G-54. Strongly Connected Components - Kosaraju's Algorithm
ฝัง
- เผยแพร่เมื่อ 20 พ.ย. 2022
- GfG-Problem Link: bit.ly/3R6LzID
C++/Java/Codes and Notes Link: takeuforward.org/graph/strong...
DP Series: • Striver's Dynamic Prog...
SDE Sheet: takeuforward.org/interviews/s...
Check out our Website for curated resources:
Our Second Channel: / @striver_79
In case you are thinking to buy courses, please check below:
Code "takeuforward" for 15% off at GFG: practice.geeksforgeeks.org/co...
Code "takeuforward" for 20% off on sys-design: get.interviewready.io?_aff=takeuforward
Crypto, I use the Wazirx app: wazirx.com/invite/xexnpc4u
Take 750 rs free Amazon Stock from me: indmoney.onelink.me/RmHC/idje...
Earn 100 rs by making a Grow Account for investing: app.groww.in/v3cO/8hu879t0
Linkedin/Instagram/Telegram: linktr.ee/takeUforward
---------------------------------------------------------------------------------------------------------------------------
![เจ้าผู้เดียว - ดิด คิตตี้ [OFFICIAL MV]](http://i.ytimg.com/vi/q0NLCYKBOTo/mqdefault.jpg)
Understood!! Thank you for this amazing explanation.
Understood! Super excellent explanation as always, thank you very much!!
I thought I was done with graphs 😅
Literally me too...😅
Your videos and the SDE sheet are helping me so much to brush up the DSA concepts and learn the ones I ignored in college. A BIG THANK YOU!!
Just amazing.
Great Explanation.
Just realised one thing. You can avoid reversing the graph if you were to put the time in a queue instead of a stack.
No you need stack cause if you use Queue it will keep going on to next SSC using the visited araay
Excellent tutorial and really helpful, thanks!!
Man, amazing explaination. Hats off to you buddy
great work sir i just watch ur few video and ur content is awesome
Finally understood kosaraju algo. Thank you striver.
Long time no see sir 😊
Thank you for posting 🥰🔥💖
Thanks for Explaining this concept :) really liked your explaination
You explain so well ❤❤
Awesome work 👏
understood, nice explanation
Going to end the graph..❤ understood
awesome explanation!
Thanks for Explaining the logic
You are a very good teacher apse sikhna easy lagta hai
Thank You So Much for this wonderful video..............🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
Honestly, this playlist deserves Millions views and comments..... Thanks for all you do Striver!!!
mazaaaaaaa aaaaaaaagyaaaaaaaaaaa
this is first time i have come across any of your videos
der aaye par durust aaye....
let me subscribe!
Understood, thanks!
Amazing content🔥
amazing content!
The above code gave me `TLE`
so, with few minor changes it will work for you.
To optimize the given code, you can make a few changes to improve its efficiency:
1. Pass the `adj` vector as a reference to the `dfs` function: Currently, the `adj` vector is passed by value, which creates a copy of the vector in each recursive call. Instead, pass it by reference to avoid unnecessary copies.
2. Use a vector of vectors (`adjT`) instead of an array of vectors for the transpose graph: Declaring `vector adjT[V]` as a variable-length array may cause a stack overflow for large values of `V`. Instead, use a vector of vectors to represent the transpose graph (`adjT`).
3. Reserve memory for the transpose graph vectors: Before pushing elements into `adjT`, reserve memory for each vector based on the size of the corresponding adjacency list in the original graph. This avoids frequent reallocations and improves performance.
4. Use vectors instead of stacks: Instead of using a stack to store the finish times of the nodes in the first DFS, you can use a vector and append nodes at the end. This eliminates the need for reversing the stack later on.
