G-41. Bellman Ford Algorithm

[@takeUforward]

Let's continue the habit of commenting “understood” if you got the entire video. Please give it a like too, you don't 😞 Do follow me on Instagram: striver_79

[@KiranKumar-sb3ti]

understood

[@KUMARBISHWAJEET-iq6ym]

Understood

[@sribalajij4386]

Understood

[@bhagatalisha]

Striver you are amazing, you need to worry, your supporters will keep on increasing, infact next gen of engineers will be brought up "only by you" 😄✨

[@aditya-at_its_lowest]

understood

[@harleenkaur7751]

Thanks Striver. Trust me, even in my paid course, they just simply explained the working of Dijkstra and Bellman without going into such detail. U r the best teacher.

[@dharmvir2330]

True , i am also here from a paid course , Someone believes it or not These explanations are way better than in a paid course.

[@harshitrautela6585]

@@dharmvir2330 So why did you take the paid course? Is there any advantage you can think of that the paid course is giving better than Striver(just curious to know)?

[@EliasEH89]

Thank u. Note: The Dijkstra's algorithm implemented in G-32 can handle any negative edges graph EXCEPT the following cases: 1- Directed graph having any negative cycle (cycle with sum < 0) 2- Undirected graph having any negative edge because the edge in undirected graph is by itself a cycle

[@Mercer80]

What if graph is disconnected and negative cycle isnt reachable from source then your first point is false.

[@shravani2922]

Your thought is right, my doubt is why not follow Dijkstras algorithm implemented in G-32 for directed graphs with no negative cycles. The time complexity is even less than Bellmanford algorithm. What is the use of Bellman ford algorithm?

[@tasneemayham974]

@@Mercer80 I think we can apply a quick visited check? Like we did in the first lectures?

[@akshaysoni3496]

Improve Time Complexity by exponential with just minor observation: Put int count =0; After the first loop & increment the value of count by 1 when the dist array will get updated and at the end of the second loop, if the value of count is not increased then directly return dist array. if no update in dist array happened that means we already calculate dist array, no need to do further iteration, In the worst case it does not have any impact, but for rest, it decreases TC a lot. It Reduce the number of iteration in Best & Average cases.

[@srimanproductions8396]

this should be pinned

[@amalkumar256]

I did the same thing.

[@shivangisehgal]

exactly!

[@nalingoyal881]

Is this case even possible? in a graph like this : a-> b-> c . if we have found smaller distance for a->b , we will surely find shorter distance for b->c in next iteration. Let me know if you think differently.

[@dank7044]

@@nalingoyal881 Bhai you would find smoller distance for c in the same iteration.

[@mashfy6314]

This guy got superpower. Can be cast as a Marvel hero "The Explainer" .

[@samarthagarwal6965]

And we all guys as watchers 😂

[@noobcoder6306]

nice one

[@noobcoder6306]

@@samarthagarwal6965 already taken by the "Watcher"

[@blutoo1363]

I think Striver is already a good enough superhero name

[@amanasrani6405]

Wahi yr his teachings skills with so much of patiencee

[@mandarbhatye17]

Thanks

[@VANSHNEGGICS-2022-2]

Bhai mrko b bej dai

[@av21015]

You explained it really well, If I was to trace this myself I would have sat for an entire day.

[@tanyacharanpahadi158]

OMG! too much hype about bellman ford algorithm and this is what it is? WOW! you made it so simple. Thanks a ton striver!

[@GauravThinks]

Question: why do we need N-1 iterations? Reason: Because we first of all set the source distance out ot all the N edges, now we have N-1 edges, to fill their distances w.r.t source, we need at max N-1 iterations for each Node.

[@srinayclg390]

when u said "yes u r correct", my confidence became infinity❤

[@imajt5]

Have watched multiple videos, but got the understanding from this explaination. Thanks

[@pranshu_awasthi]

Dijsktra's code which striver has taught works for negative edges , it just not works for Negative edge cycles. So all in all , it would work for DAG with positive / negative weights.

