VERY HARD Mock Coding Interview with STRIVER, YouTuber, Googler
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- čas přidán 22. 07. 2024
- One of the most requested videos - hope you enjoy and learn watching @striver_79 solve very-hard graph question! @takeUforward is known for his graph series, so this should be super-fun! ✌🏻
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The video contains following parts-
0:00-1:14 - Intro and Bosscoder Academy promotion
1:14-24:20 - Question and Algo Discussion
24:20-56:30 - Code with Dry Run
56:30-1:01:05 - Feedback
#mockcodinginterview #datastructures #graphs
Initially I thought it was easy, until I figured out the edge case. Probably could have been a cleaner implementation if the camera pressure was not there :P. But this is what time constraint does to you. It was an amazing experience :)
Bhaiya ❤you are my saviour ❤your sheet and videos helped alot to gain knowledge ❤
The interview plus camera pressure is next-level! Being able to do this much proves that you truly are an expert in DSA (specially Graphs!)
What would the solution be if we change the question to - Find the number of edges to be removed such that the sum of weights of removed edges is minimum?
Thank You
Nice explanation and I thoroughly enjoyed the problem solving part, one edge case I could think this is missing is the component can have both cyclic part and non cyclic part, so greedily first we have to remove non cyclic edges within a component and then start removing cyclic edges and so on. What do you say?
I am so satisfied to see Raj writing code. I wonder, how much effort it would take to reach a place where he is. It's so inspiring to see him :)
We don't write code like that in a real job though. There we have a lot more interfacing and abstractions.
The given solution misses one edge case, imagine a component with 1 - 2 - 3- 4 - 5 - 3 , in this component we have a cycle, but still if we remove just the 1-2 edge, it will split them into two components. The correct solution seems to be the one where we end up removing the nodes with the smallest degrees, and then keep doing till we get the answer. This might also have some edge cases, need to grill on it. Minimum degree edges which are not a part of cylce, is what we have to grill on, does not seems that easy..
Even this does not work,
Consider follwing case
n=9 m=12 k=4,
1-2
2-3
3-4
4-1
5-6
5-7
6-7
6-8
6-9
7-8
7-9
8-9
Now, smallest degree is 2 for nodes 1,2,3,4,5 . Now if you select 5 first, then required edges to be removed is 4 but if you select apart from 5, required edges will be 3. So it depends on which node with minimum degree you select
@@vaibhavdawra8960 Yes it has be a combination of smallest degree and not part of cycle, seems a complicated one.. Tougher than it appears..
Exactly thought the same, you need something like topological sorting and also keep track of which compos contain cycle, and calculate accordingly, I guess this solution would be a bit more trickier than the one you wrote😜
I believe this question has to do something with bridges.
@takeUforward I think solution would be to find the bridges, each bridge would give you 1 more component(as it's an undirected graph it would be connected), If we fall short, we need to maintain indexed min heap of vertex's as all the components would be cycles and remove vertexes with minimum degrees, update the remaining degrees of adjacent vertexes and repeat till we reach #of Vertexes. (Max number of components in a given graph).
I don't know how or why I always fall for striver's way of solving problems, but he is a legendd!!! Thank you keerthi
+1
william lin sitting in corner :- hold my beer!
@@yamangoyal7629 tourist sitting in corner :- hold my codeforces and codechef rank
@@yamangoyal7629 Tourist to william lin : Do u know who i am?
@@yamangoyal7629 let him sit corner . Itna kuch khass nhi he
I found Keerti also confused 🤔 in middle of this video when striver explain Keerti expression like what he doing 😅 anyone thinking same like....
He may not be able to come up with the 100% accepted solution which is probably hard for a problem like this in an interview or in front of a camera but what he said at the end is 100% correct. Interviews are just not about if you were able to solve a probem but how you approach a problem.
I kept trying to think and solve along. Now I am just glad I was able to follow along and understand 😝
Lots of respect to Striver writing so much code to guide us through.
However suppose the graph initially has only one component which is densely connected with a lot of edges and k=2 that is you have to remove some edges so that you get one more component. In this code first you would end up removing all the edges that are causing cycle in the graph and 1 more edge to make a new component. But it is possible it is not the minimum number of edges required to make 2 components but far more than the actual answer. Am I wrong somewhere?
@9:39 5-6 6-7 7-8 won't break the component , they are part of the cycle too as 8-5 is. Just find all the cycles in the graph and keep the list of edge not in any cycle. After that you can also find the edges with which you can have your prob solved in minimum cost not just minimum number.
