How Stacks Work — API Cheat Sheet

This is a reference page, not a graded exercise. A stack is last-in-first-out (LIFO): the last thing you pushed is the first thing you pop. Reach for it for matching/nesting problems (parentheses, undo history), and for turning recursion into an explicit loop.

Java — use Deque<T>, not the old Stack class

Java has a legacy Stack class, but its own documentation recommends against it (it's synchronized, which you don't need, and extends Vector). Use Deque instead:

Want to...	Method	Example
Create	new ArrayDeque<>()	Deque<Integer> stack = new ArrayDeque<>();
Push	.push(v)	stack.push(5);
Pop	.pop()	removes and returns the top; throws if empty
Peek	.peek()	returns the top without removing; returns null if empty
Check empty	.isEmpty()

C# — Stack<T>

Want to...	Method	Example
Create	new Stack<T>()	var stack = new Stack<int>();
Push	.Push(v)
Pop	.Pop()	throws if empty
Peek	.Peek()	throws if empty — check .Count first
Size	.Count	property

Python — just use a list

No dedicated stack class needed — a plain list's end is O(1) for both operations:

Want to...	Method	Example
Push	.append(v)
Pop	.pop()	pops from the end; throws IndexError if empty
Peek	lst[-1]	negative indexing reaches the last element

JavaScript / TypeScript — just use an Array

Want to...	Method	Example
Push	.push(v)	pushes to the end
Pop	.pop()	pops from the end; returns undefined if empty
Peek	arr[arr.length - 1]	no built-in .peek()

The gotcha that costs the most time: in Python and JS, you push/pop from the end of the list/array, not the front — .pop() on an array is O(1) precisely because it operates on the end. If you instead call .pop(0) (Python) or .shift() (JS), you're doing a queue operation, not a stack operation, and it's O(n) too.