[@tasneemayham974]

Of course, because Dijkstra doesn't work for negative edges in UNdirected grapsh: What if you have 0 - 1 with -1 weight? It will give TLE. But in directed, 0->1 with -1, 1 can never go back to 0 so the loop will not work.

[@pranshu_awasthi]

@@tasneemayham974 That's what I said. It works for DAG.

[@jagannathan1014]

Shortly, if N nodes, the node at the farthest level will be at

[@LokeshSharma-hm5jz]

understood, I dont know why i was afraid of this algo in explaining. You made it a butter.

[@decepticonWave]

The fact that you explained N-1 is why you are the GOAT. Please make a paid course and we will pay

[@ashutoshpandey1639]

But the one thing should also be mentioned that if a graph on N nodes have cycle then their is path exist having edges more than N - 1.

[@ashutoshpandey1639]

one thing should also be mentioned that if a graph on N nodes have cycle, then their is path exist having more than N - 1 edges from first to last node. By the way best explanation on KZbin👌

[@Curiousdev01]

Amazing content sir !! ..... if I get a job will be because of you.🙂

[@anonymousanonymous7507]

Lggyi

[@tasneemayham974]

SUBSCRIBED FROM FIRST RECURSION LIST VIDEO, SIRE!!!! UNDERRRSSTOOODDDD

[@dhanashreegodase4445]

thanks striver, you are the real gem

[@ashutoshkumargiri3194]

One of the best playlist of Graph on youtube bhaia you deserve more

[@arnabdebnath2425]

Bhaiya nind se uth ke breakfast mai apka video khaneka Maza hi kuch Alag h……. Thank you so much bhaiya ….❤

[@ayushsharma-bw5ch]

we need to relax each edge for n - 1 times but in the code we are running loop for

[@vishnubanjara5209]

if we assume it 0 index based then V-2 is right and if we take it 1 based than simply run it for 1 less as V-1

[@shreyanshagrawal3115]

for(int i = 0; i < N; i++) : runs for N times ( 0 to N-1) for(int i = 0; i < N-1; i++) : runs for N-1 times ( 0 to N-2) - which is required

[@newsbuster2194]

Thanks for putting such kind of effort for us.

[@HarshnitePrasad]

Things i figured out : 1. If you know a path to reach a node you can figure out the paths to reach nodes adjacent to it. 2. why repeat n-1 times, every time you run through the edges you find a path to reach the node and the cost to reach it, if its better you update, incase the cost doesnt update you have been able to exhaust the minimum cost path.

[@keshavprasad1017]

thank you so much for the clean and crisp explanation.

[@harshith4549]

We can also fit the negetive cycle check in the for loop with extending its range to V and checking if its the Vth iteration in relaxtion without writing repeated code. Also the best and worst cases can be improved by keeping a count of how many relaxations done in each iteration which signifies that if at any point no relaxation can be done then no further iteration is required bcoz there will be no change in distances further. Here's how its done: vector bellman_ford(int V, vector& edges, int S) { vector dist(V, 1e8); dist[S] = 0; for(int i = 1; i

[@naitikrathore3317]

2 din baad DAA ki paper hai and Bellman Ford aayega exam me, soch hi rha tha ki striver agr next video yhi upload kr de fir to mjaa hi aa jaye and aaj subh dekha BOOM!!

[@alessandrocamilleri1239]

Top notch explanation as usual. I would have included an update flag to pre-empt unnecessary iterations.

[@kaichang8186]

understood, thanks for the intuition part

[@amankush2408]

Understood. Great explanation for the intuition.

[@stith_pragya]

Thank You So Much for this wonderful video...........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

[@prateekshrivastava2802]

Best explaination of this algo till date !!