Here's the optimized code with these changes:
Code :
class Solution {
private:
void dfs(int node, vector& adj, vector& vis, vector& finishOrder) {
vis[node] = 1;
for (int it : adj[node]) {
if (!vis[it]) {
dfs(it, adj, vis, finishOrder);
}
}
finishOrder.push_back(node);
}
void dfs2(int node, vector& adj, vector& vis) {
vis[node] = 1;
for (int it : adj[node]) {
if (!vis[it]) {
dfs2(it, adj, vis);
}
}
}
public:
// Function to find the number of strongly connected components in the graph.
int kosaraju(int V, vector& adj) {
vector vis(V, 0);
vector finishOrder;
for (int i = 0; i < V; i++) {
if (!vis[i]) {
dfs(i, adj, vis, finishOrder);
}
}
// Creating the transpose graph (adjT)
vector adjT(V);
for (int i = 0; i < V; i++) {
vis[i] = 0;
for (int it : adj[i]) {
adjT[it].push_back(i);
}
}
// Last DFS using the finish order
int scc = 0;
for (int i = V - 1; i >= 0; i--) {
int node = finishOrder[i];
if (!vis[node]) {
scc++;
dfs2(node, adjT, vis);
}
}
return scc;
}
};
These optimizations should improve the efficiency of the code.
Thanks a lot man!!
yeah that improves the code a lot!
Thank you so much
I think after seeing the video if you read the striver sheet explanation for the same question it will give you more insight since it is really well written. I really liked it!
Can you provide me link of that..
It is in the description
awsm explanation thank u so much
Awesome thanks a lot
And here comes the real OG Striver.❤
Understood Sir!
amazing content
🔥🔥🔥🔥
Explanation
Thank you sir 🙏
Understood ❤
Thanks buddy
crystal clear
Understood after a little bit of effort.
thanks for the video
thank you very muchhh! You've literally changed so many lives ✌️
the first step is same as topo sort using dfs technique, will not work with bfs (khann's algo) due to directed cyclic graphs.but if you apply topo sort using dfs in directed cyclic graphs ,it will work.
Thanks🙌
This is much intutive than the previous video on kosaraju(other graph playlist)
Yes I read the comments, and then made this one.
@@takeUforward you have changed the lives of so many students, you have blessings of many. ♥️
@@takeUforward when we sort all the edges according to finishing time and store it in stack like this 0,1,2,3,4,5,6,7 , so instead of storing in stack can we use queue and then when we pop out elements from queue we start DFS from 7 instead of 0 and 7 is not connected to anyone so we found 1st SCC then we pop 6 and do DFS and found 2nd SCC then we pop 3 and do DFS and got 3rd SCC and so on. In this approach we don't need to reverse the graph. Can we do like this?
@@tanmaypal2003 Nice observation. I think this should work fine.
@@tanmaypal2003 yes
Understood 👍
maza aaala re ala striver aala
Understood❤
Big fan of your work striver
amazing..
Understooddd!!
reach++
Loved it
Understood!
Understood.
understood sir
Understood!!
mast padha dia bahiya
understood
understooood
guys just in case any one wondering why that stack of dfs calls was required that is only if u r solving for the second case of actually printing the components and not just finding number of components.
like based on the stack trace u can make a note of elements while popping and put them into groups and print
without using stack, how to do solve the first case of finding no. of compo? I mean to find no. of components also we need it in sorted way right?
Note: we can't say this as a Topological sort(as topological sort using DFS will get you stuck in Loop i.e cycle), but its a bit different. Its topological Sort for the SCC.