Yes you are correct. That's what I replied. Each edge is dependent whether the other edge has been already removed or not. Even removing 6-7 will not break the component.
@@surajmamgai102anybody please tell me which programming language striver bhai uses to solve the problem in this video..I'm completely new so ...c,c++ or java
@@maheshvommi1092 It's C++ bro. If you are new then you might find syntax different here bcz of the use of STL's and bits/stdc++.h library.
its c++@@maheshvommi1092
@@maheshvommi1092 c++
I think Edmond Karp variation of Ford Fulkerson method to find the maximum flow between source and sink nodes which would be equal to min cut will work. We probably have to do it for all the possible pairs in the graph and then pick the one with minimum cut and then remove the edges from the graph. This will lead to 1 more disconnected component in the graph with removal of minimum cost of edges. Have to do this k - 1 times to have k components. Complexity goes somewhere around k * V2*(V*E2).
yes
It was great...want some more such interviews with Striver.❤
Learnt a lot of tackling from this video. Thanks Keerti and Striver!
Assalamualaikum
Tamila bro
@@mehranmushraf5224😂😂😂
Love u striver bhaiya. ❤ You are master in DSA. Currently I am watching your graph series. Awesome 👍
What about this solution.... You count the indegree of all nodes. Subtract the 'k' from total components. That becomes your target. Now keep on visiting those nodes which has minimum indegree greedily. Disconnect them and add their indegree to the answer. Visit the immediate neighbors and decrease their indegree by 1. Seems to work on all test cases I came across.
You are on the right path, just keep improving on this. It's a good start and a lot better than the attempt made by striver in the video
n=8, m=9, k=2
1 2
2 3
3 4
4 1
3 5
5 6
6 7
7 8
8 5
Try on this test case.., you answer would be *2* (minimum indegree is 2 here), but u can just remove the 3-5 edge to make 2 companents. So, actual ans is *1*
Thanks u striver Bhaiya and keerti didi for making this video this is very helpful and interesting video how to approach problem, represent to our ideas and thought process to interviewer. 😍😍
for the loop checking and knowing whether the edge could give me separate component i was thinking of union-find algorithm.even we can check for component using union find algorithm.
We can utilize the property of bridges because as per definition "a bridge is an edge which on removal increases the number of connected components by one".
The procedure I have thought of is,
1. We have to make k connected components.
2. Let the current number of connected components be 'curr_con', now 'k-curr_con' more components to make.
3. Count the number of bridges available in the graph (using DFS), let it be 'b' and break the bridges, now ' k-curr_con-b' more components to make.
4. Now again find the number of bridges, break them and keep on repeating this until we make it to k connected components.
It wont work
@@knowcode6374 Can you please explain why ?
It will fail where we have to remove the edge in the cycle to get the required k component.
@@varungurnani4989 kosa Raju algo we can use I think, no of strongly connected components
Love you Striver Bhaiya your the King of Coding😍
He is a great problem solver how logical thinking his brain has 🔥
Subscribed! thanks for bringing striver!
my approach:
1). count no. of components - x
if x>=k return -1
2). so, we need to make the k-x components
3). calc the degree of all the node and put it into treemap, tm
4). take a variable ans =0 for min #edges need to remove
5). iterate over treemap until x
Is this approach functioning correctly?
I am trying to rewrite a big number library I wrote many years ago. I am trying to move it from 32 bit native low-level C code to 64-bit native low level C code.
I have used it to do prime number mathematics, RSA encryption, and Diffie Helman key exchange.
I use it as a driver to explain how these technologies work.
I just want to see if there will be any performance boost going native 64-bit. I will probably have to use some assembly language.
This is the approach i could think of using UNION-FIND:Let me know if I am wrong ^^
We need to find the minimum number of edges that can be removed to get exactly k-connected components.
Consider it the reverse way: Find the max number of edges that can be joined to give you exactly k connected components.
Inittially consider all the nodes as individual graphs(consider each node as the parent of itself). USE UNION -FIND to join two nodes and check the number of components thus remaining(check number of unique parent nodes- make a parent array corresponding to the ). Keep a count variable to store number of edges covered.
There will be a point you reach where there will be exactly k unique parents. Keep going forward but this time only use an edge if on using it, two components of same parent are connected(because cycles won't increase the number of components). Only then increase count.
Count will be your answer.
