What is a potential limitation of array-based stacks?*1 pointa. Limited by the fixed size of the arrayb. Can lead to stack overflowc. Offers dynamic sizingd. Requires additional memory for pointers

17. What is the role of the top variable in a stack implementation?a. It stores the value to be pushed.b. It tracks the most recently added element.c. It points to the bottom of the stack.d. It manages the display operation.18. What is a potential limitation of array-based stacks?a. Limited by the fixed size of the arrayb. Can lead to stack overflowc. Offers dynamic sizingd. Requires additional memory for pointers19. In the C++ code for pushing elements onto a stack, what is stack[top] = x; doing?a. Removing the top elementb. Initializing the stackc. Adding an element to the stackd. Checking if the stack is full20. In the C++ code for displaying elements in a stack, what does the loop for(i = top; i >= 0; i--)do?a. Adds elements to the stackb. Removes elements from the stackc. Checks if the stack is emptyd. Displays elements in reverse order

What is the key advantage of a linked list-based stack over an array-based stack?*1 pointa. Constant-time access to elementsb. Fixed size and memory efficiencyc. Dynamic sizing and flexibilityd. Faster push operation

Array implementation of Stack is not dynamic, which of the following statements supports this argument?space allocation for array is fixed and cannot be changed during run-timeuser unable to give the input for stack operationsa runtime exception halts executionimproper program compilation

Which representation is more flexible for handling an indefinite number of values in a stack?*1 pointa. Array-based stackb. Linked list-based stackc. Both have the same flexibilityd. Neither can handle an indefinite number of values

A single array A[1..MAXSIZE] is used to implement two stacks, The two stacks grow from opposite ends of the array. Variables top1 and top2 (top1 < top2) point to the location of the topmost element in each of the stacks, If the space is to be used efficiently, the condition for "stack full" is*2 points(top1 = MAXSIZE/2) AND (top2 = MAXSIZE/2 + 1top1 + top2 = MAXSIZE(top1 = MAXSIZE/2) or (top2 = MAXSIZE)top1 = top2 - 1