Test: Stack - Computer Science Engineering (CSE) MCQ

Infix: This type of string is of the form a op b. Here the operator is between the pair of operands on which the operation is to be performed.
Postfix: This type of string is of the form a b op. Here the operator is placed after the pair of operands on whom the operation is to be performed.

• To find the maximum levels of memory blocks required in operand stack for infix to postfix conversion can be found be finding the longest continuous operands in the postfix expression. As we can see, 426 is the longest sequence of operands in the final postfix expression, hence, 3 memory blocks will be required at most in the operand stack.
• Similarly, maximum levels of memory blocks in operator stack is 2 as we can see in the table that only 2 operators at most are coming together.

Conversion of infix to postfix is,​
Infix: ((a+b)*(c-d))

The postfix expression is ab+cd-*
Hence the correct answer is Stack.

Given five items P, Q, R, S, and T.
These are pushed into the stack as : 

Now stack is popped four times, and each item popped is inserted into the queue.
So, items that are popped from the stack are: T, S, R, Q
After insertion into queue, queue looks like this:

Now, two items are deleted from the queue which is: T, S
These are inserted back into the stack.
Stack will be :

Queue will be :

Now, one item is popped from the stack which is S.

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.

Initially stack is emput
Stack:
push(1)

push(2)

pop()

Output : 2
push(1)

push(2)

pop()

Output : 2 2
pop()

Output : 2 2 1
pop()

Output : 2 2 1 1
push(2)

pop()

Output : 2 2 1 1 2

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

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 / –