Question

6. In an array-based chain implementation of a Stack ADT, what is the performance of the ensureCapacity method when the array is full?

1. O(1)
2. O(n)
3. O(n log n)
4. O(n2)
5. Undefined

7. In an array-based chain implementation of a Stack ADT, what is the performance of the ensureCapacity method when the array is not full?

1. O(1)
2. O(n)
3. O(n log n)
4. O(n2)
5. Undefined

8. The efficiency for recursively traversing a chain of linked nodes is

1. O(1)
2. O(n)
3. O(n2)
4. it cannot be proven

9.

What is the output of the following program when the method is called with 3?

void unknown(int n)

   {

System.out.print("?");

if (n > 0)

   unknown(n-1);

   }

1. ???
2. ????
3. ?????
4. none of the above

10.

How many recursive calls will be made if the following method is called with 6?

void greeting(int n)

   {

if (n > 0)

{

   System.out.println("Hello!");

   greeting(n+1);

}

   }

1. infinitely
2. 7
3. 6
4. 5

Answer 1

Here is the solution to above problem

6) In a linked list based stack , the last nod represents the top node of the stack in ensure capcity method if the stack is full the size of stack is increased by creating a new array linked list of more capacity and old list is copied into the new list

Hence if the array is full we willl need to create a copy of old linked list hence it will take O(n) time to do so.

7) If the array is not full the ensure capcity method will do nothing as the number of element in array based linked list are still under capacity hence it will take only O(1) time to check the capacity

8) The efficiency of travelling a linked list recursively cannot be proven, as we can only find the time complexity of an algorithm but cannot measure its efficiency, as efficiency can be based on many factors both hardware and software

9) The output of above problem is ????

explanation

when n =3

we will get 1 "?"

then for n=2

we will get total 2 " ??"

then for n=1

we will get total 1 "???"

when for n=0

the program will print one more "?" so in total 4 "????" then when it sees n is not greater than 0 it will terminate

10) Infinitely number of recursive calls will be made since there is no base condition to stop the loop and always the value of n is increasing hence it will never stop executing