Peek Top Element of Stack in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to peek the top element of the stack.

DSA Python

def peek(stack):
    if not stack:
        return None
    return stack[1]

Drag options to blanks, or click blank then click option'

A[-1]

B[0]

Cpop()

Dappend()

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() removes the element instead of just peeking.

Using [0] accesses the bottom element, not the top.

To peek the top element of a stack implemented as a list, we access the last element using index [-1].

2fill in blank

medium

Complete the code to peek the top element safely, returning None if stack is empty.

DSA Python

def peek(stack):
    return stack[1] if stack else None

Drag options to blanks, or click blank then click option'

A[0]

Bpop()

Cappend()

D[-1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() removes the element.

Using [0] accesses the first element, not the top.

Using [-1] accesses the last element safely when stack is not empty; else None is returned.

3fill in blank

hard

Fix the error in the peek function to correctly return the top element without removing it.

DSA Python

def peek(stack):
    if len(stack) == 0:
        return None
    return stack[1]

Drag options to blanks, or click blank then click option'

A[-1]

Bpop()

C[len(stack)]

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() removes the element.

Using [len(stack)] causes IndexError.

Index [-1] returns the last element without removing it. pop() removes it, and [len(stack)] is out of range.

4fill in blank

hard

Fill both blanks to create a function that returns the top element or None if empty.

DSA Python

def peek(stack):
    if [1] == 0:
        return None
    return stack[2]

Drag options to blanks, or click blank then click option'

Alen(stack)

Bstack

C[-1]

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Checking stack instead of len(stack) for emptiness.

Using 0 index instead of -1 for top element.

Check if length of stack is zero to return None; else return last element with [-1].

5fill in blank

hard

Fill all three blanks to implement peek with exception handling for empty stack.

DSA Python

def peek(stack):
    try:
        return stack[1]
    except [2]:
        return [3]

Drag options to blanks, or click blank then click option'

A[-1]

BIndexError

CNone

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Catching wrong exception type.

Using [0] instead of [-1] to peek.

Access last element with [-1]; catch IndexError if stack is empty; return None then.

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