Inspirational and so motivational video. So much knowledgeable. Thank you.
Asking constraints for n to understand the time complexity and understand which algorithm can be fit there ...is something I loved most ....I never knew this trick can be useful in interview
Adding one more thing ....not giving up and trying to write code even if it is not correct.....In my last interview I had 3 coding questions (1 easy and 2 medium) ..I was able to solve starting two with multiple approach but in last one I was only able to give brute force and better approach and while trying to give optimal solution I gave up ...and guess what ....I gave almost every answer I still got rejected due to giving up attitude....( Lesson learnt but hurtfully)
Dayum! Just amazing
Watching striver’s interview was totally worth it
Thank u keerti and you have earned a new subscriber :)
Thank you! And welcome to the channel! Hope you like rest of the content as well 😇😇
My approach
1) create an adjacency matrix for all the nodes in the graph using DFS.
2) while DFS, find the total number of islands/components present in the graph by having the visited vector (let us assume total components = x).
3) (k-x) is the required number of extra components. and ans=0.
4) start a loop for (k-x) times.
4) parse the adjacency matrix and find the node that has the lowest degree (least number of adjacent nodes)
5) Add the degree to the ans (this separates this node into a new component)
6) also remove the current node from the adjacent nodes. (updating the adjacency matrix after removing nodes)
7) return the final ans.
@@user-rn1yb5pq8k that would be wrong as you are having a greedy approach.
Which would not work for some edge cases.
Like
Imagine you have 2 big clusters of nodes and they are connected to eachother inside the cluster now also let both of them to be identical and now let a node be added to any one cluster so that it is connected to 3 nodes
And now the time is to connect the 2 clusters select 2 nodes from both the clusters except the 3 one that we added.now make the graph a single component by connecting the 2 components that we had originally
Now you have a graph
So if k=2
All you req is to sever that final edge that you had so your answer should be one
But from your algo we should remove the 3 edge node
If you are able to understand this case then good .other wise I may answer it in more detail
For the correct solution I think initially we should try to remove the bridges as removing one gives one new component guaranteed.Then if we still need more components,we are then left with disconnected graphs in which each one of the nodes obvously belongs to a part of a bigger cycle and there maybe smaller cycles including some subset of nodes in that component also.No idea how to proceed from this...😅
we could also remove articulation points also , ya but removing of a articulation point will take upto 2 edges at once.
@@pulkitjain5159can you please tell , striver bhai solving qus in which programming language..I'm new so I want to know
c++ @@maheshvommi1092
This is the correct Approach
My Solution can be like finding all bridges in the graph using tarjan's algorithm for undirected graph along with that storing bi-connected components using stack . removing a brigde will surely give you a more connnected component. So Assuming initially there was a single connected component in the graph.If we got n bridges then we can get n+1 components . if n+1>k then we will delete only k-1 brigdes else if(n+1
How about creating minimum spanning trees (MST) from the current x components and then using k-x to find the answer? The edges which create components will be present in the MSTs.
wouldn't that have an issue with cyclic graphs?
Striver is on fire. Lately he is doing best collaborations.First kirti and now harkirat 🙌🙌
Very good mock interview, Learned so many things.
I have one doubt in your example only.
1) 1 - 2 - 3 - 1
2) 4 - 5 - 6
3) 7 - 8 - 9 - 10 - 7 - 9
k = 8 (You took 7 but I took 8 because i thing here is an edge case).
From your code it will be 3 (already given) + 2 (more from 2nd compo) + 1 (cycle removing edge in 1st compo) + 2 (more from 1st original compo) + 2 (cycle removing edge from 3rd compo) + 1 (from 3rd compo)
edges req = 3 + 2 + 1 + 3 + 2 + 3 = 11.
components = 3 + 2(from 2nd) + 2(from 1st) + 1(from 3rd) = 8.
From my code it will be 3 (already given) + 2 (more from 2nd compo) + 2 (cycle removing edges from 3rd compo) + 3 (more from 3rd compo)
edges req = 3 + 2 + 2 + 3 = 10.
components = 3 + 2 (from 2nd compo) + 3 (from 3rd compo) = 3 + 2 + 3 = 8.
At the end , I got Answer with 1 less edge required.
So, I think that last greedy approach won't work every time, We need to check that how many requirements of componenets, according to that we need to break cycles and get more components from that component.
It means At Last, "The count of cycle breaking edges and getting components from it", this thing should be minimum instead of "only minising the cycle breaking edges".
