Peek Top Element of Stack in DSA Python - Time & Space Complexity

Time Complexity: Peek Top Element of Stack

O(1)

Understanding Time Complexity

We want to understand how long it takes to look at the top item of a stack.

Specifically, how the time changes as the stack grows bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Stack:
    def __init__(self):
        self.items = []

    def push(self, item):
        self.items.append(item)

    def peek(self):
        return self.items[-1] if self.items else None

This code defines a stack and shows how to peek at the top element without removing it.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Accessing the last element of the list (stack top).
• How many times: Exactly once per peek call, no loops or recursion.

How Execution Grows With Input

Looking at the top element takes the same amount of time no matter how big the stack is.

Input Size (n)	Approx. Operations
10	1
100	1
1000	1

Pattern observation: The operation count stays the same as the stack grows.

Final Time Complexity

Time Complexity: O(1)

This means peeking the top element takes a constant time regardless of stack size.

Common Mistake

[X] Wrong: "Peeking takes longer if the stack is bigger because it has to look through all elements."

[OK] Correct: Peeking only looks at the last item directly, no need to check other elements.

Interview Connect

Knowing that peek is a constant time operation helps you explain stack efficiency clearly in interviews.

Self-Check

"What if the stack was implemented using a linked list instead of a list? How would the time complexity of peek change?"

Practice

1. What does the peek operation do in a stack?

easy

A. Returns the bottom element of the stack

B. Removes the top element from the stack

C. Adds a new element to the top of the stack

D. Returns the top element without removing it

Solution

1. Step 1: Understand stack operations

   A stack follows Last-In-First-Out (LIFO). The top element is the last added item.

2. Step 2: Define peek operation

   Peek returns the top element without removing it, unlike pop which removes it.

3. Final Answer:

   Returns the top element without removing it -> Option D

4. Quick Check:

   Peek = top element without removal [OK]

Hint: Peek shows top without removing it [OK]

Common Mistakes:

• Confusing peek with pop
• Thinking peek adds elements
• Assuming peek returns bottom element

2. Which of the following is the correct syntax to peek the top element from a stack implemented as a Python list named stack?

easy

A. stack[-1]

B. stack.peek()

C. stack.top()

D. stack.pop()

Solution

1. Step 1: Recall Python list indexing

   Python lists use negative indexing; stack[-1] accesses the last element.

2. Step 2: Differentiate peek from pop

   stack.pop() removes the element, so it is not peek. stack.peek() and stack.top() are not valid list methods.

3. Final Answer:

   stack[-1] -> Option A

4. Quick Check:

   Peek = stack[-1] [OK]

Hint: Use stack[-1] to peek top element in Python list [OK]

Common Mistakes:

• Using pop() which removes element
• Using non-existent peek() or top() methods
• Using positive index 0 instead of -1

3. What will be the output of this code?

stack = [10, 20, 30]
print(stack[-1])
stack.append(40)
print(stack[-1])

medium

A. 30\n40

B. 10\n40

C. 30\n30

D. 40\n40

Solution

1. Step 1: Check initial peek

   Initial stack is [10, 20, 30]. stack[-1] is 30, so first print outputs 30.

2. Step 2: Append and peek again

   After appending 40, stack becomes [10, 20, 30, 40]. stack[-1] is now 40, so second print outputs 40.

3. Final Answer:

   30 40 -> Option A

4. Quick Check:

   Peek before append = 30, after append = 40 [OK]

Hint: Peek shows last element; append adds new last element [OK]

Common Mistakes:

• Confusing peek with pop output
• Not updating peek after append
• Printing first element instead of last

4. Find the error in this code that tries to peek the top element of a stack implemented as a list:

stack = [5, 15, 25]
top = stack.pop[-1]
print(top)

medium

A. pop removes element, so peek is wrong

B. pop is used incorrectly with square brackets

C. stack is empty, so error occurs

D. print statement syntax is wrong

Solution

1. Step 1: Analyze pop usage

   pop is a method and should be called with parentheses, not square brackets. stack.pop[-1] is invalid syntax.

2. Step 2: Understand correct peek syntax

   To peek, use stack[-1]. To pop, use stack.pop(). Mixing them causes error.

3. Final Answer:

   pop is used incorrectly with square brackets -> Option B

4. Quick Check:

   pop() needs parentheses, not brackets [OK]

Hint: Use parentheses for pop(), brackets for indexing [OK]

Common Mistakes:

• Using pop with brackets instead of parentheses
• Confusing peek with pop
• Assuming pop[-1] is valid syntax

5. Given a stack implemented as a list, how can you safely peek the top element without causing an error if the stack is empty?

hard

A. Use stack.pop() and catch exceptions

B. Use stack[-1] directly without checks

C. Check if stack is not empty before peeking

D. Assign None to top element always

Solution

1. Step 1: Understand empty stack risk

   Accessing stack[-1] when stack is empty causes IndexError.

2. Step 2: Implement safe peek

   Check if stack length > 0 before peeking to avoid error.

3. Final Answer:

   Check if stack is not empty before peeking -> Option C

4. Quick Check:

   Peek safely by checking stack length first [OK]

Hint: Always check stack not empty before peek [OK]

Common Mistakes:

• Peeking without checking empty stack
• Using pop() which removes element
• Ignoring exceptions from empty stack access