Understood
UNDERSTOOD
We can use the same dfs again by passing a dummy stack:
public void dfs(ArrayList adj, int[] vis, Stack st, int node){
vis[node] = 1;
for(int it : adj.get(node)){
if(vis[it] == 0){
dfs(adj, vis, st, it);
}
}
st.push(node);
}
//Function to find number of strongly connected components in the graph.
public int kosaraju(int V, ArrayList adj)
{
int[] vis = new int[V];
Stack st = new Stack();
for(int i = 0; i < V; i++){
if(vis[i] == 0){
dfs(adj, vis, st, i);
}
}
ArrayList adjT = new ArrayList();
for(int i = 0; i < V; i++) adjT.add(new ArrayList());
for(int i = 0; i < V; i++){
vis[i] = 0;
for(int it : adj.get(i)){
// previously: i -> it
// Now make it: i i
adjT.get(it).add(i);
}
}
int scc = 0;
while(!st.isEmpty()){
int node = st.pop();
if(vis[node] == 0){
scc++;
dfs(adjT, vis, new Stack(), node);
}
}
return scc;
}
great
HEY STRIVER, We can also reduce the time and steps if first getting there there finishing time and store in a queue rather then stack so because of FIFO we know first element to finish so we don't need to reverse the graph either we can directly go through the finishing elements which are already reverse with respect to graph it will reduce steps and clean code too.
correct me if I m wrong
please make videos on Eulerian paths also.
sorting according to finish time can be done using toposort:)
that is toposort only. Toposort can be done using stack and that's what it is here.
@abcsumit Toposort banega hi nahi, wo sirf DAG mein work karta hai.
@@deviprasad_bal Toposort banega hi nahi, wo sirf DAG mein work karta hai.
@@DheerajDivaker Yes you are right but here since graph can be cyclic as well as acyclic , so incase of CYCLIC one edge which is causing cycle is not stored , like 1->2->3->1 , is only stored as [ 1, 2 ,3 ] .
So ultimately we should not call it as a toposort ,but implementation is exactly same.
@@Anonymous_Coder yes absolutely correct.
understood💛💙💛
Is there any relation in this algo and topo sort as we get reverse topo from stack
understood 🥶
nice
hey striver can we do bfs to sort the graph according to their finishing time ? is it possible ?
ultimate
class Solution
{
private:
void dfs(int node,vectoradj[],vector&vis,stack&st) // dfs for adj
{
vis[node]=1;
for(auto it:adj[node])
{
if(vis[it]==0)
{
dfs(it,adj,vis,st);
}
}
st.push(node);
}
void dfs2(int node,vectoradjT[],vector&vis) // dfs for adjT
{
vis[node]=1;
for(auto it:adjT[node])
{
if(vis[it]==0)
{
dfs2(it,adjT,vis);
}
}
}
public:
//Function to find number of strongly connected components in the graph.
int kosaraju(int V, vector adj[])
{
vectorvis(V,0);
stackst;
vector adjT[V]; // adj list after reversing edges
for(int i=0;i
GFG compiler giving TLE, don't know why?? :(
thought process is very tricky
Sir is sorting edges according to finishing time same as topological sort?
Will topological sort work in cyclic graphs?
How do I pick the starting node for sorting.
In the above example if we would have started sorting from Node 3, I would have never reached 0, 1, 2. So how to calculate this increasing time for all nodes efficiently.
Instate on transposing the graph, can't we just start from the last node?
someone please explain....
In given example if we just make a little change with 3->2 then after reversing the edge will be 2->3 so in step 3 dfs call 2 will also go to 3 . I feel like it should not go to 3
i have the same confusion that if we start with 7 in the start rather than 0 , we will get wrong connected components
Understood brother. Could you please upload videos on sliding window and 2 pointers?? I saw most amazon OA will be based on that only.
Already there in the channel bro
wow
16:00
choit💥💥💥
19:55 Why did you mention to write as private function. Is there any advantage of it over public. As I tested in run time and got public working faster than private. Also from outer source noone is calling that dfs function. So kindly elaborate please
Why is it important to reverse the edges, can't we just start the dfs in reverse order ?
Like in the example start from 7
can any1 please explain
we can't .try dryrun on 1st example ,
suppose u created vector which store the timing then timing will be
{1 2 4 3 0} if you try dfs accoording to this you still cannot find scc.
