Answer:
a. Radom Access Memory (RAM).
Explanation:
If a computer user modifies a document that is saved on his or her computer. This changes are stored on the Radom Access Memory (RAM) until the computer user save the document.
Radom Access Memory (RAM) can be defined as the main memory of a computer system which allow users to store commands and data temporarily.
Generally, the Radom Access Memory (RAM) is a volatile memory and as such can only retain data temporarily.
All software applications temporarily stores and retrieves data from a Radom Access Memory (RAM) in computer, this is to ensure that informations are quickly accessible, therefore it supports read and write of files.
Answer:
Using a one time password OTP and using a three factor authentication.
Explanation:
Social engineering is a form system and network attack, drafted by an attacker, to steal user credentials needed for accessing their accounts, to steal information.
Social engineering attacks like phishing, spear phishing, baiting and quid quo pro are used to fool users to giving out their user details.
One time password is an implied use of a password just once and a new password his generated to boost security. In a three factor authentication, OTP can be used as well as biometrics of a user which can not be giving away by a user to an attacker.
I’m pretty sure it’s a network hope this helps
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.
