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Katena32 [7]
2 years ago
13

This question involves the creation of user names for an online system. A user name is created based on a user’s first and last

names. A new user name cannot be a duplicate of a user name already assigned. You will write the constructor and one method of the UserName class. A partial declaration of the UserName class is shown below.
public class UserName

{

// The list of possible user names, based on a user’s first and last names and initialized by the constructor.

private ArrayList possibleNames;



/** Constructs a UserName object as described in part (a).

* Precondition: firstName and lastName have length greater than 0

* and contain only uppercase and lowercase letters.

*/

public UserName(String firstName, String lastName)

{ /* to be implemented in part (a) */ }



/** Returns true if arr contains name, and false otherwise. */

public boolean isUsed(String name, String[] arr)

{ /* implementation not shown */ }



/** Removes strings from possibleNames that are found in usedNames as described in part (b).

*/

public void setAvailableUserNames(String[] usedNames)

{ /* to be implemented in part (b) */ }

}

(a) Write the constructor for the UserName class. The constructor initializes and fills possibleNames with possible user names based on the firstName and lastName parameters. The possible user names are obtained by concatenating lastName with different substrings of firstName. The substrings begin with the first character of firstName and the lengths of the substrings take on all values from 1 to the length of firstName.

The following example shows the contents of possibleNames after a UserName object has been instantiated.

Example

UserName person = new UserName("john", "smith");

After the code segment has been executed, the possibleNames instance variable of person will contain the following String objects in some order.

"smithj", "smithjo", "smithjoh", "smithjohn"

Write the UserName constructor below.

/** Constructs a UserName object as described in part (a).

* Precondition: firstName and lastName have length greater than 0

* and contain only uppercase and lowercase letters.

*/

public UserName(String firstName, String lastName)

Write the UserName method setAvailableUserNames. The method removes from possibleNames all names that are found in usedNames. These represent user names that have already been assigned in the online system and are therefore unavailable.

A helper method, isUsed, has been provided. The isUsed method searches for name in arr. The method returns true if an exact match is found and returns false otherwise.

Assume that the constructor works as specified, regardless of what you wrote in part (a). You must use isUsed appropriately to receive full credit.

Complete the setAvailableUserNames method below.

/** Removes strings from possibleNames that are found in usedNames as described in part (b).

*/

public void setAvailableUserNames(String[] usedNames)
Computers and Technology
1 answer:
Evgen [1.6K]2 years ago
8 0

Answer:

See explaination

Explanation:

import java.util.*;

class UserName{

ArrayList<String> possibleNames;

UserName(String firstName, String lastName){

if(this.isValidName(firstName) && this.isValidName(lastName)){

possibleNames = new ArrayList<String>();

for(int i=1;i<firstName.length()+1;i++){

possibleNames.add(lastName+firstName.substring(0,i));

}

}else{

System.out.println("firstName and lastName must contain letters only.");

}

}

public boolean isUsed(String name, String[] arr){

for(int i=0;i<arr.length;i++){

if(name.equals(arr[i]))

return true;

}

return false;

}

public void setAvailableUserNames(String[] usedNames){

String[] names = new String[this.possibleNames.size()];

names = this.possibleNames.toArray(names);

for(int i=0;i<usedNames.length;i++){

if(isUsed(usedNames[i],names)){

int index = this.possibleNames.indexOf(usedNames[i]);

this.possibleNames.remove(index);

names = new String[this.possibleNames.size()];

names = this.possibleNames.toArray(names);

}

}

}

public boolean isValidName(String str){

if(str.length()==0) return false;

for(int i=0;i<str.length();i++){

if(str.charAt(i)<'a'||str.charAt(i)>'z' && (str.charAt(i)<'A' || str.charAt(i)>'Z'))

return false;

}

return true;

}

public static void main(String[] args) {

UserName person1 = new UserName("john","smith");

System.out.println(person1.possibleNames);

String[] used = {"harta","hartm","harty"};

UserName person2 = new UserName("mary","hart");

System.out.println("possibleNames before removing: "+person2.possibleNames);

person2.setAvailableUserNames(used);

System.out.println("possibleNames after removing: "+person2.possibleNames);

}

}

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Referential integrity constraints are concerned with checking INSERT and UPDATE operations that affect the parent child relation
Jet001 [13]

Answer:

a) true

Explanation:

In Computer programming, integrity constraints can be defined as a set of standard rules that ensures quality information and database are maintained.

