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Otrada [13]
3 years ago
8

Consider two vectors that are NOT sorted, each containing n comparable items. How long would it take to display all items (in an

y order) which appear in either the first or second vector, but not in both, if you are only allowed LaTeX: \Theta\left(1\right)Θ ( 1 ) additional memory? Give the worst-case time complexity of the most efficient algorithm.
Computers and Technology
1 answer:
Darina [25.2K]3 years ago
7 0

Answer:

The correct answer to the following question will be "Θ(​n​2​)

". The further explanation is given below.

Explanation:

If we're to show all the objects that exist from either the first as well as the second vector, though not all of them, so we'll have to cycle around the first vector, so we'll have to match all the objects with the second one.

So,

This one takes:

= O(n^2)

And then the same manner compared again first with the second one, this takes.

= O(n^2)

Therefore the total complexity,

= Θ(​n​2​)

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Answer:

/*********************************************************************************

* (The ComparableCircle class) Define a class named ComparableCircle that        *

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*********************************************************************************/

public class Exercise_13_06 {

/** Main method */

public static void main(String[] args) {

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 // Find and display the larger of the two ComparableCircle objects

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}

3 0
3 years ago
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The authentication mechanism start by user who sends a message to the RADIUS server. Then the RADIUS server upon receiving the message accept or denies it. It accepts if the client is configured to the server.

A large amount of additional information can be sent by the RADIUS server in its Access-Accept messages with users so we can say that RADIUS is uitable for what are called "high-volume service control applications" such as dial-in access to a corporate network.

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valentina_108 [34]

Answer:

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Explanation:

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