Please check once and correct me if i am wrong.
Thanks for Fantastic mock interview.
Everyone forgets, this line give me the hope😄
This was fun to watch xD.
I think a more interesting problem would be to find out the minimum weight of the edges to be removed instead of the number of edges.
Also, I think there might be a mathematical solution for the "number of edges" problem wherein we look at the number of edges in each component and make some greedy choices.
Let’s do another mock interview where I ask you this level of question. What say ✌🏻
@@KeertiPurswani woohoo that would be awesome. I am down for it✌️
Yaaaay! Karte baat
@@PriyanshAgarwal mostly awaiting...
same thought about the problem statement
I think, we can use union find . If both a and b belongs to same parent then breaking a----b edge won't give us new components. This way we can keep track probably.
Hope you watched the whole video and noticed that he followed same approach!
Yes I watched the video. Great efforts put in by both . Thanks for the video 👍
Nice solution, although I think there are few cases where this solution will not work, let’s say we have a fully connected component of 10 nodes and K=2, so the optimal answer would be 9, but with strivers solution we would get 37 ig.
I think this is a NP Hard problem, as if we calculate the answer for K = 1 to N, then we can solve the clique problem using that information.
Can you please share the resource of this problem and the solution given by the author if possible.
I think there is an edge case in 7 - 8 - 9 - 10 - 7 - 9. We can remove 7-8 and then 8-9 then we have 1 more component = 8. So we don't really need to remove 2 edges to get one more component. Instead I would find the bridges. If a + bridges >= k thats a return. Else keep track of the sizes of cycles like in this case 3, 3. Then individually check for those cyclic components
I have question on this. if you think of the case of : 1->3->7>5->3->8->10 .In this case there is a cycle in that component but on removing edge just 1 edge 1->3 or 3->8 or 8->10 we successfully get an extra component without affecting the cycle? Sorry Incase I am wrong. bhaiya please answer this
Two of my favourite youtubers in one video.
🧡
can i get the link to the solution i don't think the answer is correct and i don't seem to find the question on google.
Just curious to know this even though you've mentioned that this question won't be asked in an interview.
We're you able to solve it or knew the solution just like interviewers know the solution of the DSA question they ask.
we can also use Degree of a node in a graph property here .
NOTE : Here the cost is the no of edges that should be removed to make a new component
Approach :- we know that a node with a degree one can always give us a new component with a cost of 1 . A degree of node 2 give us new component with a cost of 2 and similarly for a component with a degree of n we get a new component with a cost of n .
So basically we need to spot the nodes with a degree one in the current formed graph , another thing is that when we remove a node we have to decrease the degree of all the nodes which are directly connected to it by 1 .
And then just iterate till the components are not equal to K along with adding the cost
Pls comment if there are any edges cases / logic error in the approach
Logic is wrong. If we consider this case when k = 2:
2 4
| \ / |
3 --7-- 6 -- 1
| / \ |
5 8
clearly, your logic would say the answer is 2, but the answer should be 1.
Can I find this question somewhere with testcases? Wanted to try my approach
Is it similar to minimum spanning tree(writing it just read 2 lines of question) ?
What is the edge weight for? 😅
what I have seen in almost other videos of keerti ,It was always like Keerti use to get Upperhand but here it was like Interviewer got nervosed by interviewee and Literally Striver Being Striver You knocked it Bro.Motivates me for Self-learning
Striver is a god damm monster 💀
Striver sir ❤ you are legend....pata nehi apke jaise kabhi ban pauga or nehi...🙏🙏🙏
,first we count number of connected components if connected components are greater than k,then print-1 else the solution always exist so,M-(n-k)+k will give minimum number of edges
Please correct me if I am wrong. This is a greedy approach.
We create a list.
We store all the degree of node into it and list of node connected with that node.
We declare an integer ans=0;
Now for minimum removal of edge to get k component, we iterate through array and find minimum degree of node.
while(k>0)
{
k--;
......................
}
And add that degree in ans.
After finding, we have to update array because that node is disconnected.
can you provide the code that striver coded so it will be easier to go through it.
thanku for the sharing the mock interview with striver
One simpler solution that I feel would work is, maintain degree of each node, a new component is created only if we cut a node whose degree is 1, if a node has degree more than 1, say x than x+1 edges need to be cut , so this generates the greedy algo, maintain a multiset of (degree,node) and remove the node which has min degree and accordingly adjust the degree of its neighbours. If anybody feels it wont work, feel free to mention some test
Seems an easier one, good way to think of it in this way..Am still thinking if this will give me the minimum or not, because of the following thought process.
