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MaRussiya [10]
1 year ago
14

A company is performing an analysis on the computers at its main office. The computers are spaced along a single row. The analys

is is performed in the following way: 1. Choose a contiguous segment of a certain number of computers, starting from the beginning of the row. 2. Analyze the available hard disk space on each of the computers. 3. Determine the minimum available disk space within this segment.
Computers and Technology
1 answer:
harina [27]1 year ago
5 0

Answer:

array = input("Enter the list of computer memory: ").split(',')

int_arr = [int(x) for x in array]

segment = int(input("Enter the length of segment to analyze: "))

list_len = len(int_arr)

segList = []

mini = []

for x,i in enumerate(int_arr):

   seg = int_arr[i-1:segment+i-1]

   if len(seg) == segment:

       segList.append(seg)

for x in segList:

   mini.append(min(x))

result = mini.index(min(mini))

print("Segment with the minimum memory is: ",segList[result])

Explanation:

The python program prints out the segment after analysis, with the smallest disk memory space. The code gets the array of computer rows from the user input and segments the array with the number of segment input, then the minimum is compared for all segments in the network.

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Solution :

class Employee:

   #Define the

   #constructor.

   def __$\text{init}$__($\text{self, nam}e$,  ID_number, $\text{salary}$, email):

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       $\text{self.nam}e$ = name

       $\text{self.ID}$_number = ID_number

       $\text{self.salary}$ = salary

       self.email_address = email

#Define the function

#make_employee_dict().

def make_employee_dict(list_names, list_ID, list_salary, list_email):

   #Define the dictionary

   #to store the results.

   employee_dict = {}

   #Store the length

   #of the list.

   list_len = len(list_ID)

   #Run the loop to

   #traverse the list.

   for i in range(list_len):

       #Access the lists to

       #get the required details.

       name = list_names[i]

       id_num = list_ID[i]

       salary = list_salary[i]

       email = list_email[i]

       #Define the employee

       #object and store

       #it in the dictionary.

       employee_dict[id_num] = Employee(name, id_num, salary, email)

   #Return the

   #resultant dictionary.

   return employee_dict

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