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 (topl

Question

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 (topl< top 2) point to the location of the topmost element in each of the stacks. If the space is to be used efficiently, the condition for &ldquo;stack full&rdquo; isa)(top1 = MAXSIZE/2) and (top2 = MAXSIZE/2+1)b)top1 + top2 = MAXSIZEc)(top1= MAXSIZE/2) or (top2 = MAXSIZE)d)top1= top2 -1Correct answer is option 'D'. Can you explain this answer?

Answer

And top2) are used to indicate the top elements of both stacks. Explain how this array can be used to implement two stacks and discuss the time complexity of the operations.

To implement two stacks using a single array, we can allocate the first half of the array to one stack and the second half to the other stack. For example, if the array size is MAXSIZE, we can allocate the first MAXSIZE/2 elements to the first stack and the remaining MAXSIZE/2 elements to the second stack.

To keep track of the top elements of both stacks, we can use two variables top1 and top2. top1 will initially point to the first element of the first stack, and top2 will initially point to the last element of the second stack. As elements are pushed onto or popped from either stack, these variables will be updated accordingly.

To push an element onto the first stack, we can increment top1 and store the element at A[top1]. Similarly, to push an element onto the second stack, we can decrement top2 and store the element at A[top2].

To pop an element from the first stack, we can read the element at A[top1] and then decrement top1. Similarly, to pop an element from the second stack, we can read the element at A[top2] and then increment top2.

The time complexity of these operations depends on the implementation. If we use a simple array with no additional data structures, the time complexity of push and pop operations will be O(1) as they involve simple array access and update operations.

However, if we want to ensure that both stacks have equal size and avoid overflow or underflow, we can use additional data structures such as a counter or a flag to keep track of the number of elements in each stack. This will increase the time complexity of push and pop operations to O(2) or O(3) as we may need to update the counter or flag in addition to the array access and update operations.

Answer

If we are to use space efficiently then size of the any stack can be more than MAXSIZE/2.
Both stacks will grow from both ends and if any of the stack top reaches near to the other top then stacks are full. So the condition will be top1 = top2 -1 (given that top1 < top2)

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