[@Stickman-rz9nu]

bhaiya , in the for loop terminating condition should be “ i < V “ for n-1 iterations

[@rithikraj4316]

Simple in every cycle worst can be 1 updation, so we need to update n - 1 updation as we already have 0 or src updated manually

[@120-venkataraghavareddyvar4]

I guess for the question why we need to do n-1 relaxations to each node is that ,, suppose we have 3 nodes directly connected a node and we have relaxed those nodes,, now in typical dijkstra algo,, the node with smallest value of relaxation will never get relaxed again, but if there are negative weights,, then there is a chance that the the node with smallest value of relaxation will get relaxed again. Since each node at max gets connected to n-1 nodes, so thus n-1 relaxations.

[@mdyousufgazi4030]

the thought process is insane

[@psurya3053]

thank you bhaiyya , please can you make a playlist on examples on segment trees from codeforces

[@Chirayu19]

You got so much energy, bro!

[@Kokowangmini]

understood, It was so Awesome.

[@cinime]

Understood! Super amazing explanation as always, thank you very much!!

[@coder6109]

Imp when to apply - > when have -ve cycles, idea-> relax all the edges v-1 times , tc->(O(VE))

[@VrundLeuva-g5s]

very nice explanation

[@rajatjakhar3146]

great explanation 🔥🔥🔥🔥

[@lakshyaagrawal2847]

bahi kya kar raha h

[@vakhariyajay2224]

Thank you very much. You are a genius.

[@chitranshkulshrestha485]

Thanks for the intution

[@AyushiPanday12]

Thanks Striver for these wonderful lectures. Understood.

[@gamerjain3795]

normally we say that dijkstra fails when :- 1. negative cyclic dependency 2. negative weights (not right always) as for a directed graph , if assigning negative weights to an edge , doesn't result in an negative cycle, then you are good to say , your Dijkstra algo. will work there correctly, even with negative edge weight.

[@gamerjain3795]

so, only thing to consider when choosing between dijkstra and bellman ford is , check for negative cyclic dependencies and not just negative edges. Having negative edges in an undirected graph, which will be converted into directed graph, then definitely dijkstra will fail, bcoz a negative cycle will be formed then.

[@varunakrishnani7688]

Understood!! :) Thank you! 🙏🏻😊

[@user-zn3be9ik1x]

start the relaxation loop from i=1 to i

[@CodeNow-nc4rk]

23:34 - If someone is wondering like me as to why the edges are passed by reference using & If you don't use a reference, the function would create a copy of the edges vector every time the function is called, which can be very costly in terms of memory and time, especially for large graphs

[@smile8510]

Master piece !

[@CodeMode9313]

Habibi ye ek number bideo bana di tumne toh ...baut baut danyawaad

[@dipendrapratap6495]

If we will take nodes in a increasing order then that time I think we can find the distance in 2-3 iteration as well we don't need to do n-1 iterations

[@virgarg9653]

Beautiful Explanation . Loved your content keep going 100%

[@sdfg204]

Had no idea it was this easy, damn. Obviously now that i know the logic, i don't even need to remember it.

[@YourITGuy9]

at 17.25 Edges are 5 i think u r referring to the no of nodes not edges

[@kr_ankit123]

Understood👍👍 Thanks a lot

[@rajeshj4066]

@ 16:00 : explained why it has n-1 iterations

[@UECAshutoshKumar]

Thank you sir 🙏

[@ajitpal0821]

5 nodes not edges at 17:24 @take U forward

[@kirankuyate6056]

greate explaination and with great energy while explaining that make people more creative affracting getting more..💖

[@suyashjain3223]

Amazing Explanation!!🔥🔥

[@komalkrishna7836]

Wow! very well explained, completely Understood

[@harshalgarg1676]

Amazing, thanks a lot.