We might end up reductantly removing too many edges unnecessarily just because of the cycling thing. Need to sit back and think of some cases, but seems a good way to approach.
@@takeUforward One case where it fails, I just thought is what if more than one node has same minimum degree, which should i choose and came with a test and this algo fails.
One way i thought it can be fixed is select that node whose component is lexicographically smallest when i arrange all nodes of a component in increasing order.
What do you think?
@@vaibhavdawra8960 This is nice a thought , but this would fail in cases where you have multiple nodes with same degree , decision of choosing the node to disconnect would ultimately affect your answer. For example, current node with min degree that you decided to disconnect would be connected to some other nodes which are still in the cycle. While at the same time there may be node with same indegree that is connected to nodes which are not in cycle which you missed to choose at first
@@user-le6ts6ci7h Yep, realised that and also proposed an alternative in this thread. What do you think about that?
I didn't finish the video yet but i think the algorithm is to find the connected components first, find the bridges with the low link value, and calculate the minimum edges that you need to move.
And now i will finish the video and see if i 'm wrong hhhhh.
dont we have to think in terms of the kruskals algorithm? it does almost the same thing..combine components into one repeatedly..?
Why not remove 2 edges from cycle to get 2 components, instead of removing component with 1 cycle
i think in the first question tarjans algorithm to find bridges in a graph can be used
Tarajan's algorithm for articualtion point and bridges!
The ambiguity in question, around weight was put purposefully or it was a mistake🤔
Hi Keerti, can u give the link of the question please? I have a solution in mind. Just wanted to test it.
can you please tell which live compiler is this?
i mean is it not that we find bridges in the graph using tarjans algo
Yupp I also thought of Union find after the problem statement appears and mine solution is exactly as same as striver
I need the problem link to solve this if you have it Keerti please share
I think for this question degree of vertex and alredy have components needed.
Correct me if I am wrong
Can we not use minimum spanning tree to solve this
My idea is we find the number of components and if they are already greater than k, we print -1
Else we want to start disconnecting our graph until we get a component count of k
We can use DSU to make it happen
In MST instead of choosing edges with least weight we choose all the edges max weight until we get component count =k and after that juncture, there are two options either we come across an edge that increases the components then we have to add it’s cost in answer and other edge connects nodes belonging to the same disjoint set.
Some one Please let me know if that would work
Thanks
What is the sense of weight in the graphs?
That moment, when the interviewee outshines the inteviewer \m/,
I was thinking instead of breaking the edges, we keep track of components while creating the dsu. suppose we iterate over edges and my_components are greater than k, so we take it but if our components =k then we take only edges which doesn’t effect the count of components
striver is a living legend i guess he solved problem during his sleeping time too 🤣🤣🤣🤣🤣🤣
Idk but while watching this, I felt that the interviewer is more confused than the interviewee
He can decrypt imagination into code 👌
you are the best keerthi. you are going to rock the world.
Please take a mock interview of love babber ❤
I think i have a edge case on which this approach or code should not work,
lets say n = 4,m =4,
edges are - [ [ 1,2], [2,3] , [3,4], [4,2] ]
i'll try to draw it-
4
/ \
1__2/___3
and k = 2, which means 2 components and answer should be 1 for it as if you remove edge between 1 and 2 you have 2 components,
but according to striver's approach this is cyclic component so first he will remove edges responsible for cycle that mean one edge than one more to edge removal for havig two components which means his code should give answer 2 while it should be 1.
i have more example of this edge case but this is the simplest and should work to explain what i am saying.
I think this code will work on given edge case . Please have a look :
#include
using namespace std;
mapind;
vectoradj(1001);
mapvisited;
int main()
{
int n,m,k;
cin>>n>>m>>k;
for(int i=0;i>u>>v;
adj[u].push_back(v);
adj[v].push_back(u);
ind[u]++;
ind[v]++;
}
priority_queuepq;
for(int i=0;i
Amazing!!
not able to see strivers screen completely! making it harder to follow along;
Hey, this mock is for which exam exactly. Someone please tell me
Hi , Please could you provide us some more difficult questions with solutions.
Nice content❤
I will think about answering this question once I am done with my preparation. It's going to be my mock interview at that time
Haha man ki batt chin Li.
Kas esa hota.
