t2; temp = t2; t2 = t2->next; } } // Checking if the t1 is still present or not while (t1 != NULL) { temp->next = t1; temp = t1; t1 = t1->next; } // Checking if the t2 is still present or not while (t2 != NULL) { temp->next = t2; temp = t2; t2 = t2->next; } return Dummy_node->next; } Node *sort(Node *head) { // Splitting the linked list into 2 parts Node *first = new Node(-1); Node *second = new Node(-1); Node *t1 = first; Node *t2 = second; Node *curr = head; while (curr != NULL) { t1->next = curr; t1 = t1->next; curr = curr->next; if (curr != NULL) { t2->next = curr; t2 = t2->next; curr = curr->next; } } t1->next = NULL; t2->next = NULL; Node *ptr1 = first->next; Node *ptr2 = second->next; // after the spliting part just reverse the 2nd linked list ptr2 = reverse_linked_list(ptr2); // Now there is a ptr1 pointing to 1 linked list and a ptr2 pointing to the other linked list head = merge2Sortedlist(ptr1, ptr2); return head; } };

I have a coupon for a course worth INR 1500. If I purchase a course using this coupon, will the 90% refund offer still apply? Can I get 90% of my fees back after completing the course within 90 days?

very poor explanation

worst explanation

The explanation is not that good.If necessary make the video long but don't compromise on explanation.

didn't understand a thing you were saying. please give clear explanations

explained very well 😊!!

🎉😊

for question 2nd simple approach would be to sort arr1 in ascending order and arr2 in descending order and then apply simple 2 pointers. code will be like this: long long fndMax(int n, int m, vector<int>& arr1, vector<int>& arr2) { sort(arr1.begin(),arr1.end()); sort(arr2.begin(),arr2.end(),greater<int>()); int i=0,j=n-1; int k=0,l=m-1; long long ans=0; while(i<=j&&k<=l){ if(abs(arr1[i]-arr2[k])>=abs(arr1[j]-arr2[l])){ ans+=abs(arr1[i]-arr2[k]); i++; k++; } else{ ans+=abs(arr1[j]-arr2[l]); j--; l--; } } return ans; }

This weekly contest was tough

Worst explanation 👎👎

Good explanation👌

3/10 for explanation

bro you can speak in hindi if you want

Will this code(after base change) work for any other digit?? for example - "after removing all numbers with digit 3".

Nice Logic 👍

Best explanation ever

the time complexity for the brute force is O(nlogn) right, if we use digit extraction for finding if 9 is present in the number or not?

Johnson Brian Thomas Charles Wilson Patricia

first question: c=0 while n: if n%2==0: break c+=1 n=n>>1 c=2**c return c

digit dp

Good approach we can also calculate sum of first n natural number n*(n+1)/2 and substract accumulate(are.begin(),are.end(),0) from it

Nyc

public class String_Question2 { public static int atoi(String str) { int result = 0; int i = 0; int sign = 1; if (str.charAt(i) == '-') { sign = -1; i++; } while (i < str.length()) { if (str.charAt(i) < '0' || str.charAt(i) > '9') { return -1; } result = result * 10 + str.charAt(i) - '0'; i++; } return result * sign; } public static void main(String[] args) { String str = "123"; int integerValue = atoi(str); System.out.println("Integer value: " + integerValue); } } // Another method // public class String_Question2 { // public static void main(String[] args) { // String str = "1234"; // int num = Integer.parseInt(str); // System.out.println(num); // } // }

Or simply return (n/2)+1 as for every n we can represent it in multiple to 2 like for n = 1 (1+0*2), n=2 (1+1+0*2 or 0+1*2) etc

kya ghatiya explaination tha yr,..........aisa lag raha bas bhaga rahe hai, samjhane ka man hi nahi kar raha, ......pata nahi jab nahi man karta to kyu video dalte hai faaltu fokat me,..........

not even explained how (b^-1)%mod = b^(mod-2), the main thing,...........then why you are teaching can I ask you @Yash Dwivedi,...If you have some shame remaining, then please don't do this type of mistake again..

Why the hell gfg is recruiting such fellows 💀💀💀💀💀💀

Love this mentor ..... amazing explanations (specially last problem) Thankyou GFG 🙂

Nice explanation. Cheers.

Kindly discuss Time & complexity as well.

great explanation!!

He is a little bit dumb. He just reads things and tells them to the audience.I do not understand the question solution.Please back Yash Dwivedi, sir.

Why can't we use dfs + dp? It is giving wrong result

class Solution { public: vector<int> start, end; void Euler_tour(int &times, vector<int> adj[], int node, int par) { start[node] = times++; for (auto it : adj[node]) { if (it != par) { Euler_tour(times, adj, it, node); } } end[node] = times - 1; } vector<vector<int>> twoMax(int n, int m, int rootServer, vector<vector<int>> &edges, vector<int> &requests) { start.resize(n + 1, 0); end.resize(n + 1, 0); vector<int> adj[n + 1]; for (auto it : edges) { adj[it[0]].push_back(it[1]); adj[it[1]].push_back(it[0]); // Assuming undirected graph } int times = 1; Euler_tour(times, adj, rootServer, -1); vector<int> temp(n + 1, 0); for (int i = 1; i <= n; i++) { temp[start[i]] = i; } vector<pair<int, int>> prefix(n + 1, {INT_MIN, INT_MIN}); vector<pair<int, int>> suffix(n + 1, {INT_MIN, INT_MIN}); int maxi1 = INT_MIN; int maxi2 = INT_MIN; for (int i = 1; i <= n; i++) { if (temp[i] > maxi1) { maxi2 = maxi1; maxi1 = temp[i]; } else if (temp[i] > maxi2) { maxi2 = temp[i]; } prefix[i] = {maxi1, maxi2}; } maxi1 = INT_MIN; maxi2 = INT_MIN; for (int i = n; i >= 1; i--) { if (temp[i] > maxi1) { maxi2 = maxi1; maxi1 = temp[i]; } else if (temp[i] > maxi2) { maxi2 = temp[i]; } suffix[i] = {maxi1, maxi2}; } vector<vector<int>> result; for (int i = 0; i < m; i++) { int node = requests[i]; int start_time = start[node]; int end_time = end[node]; int res_maxi1 = INT_MIN; int res_maxi2 = INT_MIN; if (start_time > 1) { res_maxi1 = prefix[start_time - 1].first; res_maxi2 = prefix[start_time - 1].second; } if (end_time < n) { res_maxi1 = max(res_maxi1, suffix[end_time + 1].first); res_maxi2 = max(res_maxi2, suffix[end_time + 1].second); } result.push_back({res_maxi1, res_maxi2}); } return result; } }; where i made the mistake plz help me guys

Thank you for amazing content ❤❤❤❤🎉🎉😊😊😊

In the second the problem why we are dividing the total number with 3 ,is there any specific intuation behind that?