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Stacks in Python: LIFO, Balanced Brackets and Expression Evaluation

Learn the stack data structure: last in, first out, push and pop in O(1), stacks with Python lists and linked lists, checking balanced brackets, evaluating postfix expressions and the min stack.

Beginner10 min readUpdated
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A stack is a collection where the last item added is the first one removed: last in, first out (LIFO). Think of a stack of plates: you put plates on top and take them from the top.

OperationMeaningCost
push(x)put x on topO(1)
pop()remove and return the topO(1)
peek()look at the top without removing itO(1)
is_empty()is there anything left?O(1)

Your browser's Back button, an editor's Undo, and the call stack that runs recursive functions all work this way.

A Python list is a stack#

append and pop() both work at the end of a list, which is O(1):

Python
stack = []
stack.append("page1")     # push
stack.append("page2")
stack.append("page3")
print(stack[-1])          # peek
print(stack.pop())        # pop
print(stack.pop())
print(stack, len(stack) == 0)
Output
page3
page3
page2
['page1'] False

Never use the front of a list as the top: insert(0, x) and pop(0) are O(n).

A stack built on a linked list#

The same operations on a linked list: push and pop at the head, both O(1), and no resizing ever.

Python
class Node:
    def __init__(self, value, next=None):
        self.value, self.next = value, next

class Stack:
    def __init__(self):
        self.top = None
        self.size = 0

    def push(self, value):
        self.top = Node(value, self.top)
        self.size += 1

    def pop(self):
        if self.top is None:
            raise IndexError("pop from empty stack")
        value = self.top.value
        self.top = self.top.next
        self.size -= 1
        return value

    def peek(self):
        return None if self.top is None else self.top.value

s = Stack()
for x in [1, 2, 3]:
    s.push(x)
print(s.pop(), s.pop(), s.peek(), s.size)
Output
3 2 1 1

Balanced brackets#

Is every opening bracket closed by the right kind, in the right order? A stack remembers which brackets are still open; the most recent one must close first.

Python
def balanced(text):
    pairs = {")": "(", "]": "[", "}": "{"}
    stack = []
    for ch in text:
        if ch in "([{":
            stack.append(ch)
        elif ch in pairs:
            if not stack or stack.pop() != pairs[ch]:
                return False
    return not stack            # anything left open is unbalanced

for t in ["(a[b]{c})", "([)]", "((", "print(len(x))"]:
    print(t, balanced(t))
Output
(a[b]{c}) True
([)] False
(( False
print(len(x)) True

Evaluating postfix expressions#

In postfix (reverse Polish) notation the operator comes after its operands: 3 4 + 2 * means (3 + 4) × 2. A stack evaluates it in one pass: push numbers; on an operator, pop two numbers, apply it, and push the result.

Python
def eval_postfix(expr):
    stack = []
    for token in expr.split():
        if token in "+-*/":
            b = stack.pop()          # the second operand is on top
            a = stack.pop()
            if token == "+": stack.append(a + b)
            elif token == "-": stack.append(a - b)
            elif token == "*": stack.append(a * b)
            else: stack.append(a / b)
        else:
            stack.append(float(token))
    return stack.pop()

print(eval_postfix("3 4 + 2 *"))
print(eval_postfix("5 1 2 + 4 * + 3 -"))
Output
14.0
14.0

The order of a and b matters for - and /: 10 2 - is 10 − 2, so the item popped first is the right-hand operand.

Min stack: the minimum in O(1)#

Support push, pop and get_min(), all in O(1). Store, next to each value, the minimum of everything at or below it:

Python
class MinStack:
    def __init__(self):
        self.items = []          # pairs of (value, min so far)

    def push(self, x):
        current_min = min(x, self.items[-1][1]) if self.items else x
        self.items.append((x, current_min))

    def pop(self):
        return self.items.pop()[0]

    def get_min(self):
        return self.items[-1][1]

m = MinStack()
for x in [5, 3, 7, 2]:
    m.push(x)
print(m.get_min())
m.pop()
print(m.get_min())
m.pop()
m.pop()
print(m.get_min())
Output
2
3
5

Common mistakes#

Practice#

Practice exercise

Write simplify_path(path) that simplifies a Unix-style path: . means "this folder", .. means "go up one folder" (ignored at the root), and repeated slashes count as one. "/home//user/./docs/../photos/" becomes "/home/user/photos".

Show hint

Split on /. Use a stack of folder names: skip empty parts and ., pop on .. (if the stack is not empty), and push anything else. Join with / at the end.

Show solution
Python
def simplify_path(path):
    stack = []
    for part in path.split("/"):
        if part in ("", "."):
            continue
        if part == "..":
            if stack:
                stack.pop()
        else:
            stack.append(part)
    return "/" + "/".join(stack)

print(simplify_path("/home//user/./docs/../photos/"))
print(simplify_path("/../"))
print(simplify_path("/a/b/../../c"))
Output
/home/user/photos
/
/c

Key takeaway#

Next: first in, first out, in Queues and Deques.

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