Test: Stack - Computer Science Engineering (CSE) MCQ

8 push operation is equivalent to 8 EQ operation.
2 pop operation is equivalent = 1st pop and 2nd pop one after the other
1st pop operation = (7 DQ(q1) + 7 EQ(q2) + 1 DQ(q1) )
2nd pop operation = (6 DQ(q2) + 6 EQ(q1) + 1 DQ(q2) )
Total =8EQ+ 15 DQ + 13 EQ = 36

Hence the correct answer is 36.

Let S1 and S2 be two stacks and let pop(), push (), and top() be the operations on the stack.

The element 1,2,3,4,5 in order.
Option 1:
To obtain 3 1 2 4 5
S1.push(1) ;
S2.push(2);
S1.push(3);
display(S1.top());
S1.pop();
display(S1.top())
S1.pop();
display(S2.top())
S2.pop();
S1.push(4)
display(S1.top())
S1.pop();
S1.push(5);
display(S1.top())
S1.pop();

Option 2 : 4 1 5 2 3
There can be no way to generate option 2.

Option 3:
To generate  4 2 1 3 5
S1.push(1);
S1.push(2);
S2.push(3);
S2.push(4);
display(S2.top());
S2.pop();
display(S1.top());
S1.pop();
display(S1.top());
S1.pop();
display(S2.top());
S2.pop();
S1.push(5);
display(S1.top());
S1.pop();

Option 4:
To obtain option 5 4 3 2 1.
Push all elements onto the  stacks, The stack will be like:
1 2 3 4 5, where 5 is present at the top of the stack and 1 is at the bottom of the stack. Now popping those elements and printing will output the sequence 5 4 3 2 1.

Hence the correct answer is 4 1 5 2 3.

(A + B) * C – D * E / F
A B + * C – D * E / F
A B + C * – D * E / F
A B + C * – D E * / F
A B + C * – D E * F /
A B + C * D E * F / –

Postfix expression 10 5 + 60 6 / * 8 −

PUSH: 10 and 5(TOS)

Add: 10 + 5 = 15
PUSH: 15 60 6 (TOS)

Divide: 60/6 = 10
PUSH: 10

Multiply: 15 × 10 = 150
PUSH: 150 and 8

Subtract: 150 - 8 = 142

Diagram:

Evaluate Infix notation:
(((10 + 5) * (60 / 6)) - 8)  = ((15 * 10) - 8) = 150 - 8 = 142

A stack is a linear data type that acts as a collection of elements. A stack is a linear data structure that operates memory allocation and deallocation on the Last-In-First-Out (LIFO) principle. In linear data structures, the elements are accessed in sequential order but it is not compulsory to store all elements sequentially. It has two main operations:

• Push, which adds a new element to the collection, and
• Pop, which removes the most recently added element that hasn't been removed yet.

Hence the correct answer is Last­-in-­first­-out (LIFO) manner.

A stack can be implemented using two queues. Let stack to be implemented be ‘x’ and queues used to implement be ‘a’ and ‘b’.

Method 1 (By push operation)
This method makes sure that the newly entered element is always at the front of ‘a’, so that pop operation just dequeues from ‘a’. ‘b’ is used to put every new element at front of ‘b’.

Method 2 (By making pop operation costly)
In a push operation, the new element is always enqueued to a. In pop() operation, if b is empty then all the elements except the last, are moved to b. Finally, the last element is dequeued from a and returned.

Therefore Option 2 is correct

The given data:

Infix: (9 - 5) + 2 
Postfix: 95-2+

Hence the correct answer is 95-2+.

For example, if we take an array with N = 4 to represent our stack. It will be A [0, 1, 2, 3].

Now, in order to push four elements, say x, y, z, w

• First we will push at x index 3 which is 4 – 1
• Then, we will decrement pos and push y at index 2 and so on.

Note that, the pop operation is also behaving inversely here.

• In usual stack, pop would decrement value of pos.

In our case, pop is incrementing the value of pos.

A recursive problem like the Tower of Hanoi can be rewritten using system stack or user-defined stack
Recurrence relation of tower of Hanoi:  T(n) = 2T(n - 1) + 1 

Method 1 (efficient push):

• push:
  • enqueue in queue1
• pop:
  • while the size of queue1 is bigger than 1, pipe dequeued items from queue1 into queue2
  • dequeue and return the last item of queue1, then switch the names of queue1 and queue2

Method 2 (efficient pop):

• push:
  • enqueue in queue2
  • enqueue all items of queue1 in queue2, then switch the names of queue1 and queue2
• pop:
  • deqeue from queue1