Answer:
Question is incomplete.
Assuming the below info to complete the question
You have a collection of n lockboxes and m gold keys. Each key unlocks at most one box. Without a matching key, the only way to open a box is to smash it with a hammer. Your baby brother has locked all your keys inside the boxes! Luckily, you know which keys (if any) are inside each box.
Detailed answer is written in explanation field.
Explanation:
We have to find the reachability using the directed graph G = (V, E)
In this V are boxes are considered to be non empty and it may contain key.
Edges E will have keys .
G will have directed edge b1b2 if in-case box b1 will have key to box b2 and box b1 contains one key in it.
Suppose if a key opens empty box or doesn’t contain useful key means can’t open anything , then it doesn’t belongs to any edge.
Now, If baby brother has chosen box B, then we have to estimate for other boxes reachability from B in Graph G.
If and only if all other boxes have directed path from box B then just by smashing box B we can get the key to box b1 till last box and we can unlock those.
After first search from B we can start marking all other vertex of graph G.
So algorithm will be O ( V +E ) = O (n+m) time.
I’m pretty sure it’s either A or D
Answer: True
Explanation:
TCP is the transmission control protocol which and TCP ACK is the transmission control handshake method which are used by TCP. It basically indicates the next sequence number in the method when the flag is set.
Firstly, the ACK (Acknowledge) send by the each end for the initial sequence number itself but it does not contained any type of data. In the TCP segment header ACK contain 32 bit field.
The acknowledgement is just a proof to clients that ACK is a specific to the SYN when the clients initiate.
Explanation:
The option to delete your account can be found in your Profile Settings under Privacy. Click on the box labeled I want to delete my account, and the request will be sent for the account to be deleted.
