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vladimir2022 [97]
2 years ago
8

Algorithmic Complexity: what is the asymptotic complexity (Big-O) of each code section? Identify the critical section of each.\

Line 1: for (int i=0; i<532; i++) { f(n) = O( ) Line 2: for (int j=1; j
Computers and Technology
1 answer:
AleksandrR [38]2 years ago
4 0

Answer:

Check the explanation

Explanation:

1) f(n) = O( 1 ), since the loops runs a constant number of times independent of any input size

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

2) f(n) = O( log n! ), the outer loop runs for n times, and the inner loop runs log k times when i = k,ie the total number of print will be – log 1 + log2 +log3 +log4+…...+ log n = log (1 . 2 . 3 . 4 . ……. . n ) =log n!

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

Note : Log (m *n) = Log m + Log n : this is property of logarithm

3) f(n) = O( n^2 ), since both outer and inner loop runs n times hence , the total iterations of print statement will be : n +n+n+…+n

for n times, this makes the complexity – n * n = n2

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

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c++ You are given an array A representing heights of students. All the students are asked to stand in rows. The students arrive
Lilit [14]

The below code will help you to solve the given problem and you can execute and cross verify with sample input and output.

#include<stdio.h>

#include<string.h>

 int* uniqueValue(int input1,int input2[])

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   int left, current;

   static int arr[4] = {0};

   int i      = 0;

     for(i=0;i<input1;i++)

      {

         current = input2[i];

         left    = 0;

         if(current > 0)

         left    = arr[(current-1)];

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       {

       arr[current] = input1-current;

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       else

   {

       for(int j=(i+1);j<input1;j++)

       {

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4 0
2 years ago
Assume there is a class AirConditioner that supports the following behaviors: turning the air conditioner on and off. The follow
Dafna11 [192]

Answer:

1)

officeAC = new AirConditioner();

officeAC.turnOn();

2)

officeAC = new AirConditioner();

officeAC.turnOn();

officeAC.setTemp(69);  

Explanation:

1)

In the first statement a new object of the class AirConditioner whose reference is assigned to the officeAC. new is a keyword which creates an object of the class.

Next statement uses the method turnOn(). Reference to the new object officeAC is used to invoke this method turnOn().

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6 0
2 years ago
Constantine forms the following hypothesis. Let n be any non-negative number that meets the following condition: when n is divid
notka56 [123]

Answer:

n=5

Explanation:

if n= 5

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8 0
2 years ago
Create a single list that contains the following collection of data in the order provided:
ivann1987 [24]

Answer:

A code was created to single list that contains the following collection of data in the order provided

Explanation:

Solution

#list that stores employee numbers

employeee_numbers=[1121,1122,1127,1132,1137,1138,1152,1157]

#list that stores employee names

employee_name=["Jackie Grainger","Jignesh Thrakkar","Dion Green","Jacob Gerber","Sarah Sanderson","Brandon Heck","David Toma","Charles King"]

#list that stores wages per hour employee

hourly_rate=[22.22,25.25,28.75,24.32,23.45,25.84,22.65,23.75]

#list that stores salary employee

company_raises=[]

max_value=0

#list used to store total_hourly_rate

total_hourly_rate=[]

underpaid_salaries=[]

#loop used to musltiply values in hourly_wages with 1.3

#and store in new list called total_hourly_rate

#len() is used to get length of list

for i in range(0,len(hourly_rate)):

#append value to the list total_hourly_rate

total_hourly_rate.append(hourly_rate[i]*1.3)

for i in range(0,len(total_hourly_rate)):

#finds the maximum value

if(total_hourly_rate[i]>max_value):

max_value=total_hourly_rate[i] #stores the maximum value

if(max_value>37.30):

print("\nSomeone's salary may be a budget concern.\n")

for i in range(0,len(total_hourly_rate)):

#check anyone's total_hourly_rate is between 28.15 and 30.65

if(total_hourly_rate[i]>=28.15 and total_hourly_rate[i]<=28.15):

underpaid_salaries.append(total_hourly_rate[i])

for i in range(0,len(hourly_rate)):

#check anyone's hourly_rate is between 22 and 24

if(hourly_rate[i]>22 and hourly_rate[i]<24):

#adding 5%

hourly_rate[i]+=((hourly_rate[i]/100)*5)

elif(hourly_rate[i]>24 and hourly_rate[i]<26):

#adding 4%

hourly_rate[i]+=((hourly_rate[i]/100)*4)

elif(hourly_rate[i]>26 and hourly_rate[i]<28):

#adding 3%

hourly_rate[i]+=((hourly_rate[i]/100)*3)

else:

#adding 2% for all other salaries

hourly_rate[i]+=((hourly_rate[i]/100)*2)

#adding all raised value to company_raises

company_raises.extend(hourly_rate)

#The comparison in single line

for i in range(0,len(hourly_rate)):

 For this exercise i used ternary

operator hourly_rate[i]= (hourly_rate[i]+((hourly_rate[i]/100)*5)) if (hourly_rate[i]>22 and hourly_rate[i]<24) else (hourly_rate[i]+((hourly_rate[i]/100)*4)) if (hourly_rate[i]>24 and hourly_rate[i]<26) else hourly_rate[i]+((hourly_rate[i]/100)*2)

8 0
2 years ago
Here is a super challenge for you if you feel up for it. (You will need to work this out in Excel.) In 2017 the Islamic month of
Ann [662]

Answer:

7/73

Explanation:

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4 0
2 years ago
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