Peek Top Element of Stack in DSA Python - Execution Trace

Concept Flow - Peek Top Element of Stack

Start

↓

Check if stack is empty?

Yes→Return None or Error

No↓

Access top element

↓

Return top element

↓

End

The flow checks if the stack is empty. If not, it accesses and returns the top element without removing it.

Execution Sample

DSA Python

stack = [10, 20, 30]
top = stack[-1]
print(top)

Peek the top element of the stack and print it without removing it.

Execution Table

Step	Operation	Stack State	Pointer Changes	Visual State
1	Start with stack = [10, 20, 30]	[10, 20, 30]	top pointer not set	10 -> 20 -> 30 (top)
2	Check if stack is empty	[10, 20, 30]	No change	10 -> 20 -> 30 (top)
3	Access top element stack[-1]	[10, 20, 30]	top pointer points to 30	10 -> 20 -> 30 (top)
4	Return top element 30	[10, 20, 30]	top pointer unchanged	10 -> 20 -> 30 (top)
5	Print top element	[10, 20, 30]	No change	10 -> 20 -> 30 (top)
6	End	[10, 20, 30]	No change	10 -> 20 -> 30 (top)

💡 Peek operation ends after returning the top element without modifying the stack.

Variable Tracker

Variable	Start	After Step 3	Final
stack	[10, 20, 30]	[10, 20, 30]	[10, 20, 30]
top	undefined	30	30

Key Moments - 3 Insights

Why do we check if the stack is empty before peeking?

Because peeking an empty stack has no top element to return, so we avoid errors by checking first (see execution_table step 2).

Does peeking remove the top element from the stack?

No, peeking only reads the top element without removing it, so the stack stays the same (see execution_table steps 3 and 4).

What happens if we try to peek an empty stack?

The operation returns None or an error, depending on implementation, because there is no element to return (refer to concept_flow step 'Return None or Error').

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'top' after step 3?

A 30

B 20

C 10

D undefined

💡 Hint

Check the 'Pointer Changes' column at step 3 in the execution_table.

At which step does the code confirm the stack is not empty?

A Step 1

B Step 4

C Step 2

D Step 5

💡 Hint

Look at the 'Operation' column for the check condition in the execution_table.

If the stack was empty, what would the execution_table show at step 2?

A Stack state with elements

B Return None or Error

C Pointer changes to top element

D Access top element

💡 Hint

Refer to the concept_flow where empty stack leads to returning None or Error.

Concept Snapshot

Peek Top Element of Stack:
- Check if stack is empty first
- Access top element without removing
- Return the top element
- Stack remains unchanged
- Commonly use stack[-1] in Python

Full Transcript

This visual trace shows how to peek the top element of a stack. First, we check if the stack is empty to avoid errors. If not empty, we access the last element (top) without removing it. The stack stays the same throughout. The top element is then returned and printed. This operation is safe and does not modify the stack.

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