Reconstruct itinerary || Leetcode

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[@Beingyash7]

what a command over graph ! 👑

[@anulrajeev7891]

++🔥🔥

[@devraj-2310]

Best explaination for this problem.

[@LegitGamer2345]

oh so we are making visited of edges! thank you so much sir crystal clear hogaya

[@ADNANAHMED-eo5xx]

Thanks... gr8 explanation

[@heisenberg1844]

Well explained. Thanks.

[@SagarKumar-rn4cq]

Bhaiyo topological sort,mother vertex,kosaraju algo and yee sab connected Hai

[@vaibhavis1]

It will fail a case, where the lexographically smallest result is not valid, i.e. doesn't visit all the edges. Example: [["JFK","KUL"],["JFK","NRT"],["NRT","JFK"]]
So in this case we will have to go ahead with Linked List instead of prioirty Queue, maintain a overall edges, and number of edge removed count,
and add back the string to LL if they don't match.

[@MGtvMusic]

It will work!

[@MGtvMusic]

It is working!!

[@vishalgaurav4411]

yes. but how his code s working?

[@MGtvMusic]

@@vishalgaurav4411 Just do a dry run ,It covers this case too

[@kumarankit5726]

@@vishalgaurav4411 see with this approach kul will be added first in last position because no dfs call can be done from kul as there is no ticket from kul.And this is what we need .We need the location that will be visited at last in the last position and the place from where no other places can be visited shall be last.

[@rupdeepthey]

King.

[@geetanjaligoyal7291]

[["JFK","KUL"],["JFK","NRT"],["NRT","JFK"]] ye wala case kaise handle ho rha hai? can you pls walkthro' this once?

[@vishalgaurav4411]

i am also stuck in same. you got the answer?

[@raunaknarayan4092]

Time complexity tow bataye heen nhi

[@mohammadkaif8143]

ELogE

[@devir9674]

Today loudly bol lijiye bhai pls