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solmaris [256]
3 years ago
8

Suppose that you are asked to modify the Stack class to add a new operation max() that returns the current maximum of the stack

comparable objects. Assume that pop and push operations are currently implemented using array a as follows, where item is a String and n is the size of the stack. Note: if x andy are objects of the same type, use x.compareTo(y) to compare the objects x and y public void push String item ) { [n++] = iten; } public String pop { return al--n]; } Implement the max operation in two ways, by writing a new method using array a (in 8.1), or updating push and pop methods to track max as the stack is changed (in 8.2). Q8.1 Implement method maxi 5 Points Write a method max() using Out) space and Oin) running time. public String max() {...} Enter your answer here Q8.2 Update push() and popo 5 Points Write a method max() using On) space and 011) run time. You may update the push and pop methods as needed public void push {...} public String pop() {...} public String max() {...}
Computers and Technology
1 answer:
Advocard [28]3 years ago
7 0

Answer:

Following are the code to the given points:

Explanation:

For point 8.1:

public String max()//defining a method max

{

   String maxVal=null;//defining a string variable that holds a value

   for(int x=0;x<n;x++)

   {

       if(maxVal==null || a[i].compareTo(maxVal)>0)//defining if blok to comare the value

       {

           maxVal=a[i];//holding value in maxVal variable

       }

   }

   return maxVal;//return maxVal variable value

}

For point 8.2:

public void push(String item)//defining a method push that accepts item value in a parameter

{

       a[n]=item;//defining an array to hold item value

       if(n==0 || item.compareTo(maxVals[n-1])>0)//use if to comare item value

       {

               maxVals[n]=item;//holding item value in maxVals variable

       }

       else

       {

               maxVals[n]=maxVals[n-1];//decreasing the maxVals value

       }

       n++;//incrementing n value

}

public String pop()//defining a method pop

{

       return a[--n];//use return value

}

public String max()//defining a method max

{

       return maxVals[n-1];//return max value

}

  • In the first point, the max method is declared that compares the string and returns its max value.
  • In the second point, the push, pop, and max method are declared that works with their respective names like insert, remove and find max and after that, they return its value.
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What specific job usually includes the important pre-production step of making the
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Answer:

B. assistant director

Explanation:

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Illustrate that the system is in a safe state by demonstrating an order in which the threads may complete.If a request from thre
Kamila [148]

Answer:

a. safe sequence is T2 , T3, T0, T1, T4.

b. As request(T4) = Available, so the request can be granted immediately

c. As request(T2) < Available, so the request can be granted immediately

d. As request(T3) < Available, so the request can be granted immediately.

Explanation:

It will require matrix

[i, j] = Max [i, j] – Allocation [i, j]

A B C D

T0 3 3 3 2

T1 2 1 3 0

T2 0 1 2 0

T3 2 2 2 2

T4 3 4 5 4

Available = (2 2 2 4)

1. Need(T2) < Available so, T2 can take all resources

Available = (2 2 2 4) + (2 4 1 3) (Allocation of T2) = (4 6 3 7)

2. Need(T3)<Available so, T3 will go next

Available = (4 6 3 7) + (4 1 1 0) = (8 7 4 7)

Like wise next T0, T1, T4 will get resources.

So safe sequence is T2 , T3, T0, T1, T4.

(Note, there may be more than one safe sequence).

Solution b.

Request from T4 is (2 2 2 4) and Available is (2 2 2 4)

As request(T4) = Available, so the request can be granted immediately.

Solution c.

Request from T2 is (0 1 1 0) and Available is (2 2 2 4)

As request(T2) < Available, so the request can be granted immediately.

Solution d.

Request from T3 is (2 2 1 2) and Available is (2 2 2 4)

As request(T3) < Available, so the request can be granted immediately.

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Answer:

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