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Juli2301 [7.4K]
2 years ago
8

Recall that a Set is an abstract data type somewhat similar to a Bag, they can store a finite collection of objects without any

particular order. However, unlike a Bag, a Set cannot contain duplicates. The add() method for a Set is therefore very similar to the add() method for a Bag, with the additional requirement that it must first confirm that the item being added is unique. Write the member method addLikeASet that implements the following logic: When invoked the method takes the T parameter 'anEntry' and determines if 'anEntry' already exists in the bag. If the bag does not already contain 'anEntry' then the method attempts to add it to the bag, returning true when successful. If the bag already contained 'anEntry' the method does not attempt to add 'anEntry' and returns false. Your implementation code for this problem may NOT access/invoke any of the Bag API methods but, since it is a member method, your code may access the fields numberOfEntries and contents. Your solution code may also include helper methods.
Computers and Technology
1 answer:
lina2011 [118]2 years ago
3 0

Answer:

Here is the addLikeASet() method:

public boolean addLikeASet(T anEntry){  // method takes the T parameter anEntry

       boolean res = true;  //sets the result to true

       if(Arrays.asList(bag).contains(anEntry);  {  //checks if the bag already contains anEntry

           res=false;         }  //returns false

       else         {   if the bag does not already contain anEntry

           bag[numberOfEntries]=anEntry;  //enters the anEntry in bag array

           numberOfEntries++;         }  //increments the  numberOfEntries by 1

       return res;     } //returns true

Explanation:

You can also implement this method like this:

public boolean isFilled(){ // the function to check if the bag is full

       return numberOfEntries == bag.length;  } // number of entries equals length of the bag which shows that the bag is full

public boolean addLikeASet(T anEntry){

       boolean res=true;

       if (!isFilled() || (Arrays.asList(bag).contains(anEntry));  //checks if the bag is full or contains anEntry already

           res=false;         //returns false

       else         { if bag does not already contain anEntry

           bag[numberOfEntries]=anEntry; //add anEntry to the bag

           numberOfEntries++;         } add 1 to the count of numberOfEntries

       return res;     } //returns true

You can also use a loop to check if anEntry is already present in bag:

for(int i=0; i< bag.length; i++)  {

    if(anEntry.equals(bag[i])) {

      return false;      }    }  

  return true;

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Create an array w with values 0, 0.1, 0.2, . . . , 3. Write out w[:], w[:-2], w[::5], w[2:-2:6]. Convince yourself in each case
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Answer:

w = [i/10.0 for i in range(0,31,1)]

w[:]  = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3.0]

w[:-2] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8]

w[::5]= [0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0]

w[2:-2:6] = [0.2, 0.8, 1.4, 2.0, 2.6]

Explanation:

List slicing (as it is called in python programming language ) is the creation of list by defining  start, stop, and step parameters.

w = [i/10.0 for i in range(0,31,1)]

The line of code above create the list w with values 0, 0.1, 0.2, . . . , 3.

w[:]  = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3.0]

since start, stop, and step parameters are not defined, the output returns all the list elements.

w[:-2] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8]

w[:-2] since  stop is -2, the output returns all the list elements from the beginning to just before the second to the last element.

w[::5]= [0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0]

w[::5] since  step is -2, the output returns  the list elements at index 0, 5, 10, 15, 20, 25,30

w[2:-2:6] = [0.2, 0.8, 1.4, 2.0, 2.6]

the output returns the list elements from the element at index 2  to just before the second to the last element, using step size of 6.

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2 years ago
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