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Ede4ka [16]
3 years ago
14

Suppose you are implementing the address book feature for a cellphone. The address book needs to be kept sorted by person’s last

name and support fast access when queried by last name. Which of the following data structures would be a good choice to use for storing the address book? Justify your answer for both of the following data structures?
Computers and Technology
1 answer:
NemiM [27]3 years ago
8 0

Answer:

Sorted linked list

Explanation:

  • A sorted linked list can be used for sorting the last name and support fast access while queried by the last name.
  • The hash table cannot be used because it could have a problem if there would be more buckets to handle than the expected.
  • The binary search table is implemented as the hash table hence both will give the same results.
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CRM software programs are designed to help companies gather all customer contact information into multiple data management progr
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4 years ago
Five Star Retro Video rents VHS tapes and DVDs to the same connoisseurs who like to buy LP record albums. The store rents new vi
77julia77 [94]

Answer:

here is code in C++.

#include <bits/stdc++.h>

using namespace std;

int main()

{

   // variables to store number of each video

   int new_video,old_video;

   double total_charge;

   cout<<"Please enter number of new video:";

   // read number of new video

   cin>>new_video;

   cout<<"Please enter number of old video:";

   // read number of old video

   cin>>old_video;

   // total change

   total_charge=(3.0*new_video)+(2.0*old_video);

   cout<<"total cost is: "<<total_charge<<endl;

return 0;

}

Explanation:

Declare two variables "new_video" and "old_video" to store the number of each type of video.Read the value of both from user.Then calculate total charge by multiply 3.0 to number of new video & 2.0 to number of old video and add both of them. This will be the total charge.

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Please enter number of old video:6

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Answer:

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Answer: print("THIRD")

for i in range(10):

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Explanation:

You're Welcome yall

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for i in range(10):

   for j in range(10-i):

       print(" ", end=" ")

   for j in range(i):

       print("*", end=" ")

   print("")

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