Basically, there are four (4) types of integrity constraints and these are;

1. Key constraints.

2. Domain constraints.

3. Entity integrity constraints.

4. Referential integrity constraints.

Referential integrity is a property of data which states that each foreign key value must match a primary key value in another relation or the foreign key value must be null.

For instance, when a foreign key in Table A points to the primary key of Table B, according to the referential integrity constraints, all the value of the foreign key in Table A must be null or match the primary key in Table B.

Hence, the referential Integrity constraints ensures that the relationship between the data in a table is consistent and valid.

Hence, referential integrity constraints are concerned with checking INSERT and UPDATE operations that affect the parent child relationships.

<em>This ultimately implies that, referential Integrity are rules used in database management systems (DBMS) to ensure relationships between tables when records are changed is VALID (INSERT and UPDATE).</em>

<em>In a nutshell, it always ensures a primary key must have a matching foreign key or it becomes null. </em>

6 0
2 years ago
a.Write a Python function sum_1k(M) that returns the sum푠푠= ∑1푘푘푀푀푘푘=1b.Compute s for M = 3 by hand and write
stealth61 [152]

Answer:

def sum_1k(M):

   s = 0

   for k in range(1, M+1):

       s = s + 1.0/k

   return s

def test_sum_1k():

   expected_value = 1.0+1.0/2+1.0/3

   computed_value = sum_1k(3)

   if expected_value == computed_value:

       print("Test is successful")

   else:

       print("Test is NOT successful")

test_sum_1k()

Explanation:

It seems the hidden part is a summation (sigma) notation that goes from 1 to M with 1/k.

- Inside the <em>sum_1k(M)</em>, iterate from 1 to M and calculate-return the sum of the expression.

- Inside the <em>test_sum_1k(),</em> calculate the <em>expected_value,</em> refers to the value that is calculated by hand and <em>computed_value,</em> refers to the value that is the result of the <em>sum_1k(3). </em>Then, compare the values and print the appropriate message

- Call the <em>test_sum_1k()</em> to see the result

8 0
2 years ago
A government agency is getting rid of older workstations. The agency will donate these workstations, along with other excess com
Basile [38]

Answer:

The answer is "Using the DoD 5220.22-M method and  Degauss media with a magnet".

Explanation:

The "DoD Standard" is a term used during the data sanitizing industry and refers to DoD 5220.22-M. The simplest ways are being used to help eliminate the previously stored data, by deleting hard disc storage facilities with the same data wherever that used a sequence of all zeros.

The sparging eliminates statistics to entirely delete the gravitational flux from electronic media. Hard drives as well as other data storage devices, for example, computer tapes, retain magnetic data. It could no longer be seen as storage after a disk is degaussed.

6 0
2 years ago
The problem solving process begins by first ______ the problem.
alexandr402 [8]

Answer:

identifying/acknowledging

Explanation:

7 0
2 years ago
In a Diffie-Hellman Key Exchange, Alice and Bob have chosen prime value q = 19 and primitive root a = 10. If Alice's secret key
34kurt

Answer:

a) 6

b) 11

c) 11

Explanation:

Given

Prime value q =n= 19

primitive root (a) =10

Alice secret key = 4

Bob secret key = 6

a) The value Alice Sends to Bob

a​​​​​Private key of Alice mod n

= 10^4 mod 19

= 10000 mod 19

=6

b)The value Bob sends to Alice

a​​​​​Private key of Bob mod n

= a​​​​​​3​​​​​ mod n

= 10^6 mod 19

=1000000 mod 19

= 11

c)

Both the parties calculate the value of secret key at their respective side.

secret key obtained by Alice

= 11

secret key obtained by Bob

= 11

Finally, both the parties obtain the same value of secret key.

The value of common secret key = 17

6 0
2 years ago
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