Yes, you can do that. But, this video is about Kosaraju's algorithm, which is implemented this way.
In the first step.sorting all the edges ...looks like topological sorting..... Is it the same I am thinking....
at 13:00 while traversing at scc1 from the node 0 using dfs you have taken the edges in the already present way(or say initial way) but since we have reversed the edges it should move in another way and you are also using the fact of of the edges being reversed at 13:05 when you are trying to access 3 from the dfs(2) you have said that . Then again same thing when doing same thing for dfs(3) we are not using the fact that edges are reversed. Can anyone here clear my doubt that where I am understanding it wrong?
I get ur point , Striver used scc1->scc2->scc3->scc4 , Iniital config something like scc1scc3->scc4 can be used for better understanding
Hey Striver.. I had a question.. If we store the nodes according to their finishing time in a queue and rather than reversing all the nodes can't we simply do tranversal from the nodes in the queue in the similar fashion as you did.
Can't we use queue and then not reversing the edges and just count no of scc of back, but it's not working idk
edges ko reverse karne se SCCs separate ho rahe hai..
👏👏
Isn't the first step same as topological sort?
♥️♥️✌️
🎉🎉
Why sorting is needed. Can't be simply reverse the edges and then do dfs on all the nodes. We can get group of nodes which are there in cycle which is ultimately our strongly connected components.
By doing that you've completely ignored the "Finish time" concept. While doing the dfs in the transpose graph( with reverse edges) you should start with the component which does not have an edge going to the other scc.. that's why we use a stack and pop the top most element to perform dfs. If u want, I'll post my python code
I am confused about the first step of arranging nodes on their finishing times. I thought that instead just doing normal DFS on reversed graph skipping the first step might also work but it doesnt, Can anyone explain this with an example
if you reversed the graph and then perform the dfs how will you be able to reach form[0,1,2] --> [3] so to tackle this we store them in a stack to remember their order form start to finish .
can't we just use visited array to count the scc in reversed graph , similar to counting different components
starting vertex kon hoga, fir ye kaise pta lagaoge? jiska finishing time sabse jyada hoga wahi starting point hoga aur iske liye hume stack mein Finishing time ke according vertices rakhne padenge.
Can we skip the step 1, and just reverse all the edges ? Because after reversing edges, components will be logically separated and we can run DFS on it separately, and can find the nodes in each component.
Please can anyone guide me on this ?
Ohh, got it. It's for the efficiency
for eg, in the same example as striver's, if you reverse the edge between 2 & 3, with your process, when the graph is reversed, the dfs(0) has 3 as well in it, this is the loop hole. Whereas, in the algo, when we do dfs first and calculate finishing times, we get 3 at the top of the stack, so now we can make different dfs calls for each scc. Hope you understand. Have a great day!
@@Jinxed192 Thanks for explaination.
That means we must know from where to start, to void futher conflicts.
This code is giving tle in gfg, the below code is same as what you thought
class Solution
{
private:
void dfs(int node, vector &vis,vector adj,stack &st)
{
vis[node]=1;
for(auto it: adj[node])
{
if(vis[it]==0)
{
dfs(it,vis,adj,st);
}
}
st.push(node);
}
void dfs1(int node, vector &vis, vector adjT[])
{
vis[node]=1;
for(auto it: adjT[node])
{
if(vis[it]==0)
{
dfs1(it,vis,adjT);
}
}
}
public:
//Function to find number of strongly connected components in the graph.
int kosaraju(int V, vector& adj)
{
//code here
vector vis(V,0);
stack st;
for(int i=0;i
To resolve this issue, make sure that adjT is declared as a vector of vectors
vectoradjT(V);
isn't the timestamp same as toposort ? or there is any difference
Toposort possible in DAG only
bhaiya can,t we solve it by reversing the stack instead of reversing the graph
stack reversal will reverse the finish time but reversing the graph does not allow us to go from one scc to another scc