[@saniyaverma914]

Understood, thank you bhaiya

[@sallaklamhayyen9876]

great explanation thank you so much and please continue😘😍

[@kasamadhu3509]

striver: Directed Acyclic Graph with negative edges we can use Dijkstra's right? Example: vertices 3, edge 3 node node weight; 0 1 10 1 2 -8 0 2 5 source 0 ; it is giving 0, 10, 2 correct, bellman ford is used for only detecting negative cycles that can't be done using dijkstra's as you explained in series.

[@akashkumarsingh8369]

same confusion hai bhai. smjh nhi aa rha agar cycle nhi hua aur negative weight hai to dijkstra glt ans dega ya shi. Dry run krne pr shi hi aa rha hai.

[@lakshsinghania]

@@akashkumarsingh8369 it shouldn't work as per the real meaning of dijkstra's whether its 1. having neg weights 2. having neg cycle becoz this already includes negative weight in the path

[@lakshsinghania]

@@akashkumarsingh8369 koi aur tc lo and then check

[@dhruvgupta3879]

it won't work with dijkstra. We mark a node as visited in dijkstra and do not visit it again, which should not be done in case of negative weights.

[@jiyabhagat6510]

Yes, Dijkstra can be used for Directed Acyclic Graph with negative weights. Since it doesn't have a negative cycle, it will not get caught in an infinite loop

[@UtpalPodder-b4r]

Presense of negative edge does not always result incorrect result ...that could be easily proved in the second example if there is a path say from 4 to 3 with -1 edge weight ....but if the negative edge cycle change the previous path length which we got after n-1 iteration then it cause problem.

[@herculean6748]

lots of love and respect🙌

[@Kparkhade]

Maja aa gaya Itna khatarnak explanation bro 🎉

[@g51661]

Thank you, Striver! Understood

[@harishnaik8164]

why adjacent list is not created? why iterate through edges list?

[@vyomyadav6497]

Good, well explained.

[@MojnoSardar]

Very well explained.

[@ankitpandey7262]

Understood !! Amazing as always

[@aryashjain7893]

watching it at 4 a.m. and when u say , I got a better guy, it really hurts :)🤣🤣

[@auroshisray9140]

Well explained bhai!

[@stark2461]

In Case of Negative weight cycles, Dijkstra don't run forever. It produces wrong answer

[@redgreenblue_rgb]

Understood!!! After striver Explanation my reaction So easy

[@Alwaysalearner123]

In for loop i

[@divyareddy7622]

takeaway: having n-1 iterations, helps bellman ford to avoid infinite loop. It won't again try to find a better distance.

[@NAYANPATELTECH]

Super explanation😀

[@krishnashah6654]

nicely explained, thanks alot

[@lokeshroy3944]

intution was just 🔥🔥🔥🔥🔥🔥🔥🔥

[@Stickman-rz9nu]

bhaiya , i want to thank you for this playlist, yesterday i attempted a codechef contest and there was a question related to graph and i solved it correctly using dijkstra, though i got a TLE 💀 but the code was correct. it was my first graph question on codechef. thank you so much 🥳🥳

[@udaytewary3809]

Understood bhaiya 🙏❤️

[@doingsneakypeakylike]

Solid explanation man! Thanks!

[@Shivanai_h]

👏 understood...very well explained..

[@vaalarivan_p]

5:28 - 22:00

[@soninirav]

Amazing , very well explained 🔥🔥

[@sanchitdeepsingh9663]

thanks, understood

[@manasranjanmahapatra3729]

Whatt an explanation amazing. understood.

[@satyasantoshkumarpadavala4230]

Thank you so much

[@bhargavreddy5938]

Well explained man ❤️

[@original_gangsta_]

Understood 🔥🔥

[@kushalkollu8628]

striver bhai is goated

[@nazimhussainmazumder4750]

understood very well!

[@juniorboy1903]

Understood😉 bhaiya

[@asktostranger8296]

Bahi abhi graph ki kitni video he Aur bachi He Big fan 😍🙏

[@cricketguru7596]

clear explanation