Stiver's initial case k-x fails when there is a circular graph we need to make a additional cut , Again commenting in the middle of the video since it says to think of a case.
mam can you please comment out the problem link?
U know u r done for if striver feels the pressure
Want one more session with Striver
Hey, can you also post the actual solution to this problem? I wonder how you even satisfied with his solution because in my opinion two things actually matter for any algorithmic interview:
1. Correctness of the Algorithm
2. Complexity Analysis.
First of all, where is the correctness? I wonder, that you asked for an algorithm, and you didn't even ask whether it's correct or not? It will also be helpful for the viewers who watch this video if you include the above two points.
is this also for Frontend Developer ????
Can someone please comment on below algorithm that I thought:
1) find no. of connected components. Let it be x.
2) find degrees of each vertices and store it in hashmap. also create a treemap.
3) remove any smallest degree vertex from treemap (that is why I took treemap to easily remove vertex with smallest degree). Lets say vertex v has smallest degree d. add d to the ans because this are the edges we are gonna remove to separate vertex v from the connected component. after this update degree of removed vertex and its neighbour vertex in hashmap and treemap. So,
ans += d
x++
4) repeat step 3) till x becomes K and ans is our answer.
How will you ensure that someone with the smallest degree is not form of a cylce, this is what I am thinking
@@takeUforward I think,even if it has a cycle this algo should work.
i.e. 1-2-3-1
1 -> 2 degree
2 -> 2 degree
3 -> 2 degree
So smallest degree is 2, we can remove any vertex if k=2 then ans will be 2. (Ans += 2)
For k = 3, now that degree of vertex 2 and 3 is 1 if we remove vertex 1. Ans += 1. Which is 3 edges.
i thought of exactly same solution, If you find any test on which it does not work , add it in ypur comment
@@antrapurohit8010 Seems to be correct. Remove the ones with smallest degrees, and remove the ones with degree 1, which will lead to one new component. But might fail in some cases where the minimum edges and part of the cyclic ones are same.
@@takeUforward thank you for your reply!!
I did think of the edge case:)
can' we get solution of this tough problem
This example
1-2-3-1
4-5-6
7-8-9-10-7-9
What if k is 8?
It is better to remove the edges from square(3) instead of triangle (1)
Because 1st component and 2 nd component will endup giving only 7 but we need another to get that we have to remove 3 edges from square and totally it ends up removing 8
But if we consider square rather than triangle it ends up only removing 7 edges
I think we also should consider what is max number of components we can get by removing the edges from cycle instead only checking for minimum edges to remove
Correct me if my solution is wrong 😅
you are right
He was like the boss of interview
Striver is legend in coding community.
This solution would fail for this case
N = 7 , M = 8, K=2
Edges -> 1-2 , 2-3 , 3-4, 4-5, 5-6, 2-7, 3-7
your algo would give 3 but the right answer is 2 (last 2 edges can be removed)
mam how to avoid dark circles plz..
Was asked the same question in the interview but was given 20 minutes to solve it :D. Unfortunately couldn't make it.
MORE ON MY FEEEEEFDDD
Here is an algorithm I came up with under the assumption that the graph is simple:
Let us first decompose the graph into connected components, now we may observe that If a connected component with x nodes has more than x-1 edges , deleting a single edge will not lead to an increase in the number of components. In other words we can say that for each connected component to really contribute to our result i.e cause an increase in number of connected components it must be in the form of a tree.
So This leads us to greedily do the following:
For each connected component we first figure out the minimum number of edges to delete such that it forms a tree
Then we just keep disconnecting edges from each of the components in order of the number of edges required to make the component in form of a tree until we make k components
@takeUforward is this algorithm correct?
I thought of the same solution
I am very much convinced that greedy solution will work
Any test case in mind it might fail on?
how some people prepare their mind for such hard questions? ? I think the age old saying is true that Some are god-gifted.
Because nobody can by himself be so keen on solving hard questions with a smile on his face
This might not work , for some edge case, say you have exhausted removing all edges in the components wiith no cycle. Now in components with cycle , you might have nodes that are not part of the cycle , in that case removing one edge would give you one more component. So basically the point is, to consider all edges that are part of the cycle rather than assuming every edge in the component will be part of the cycle..
Noted! Its better to remove nodes with smallest indegrees, that will make more sense.
@@takeUforward yeah, that's more sounding...
Hard question for me..
And plus keerti g in dominating mode 🥺
Goat goat and 🐐