Video

Better than my prof at ucsd

Practice problems on this topic?

improve the explanation to be more beginner friendly

great

0:15 correct 0:27 efficient 1:07 modelling 1:50 techniques (3:09 pre reqs)

i found the root array to be unnecessary. node[k] is equal to root[k] if k is a component root.

This implementation of merge sort throws an out of bounds exception, because it attempt to access A[i] (in case 1) before checking i==m (in case 2). Here's the fixed version: int Apos = 0, Bpos = 0; for (int i = 0; i < C.size(); i++) { if (Apos == A.size() || ((Bpos < B.size()) && (B[Bpos] <= A[Apos]))) { C[i] = B[Bpos]; Bpos++; } else { C[i] = A[Apos]; Apos++; } } look up short circuit evaluation if you don't know what that is.

One quick question, as you have mentioned if i use sorted array as priority queue i would need O(n) time to insert and O(1) time to delete_max, so overall to process n jobs its O(n^2), why can't i insert in a binary search way needing O(logn) to insert and take out in O(1) time making it O(nlogn) and we do not have to move to heaps in that case. Am i missing something?

Thank you sir.

Consider given array is A={2,6,8,9,11}, r = 5 & l = 0; And searching element LET K= 9; //-------fisrt call : ------ bsearch(K=9,A,l=0,r=5) { //for 1st if(r-l == 0) r -l = 5-0 not equal to 0 // if statement will not execute mid = (l+r)/2 = (0 + 5)/2 =2 //2.5 but will take integer division //For 2nd if( K == A[mid]) K= 9 and A[mid] = A[2]= 8 which is not equal to K Therefore if statement will not execute //For 3rd if(K <A[mid]) (9 is not less than 8) Therefore if statement will not execute //For else //------Second call : ------- bsearch(K= 9, A, mid+1= 3, r=5) {// l is mid+1 that is 3 r -l = 5-3=2 not equal to 0 // if statement will not execute mid = (l+r)/2 = (3 + 5)/2 =4 //For 2nd if( K == A[mid]) K= 9 and A[mid] = A[4]= 11 which is not equal to K Therefore if statement will not execute //For 3rd if(K< A[mid]) K = 9 and A[mid] = 11 therefore if statement will execute and //------Third call : -------- bsearch(K= 9, A, l= 3, mid=4) { // here r = mid = 4; r -l = 4-3=1 not equal to 0 // if statement will not execute mid = (l+r)/2 = (3 + 4)/2 =3 //3.5 but will take integer division //For 2nd if( K == A[mid]) K= 9 and A[mid] = A[3]= 9 which is equal to K Therefore if statement will execute And * return True* to second call And that will return true to 1st call of function And overall element is found }//3rd call end here }//2nd call end here }// 1st call end here

Where to find presentation?

WELL I HAVE ALWAYS BEEN A FAN THE WAY MADHAV SIR EXPLAINS THE CONCEPT...BEST AMAZING,, EVEN HIS PROGRMAMMING DATASTRUCTRES USING PYTHON ALSO AMAZING..

I just hate it when the course about algorithms has algorithms wrong

Does anyone have slides for the lecture

Language should be simple ,,, difficult to understand

Thank You So Much for This Amazing Lecture... 😊

Thank You So Much for This Amazing Lecture... 😊

Thank You So Much for This Amazing Lecture... 😊

Thank You So Much for This Amazing Introduction, looking forward to completing this course... 😊

In the beginning of the session, example of air traffic controller was stated and the issues with respect to heap were observed when checking the predecessor and successor. Thus, can we say Binary Search Tree overcomes disadvantage of Heaps for finding successor and predecessor?

Hi sir muje input size nahi samaj ra hai

we say to it ratalogist it means to memorize something but in practical quality is zero please don't waste time in such video

anyone can share the code of this problem

amazing explanation

Thank you very much for these applications. It helped me solve many questions .

Why sir you are making such an easy thing so complex?

Can you simplify the proofing part. It's getting very complex

Thank you sir for these videos your channel is really helpful !

Sir please try explaining on greenboard with chalk. Really helps in understanding things since we become a part of calculation with it and can understand steps. And gets us involved . Please consider

I think, at 13:21, the very last update statement should be Nbr_TV [ v ] = u ;

Hii i am btech IT student...but there is no option for IT dept in gate exam training session...should we follow computer science engineering videos ?

at 9:47 The indentation for the last statement is incorrect. The post count for vertex 'i' must be assigned a value once the For loop is completed and not inside it.

Thanks for the Malayalam subtitle

Seems bellman ford only works well for directed graphs with negative edges. If you have an undirected graph with a negative edge, that will be interpreted as a negative cycle by this algorithm, as you can go to-fro as many times as you want over the negative edge. For proof, consider the adjacency matrix representation of an undirected graph. Let's say edge from U to V has weight -W. then graph[U][V] = -W graph[V][U] = -W This will definitely be considered as two separate edges by the algorithm, hence the negative cycle.

Vera level Madhavan Sir

Anyone pursuing the same course in july 2022 here

👍

hi

Why + 1?

Thanks a lot ! Simple and efficient teaching techniques. Much appreciate that

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loved the proof man!!

This is some next level explanation. I am just loving all his videos. I wish my college teacher could teach like this. Such proper sequential explanation never seen before. Love you sir.

at the last step of the algorithm.. it should be quicksort(A,yellow,r)

amazing lecture!!

awesome lecture

I appreciate this. Of several explanations of the Proof and the concept of the Exchange Argument this is the best and for me, the one that enabled me to understand the proof. Thank you and well done sir!

can left child or right child equal to parent in case of BST?

Can anyone please help me where I can find the lecture slides of all the sessions on the channel.

Can you please help me to write c++ source codes for different programs