All categories

3 years ago

Time complexity of merge sort

Engineering

1 answer:

vovangra [49]3 years ago

6 0

Answer:

The correct answer is " $O (n\times Log n)$ ". A further explanation is given below.

Explanation:

Throughout all the three instances (worst, average as well as best), the time complexity including its Merge sort seems to be $O (n\times Log n)$ as the merge form often splits the array into two halves together tends to linear time to combine multiple halves.
As an unsorted array, it needs an equivalent amount of unnecessary capacity. Therefore, large unsorted arrays are not appropriate for having to search.

You might be interested in

You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you

telo118 [61]

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l = 35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L = 0.00841549 / 1.4

L = 6 × 10⁻³ H

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV

Therefore, the emf generated in the coil is 18 mV

7 0

2 years ago

Who is web developer?????<br>plez answer..

vredina [299]

Web developers design and create websites. They are responsible for the look of the site. They are also responsible for the site's technical aspects, such as its performance and capacity, which are measures of a website's speed and how much traffic the site can handle. In addition, web developers may create content for the site

8 0

3 years ago

Read 2 more answers

For this question you must write a java class called Rectangle and a client class called RectangleClient. The partial Rectangle

Alex Ar [27]

Answer:

Java program is given below. You can get .class after you execute java programs, You can attach those files along with .java classes given , Those .class files are generated ones.

Explanation:

//Rectangle.java class

public class Rectangle {

private int x;

private int y;

private int width;

private int height;

// constructs a new Rectangle with the given x,y, width, and height

public Rectangle(int x, int y, int w, int h)

{

this.x=x;

this.y=y;

this.width=w;

this.height=h;

}

// returns the fields' values

public int getX()

{

return x;

}

public int getY()

{

return y;

}

public int getWidth()

{

return width;

}

public int getHeight()

{

return height;

}

// returns a string such as “Coordinate is (5,12) and dimension is 4x8” where 4 is width and 8 is height. public String toString()

public String toString()

{

String str="";

//add x coordidate , y-coordinate , width, height and area to str and return

str+="Coordinate is ("+x+","+y+")";

str+=" and dimension is : "+width+"x"+height;

str+=" Area is "+(width*height);

return str;

}

public void changeSize(int w,int h)

{

width=w;

height=h;

}

======================

//main.java

class Main {

public static void main(String[] args) {

//System.out.println("Hello world!");

//create an object of class Rectangle

Rectangle rect=new Rectangle(5,12,4,8);

//print info of rect using toString method

System.out.println(rect.toString());

//chamge width and height

rect.changeSize(3,10);

//print info of rect using toString method

System.out.println(rect.toString());

}

==========================================================================================

//Output

Coordinate is (5,12) and dimension is : 4x8 Area is 32

Coordinate is (5,12) and dimension is : 3x10 Area is 30

========================================================================================

6 0

3 years ago

Hello, I want to introduce you to our hosting - VPSDOM

Solnce55 [7]

Answer:

pogchamp

Explanation:

sussy balls

4 0

2 years ago

I. The time till failure of an electronic component has an Exponential distribution and it is known that 10% of components have

drek231 [11]

Answer:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(d) 2.63 × 10⁻⁵

Explanation:

(a) We put time to fail = t

∴ For an exponential distribution, we have f(t) = $\lambda e^{-\lambda t}$

Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

$P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1$

e^(1000·λ) - 0.1·e^(1000·λ) = 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

b) Here we have to integrate from 5000 to ∞ as follows;

$p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [ -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905$

$P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!} = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!} = 3.915\times 10^{-12}$

p(x = 3) = 3.915 × 10⁻¹²

d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours

The Cumulative Distribution Function is given as follows;

p( t ≤ 1/4) $CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}$ .

4 0

3 years ago