**π‘ Problem Formulation:** We are often faced with problems that require validating the output of stack-based operations. The goal is to determine the final state of a stack after performing a series of push and pop operations, with the final “answer” being the top element of the stack or some other property of the final stack. For instance, given a sequence of operations [“PUSH 5”, “PUSH 10”, “POP”, “PUSH 2”], we want to check the final top value which should be “2”.

## Method 1: Using a List as a Stack

In Python, a list can be easily used as a stack, where the `append()`

method is analogous to the stack “push” operation, and the `pop()`

method without an index removes and returns the last item, emulating the stack “pop” operation. This method is native and straightforward for implementing stack behavior.

Here’s an example:

def check_stack_operations(ops): stack = [] for operation in ops: command, *value = operation.split() if command == "PUSH": stack.append(int(value[0])) elif command == "POP" and len(stack) > 0: stack.pop() return stack[-1] if stack else None # Test the program operations = ["PUSH 5", "PUSH 10", "POP", "PUSH 2"] print(check_stack_operations(operations))

Output:

2

This code snippet defines a function `check_stack_operations`

that takes a list of operations. It emulates a stack using a Python list. By looping over each operation, it modifies the stack accordingly and returns the final top element of the stack or `None`

if the stack is empty.

## Method 2: Using collections.deque for Efficiency

The `collections.deque`

is a double-ended queue and can be used more efficiently than a list when dealing with stack operations, especially for larger datasets. It optimizes append and pop operations from both ends and is especially useful in cases where performance matters.

Here’s an example:

from collections import deque def check_stack_operations(ops): stack = deque() for operation in ops: command, *value = operation.split() if command == "PUSH": stack.append(int(value[0])) elif command == "POP" and len(stack) > 0: stack.pop() return stack[-1] if stack else None # Test the program operations = ["PUSH 5", "PUSH 10", "POP", "PUSH 2"] print(check_stack_operations(operations))

Output:

2

The function demonstrated uses Python’s `collections.deque`

to maintain stack operations. Similar to Method 1, this version offers better performance without altering the core logic, providing a speed advantage when working with larger data sets.

## Method 3: Object-Oriented Stack Approach

This method involves creating a Stack class with methods for push and pop operations. This encapsulates the stack logic and allows the stack to be used with a clear interface, making the code more maintainable and reusable.

Here’s an example:

class Stack: def __init__(self): self._elements = deque() def push(self, value): self._elements.append(value) def pop(self): if not self.is_empty(): return self._elements.pop() def top(self): if not self.is_empty(): return self._elements[-1] def is_empty(self): return not self._elements def check_stack_operations(ops): stack = Stack() for operation in ops: command, *value = operation.split() if command == "PUSH": stack.push(int(value[0])) elif command == "POP": stack.pop() return stack.top() # Test the program operations = ["PUSH 5", "PUSH 10", "POP", "PUSH 2"] print(check_stack_operations(operations))

Output:

2

This code snippet demonstrates an object-oriented approach to stack operations. It defines a Stack class that encapsulates the stack and its operations. The `check_stack_operations`

function utilizes this class to process operations and determine the final state.

## Method 4: Using Function Recursion

Recursion can also be used to emulate stack operations though it’s not conventional for this purpose. The primary function recurses over the operations, using its call stack as the stack mechanism. This method is more of an academic interest and less practical due to limitations in recursion depth for large numbers of operations.

Here’s an example:

def check_stack_operations(ops, stack=None): if stack is None: stack = [] if not ops: return stack[-1] if stack else None command, *value = ops[0].split() return check_stack_operations(ops[1:], stack + [int(value[0])] if command == "PUSH" else stack[:-1]) # Test the program operations = ["PUSH 5", "PUSH 10", "POP", "PUSH 2"] print(check_stack_operations(operations))

Output:

2

The provided code uses a recursive function to process stack operations. The function calls itself, modifying the passed stack list according to the operations performed. It is concise and elegant but not suitable for long lists of operations due to Python’s recursion limit.

## Bonus One-Liner Method 5: Using Python’s reduce Function

The `functools.reduce`

function can be used to accumulate results from a sequence of operations. This one-liner method relies on `reduce`

to apply push and pop operations sequentially on the stack, returning the final result as a one-liner, but could be less readable.

Here’s an example:

from functools import reduce operations = ["PUSH 5", "PUSH 10", "POP", "PUSH 2"] result = reduce(lambda acc, op: acc[:-1] if "POP" in op else acc + [int(op.split()[1])], operations, []) print(result[-1] if result else None)

Output:

2

This one-liner uses the `reduce`

function to apply each operation on an accumulator, which represents the stack. It’s a compact and functional-programming oriented solution, but may sacrifice some readability and clarity.

## Summary/Discussion

**Method 1:**Python List as Stack. Simple implementation. Easy to understand. Slower for large datasets.**Method 2:**Using collections.deque. More efficient for large datasets. Very similar to method 1 in usage.**Method 3:**Object-Oriented Stack. Provides an encapsulated solution making code reusable. Might be an overkill for simple stack operations.**Method 4:**Function Recursion. Interesting and elegant solution. Not practical for deep stacks due to Python’s recursion depth limitation.**Bonus Method 5:**Python’s reduce Function. Compact and functional. Possibly less readable for those unfamiliar with reduce.