Mahindra Comviva Stacks Problems with Solutions Quiz- 1

Evaluation of infix expression algorithm is linear and makes only one pass through the input.

The circular queue will overflow when front = 0 and rear = capacity-1.

In Indirect  addressing mode the instruction does not have the address of the data to be operated on,but the instruction points where the address is stored(it is indirectly specifying the address of memory location where the data is stored or to be stored)

In immediate addressing mode the data is to be used is immediately given in instruction itself;so it deals with constant data.

In Indexed Addressing, elements can be accessed using their index value. In array we have index addressing.

In Autodecrement addressing mode,  Before determining the effective address, the value in the base register is decremented by the size of the data item which is to be accessed.

Within a loop, this addressing mode can be used to step backwards through all the elements of an array or vector.

In recursion last function called needs to be completed first. Now Stack is a LIFO data structure i.e. ( Last In First Out) and hence it is used to implement recursion.

Queue implemented in linear array

in queue of size 10

1) after inserting 10 element Front = 1 ,Rear = 10
2) then delete 9 element ater this Front = 10 , Rear = 10

array has 9 empty location but can not insert even a 1 new element because at the insertion it cheak Rear==Size of the queue

PUSH(1)     // Status of stack : 1
PUSH(2)    // Status of stack : 1 2
POP             // Status of stack : 1
PUSH(5)    // Status of stack : 1 5
PUSH(6)    // Status of stack : 1 5 6
POP            // Status of stack : 1 5

Hence, stack will contain 1 and 5.

If top >= 0 then only the element can be popped from the stack. Otherwise we need to print that stack is underflow.

A queue can be implemented using two stacks.

in queue initially the both Front and Rear assigned by the value -1 means the queue is empty

the size of the queue is 10 (array index no from 0 to 9)

ADD 1 it will increment both Front and Rear (in case of first element)
DELETE deletion of element set the Front and Rear to -1(queue is empty)
ADD 2 Front = 0 ,Rear = 0
ADD 3 Front = 0 ,Rear = 1
ADD 4 Front = 0 ,Rear = 2
DELETE Front = 1 ,Rear = 2
DELETE Front = 2 ,Rear = 2

now the empty location in the queue is 7 so the maximum capacity of queue is b)7

Priority queues r applied using 2-D array where it has two rows one for element and second for priority ,so minimum numbers of queues are needed to implement are two.