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Understood!!

I have a doubt: in the first tree, what will be the output if the left of 3 had a node say "12" and the left of node 5 had a node "13". In this case, will the output be 1,2,3,12,5,13,6,10,11,9,8,7

Aaj smjh aaya why this works u explained so beuatifully ki feel aagi.

Absolutely brilliant explanation. You are making one fall in love with coding. Thank you for all your efforts. We are so grateful to u.

UNDERSTOOD ❤ FROM NIT SILCHAR...

UNDERSTOOD❤

Superb explanation.

Understood❤❤🔥🔥

Tq for the Hard work

thanks sir

thank you striver

"US"

Thanku

US

Understood!

UNDERSTOOD

understood

Thankyou for great solution Striver

What if there are multiple LCS and we have to print all of them ?

Understood

US

Understood

Understood.

understood

Thank you for this exceptional course..👍🙂

our teacher taught this as if it were a problem about theory of relativity and you taught it like the alphabet. thank you so much

when I saw this question, there was in my mind that it is saying to insert, ofcourse it will be inplace and end up doing inplace which beats 5% people in runtime😂 BUT beats 98% people in terms of Memory Here is the code, if anyone wanna see this approach class Solution { public: vector<vector<int>> insert(vector<vector<int>>& inter, vector<int>& newinter) { int n=inter.size(),low=0,high=n-1,ans=-11; while(low<=high){ int mid=low+(high-low)/2; if(newinter[0]==inter[mid][0]){ ans=mid; break; } else if(newinter[0]<inter[mid][0]) high=mid-1; else low=mid+1; } if(ans==-11) ans=high; if(ans==-1){ int last=-111; while(!inter.empty() && inter[0][0]<=newinter[1]){ last=inter[0][1]; inter.erase(inter.begin()); } newinter[1]=max(newinter[1],last); inter.insert(inter.begin(),newinter); } else{ if(inter[ans][1]>=newinter[0]){ int last=inter[ans][1]; while(inter.size()>=ans+2 && inter[ans+1][0]<=newinter[1]){ last=inter[ans+1][1]; inter.erase(inter.begin()+ans+1); } inter[ans][1]=max(newinter[1],last); } else{ int last=inter[ans][1]; while(inter.size()>=ans+2 && inter[ans+1][0]<=newinter[1]){ last=inter[ans+1][1]; inter.erase(inter.begin()+ans+1); } newinter[1]=max(newinter[1],last); inter.insert(inter.begin()+ans+1,newinter); } } cout<<ans; return inter; } };

Amazing explanation bhaiya💯

i thought the logic all by myself . and coded it up .. thanks striver

Understood

uderstood the algo thanks bro for the intution behind the algorithm❣❣

If we add -2 to the stack, pop it and then try to access the value of minimum, it will return as 2, but shouldn't it return us 0 or nothing in this case?

US

understood❤

What an explanation !!

Understood.

understood

understood ig;

#include <bits/stdc++.h> using namespace std; void f(int i,int n){ if(i>n) return; f(i+1,n); cout<<i<<" "; } int main() { cout<<"Enter a number: "; int n; cin>>n; int i = 1; f(i,n); return 0; }

I haven't learned about graph yet

class Node { int data; Node next; Node(int data) { this.data = data; this.next = null; } } public class Main { // converting array to linked list private static Node convertArray2LinkedList(int[] arr) { if (arr.length == 0) return null; Node head = new Node(arr[0]); Node curr = head; for (int i = 1; i < arr.length; i++) { Node temp = new Node(arr[i]); curr.next = temp; curr = curr.next; } return head; } // traversing linked list public static void display(Node head){ Node temp = head; while (temp != null) { System.out.println(temp.data); temp = temp.next; } } // length of linked list public static int lengthOfLinkedList(Node head){ int len = 0; Node curr = head; while(curr != null){ len++; curr = curr.next; } return len; } // checking if element is present in linked list or lengthOfLinkedList public static boolean check(Node head,int val){ Node curr = head; while(curr != null){ if(curr.data == val)return true; curr = curr.next; } return false; } public static void main(String[] args) { int[] arr = {11, 2, 3, 4, 5}; Node head = convertArray2LinkedList(arr); System.out.println("Displaying linked list: "); display(head); System.out.println("Length of linked list is: "); System.out.println(lengthOfLinkedList(head)); System.out.println("checking if value exist in linked list: "); if((check(head,9) == true)){ System.out.println("yes"); }else{ System.out.println("no"); } } }

Understood bro (July 7 2024)

thank you

US

understood

understood

Wow didn't know this was a 50 minute video!! Never made me look at the time ❤❤❤ this is the best video on merge sort❤ thank you very much 😄

Understood :)

Understood!!! Thanks a lot!!

What if the input arr is [[1,2,3],[1,2,3],[1,2,3]] . then (n * m) /2 will not hold. Can someone please clarify ?