Peek Top Element of Stack in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the 'peek' operation do in a stack?

The 'peek' operation shows the top element of the stack without removing it. It lets you see what's on top without changing the stack.

beginner

In Go, how do you check the top element of a stack implemented as a slice?

You get the last element of the slice using stack[len(stack)-1]. This is the top element of the stack.

beginner

Why should you check if the stack is empty before peeking?

If the stack is empty, peeking will cause an error because there is no top element. Always check length > 0 before peeking.

beginner

What is the time complexity of the peek operation on a stack?

The peek operation runs in O(1) time because it directly accesses the top element without looping.

beginner

Show a simple Go function to peek the top element of a stack represented by a slice of integers.

func peek(stack []int) (int, bool) { if len(stack) == 0 { return 0, false // stack empty } return stack[len(stack)-1], true }

What does the peek operation return when the stack is empty?

A The bottom element of the stack

B An error or indication that stack is empty

C The first element pushed onto the stack

D The entire stack

Peeking an empty stack should return an error or a special value indicating the stack is empty.

In Go, how do you access the top element of a stack implemented as a slice named 'stack'?

A stack[len(stack)-1]

B stack[-1]

C stack[len(stack)]

D stack[0]

The top element is the last element in the slice, accessed by stack[len(stack)-1].

What is the time complexity of the peek operation on a stack?

A O(n)

B O(n^2)

C O(log n)

D O(1)

Peek operation is O(1) because it directly accesses the top element.

Which of these is NOT true about the peek operation?

A It requires checking if the stack is empty first

B It shows the top element without removing it

C It removes the top element from the stack

D It runs in constant time

Peek does not remove the top element; it only shows it.

What should you do before peeking to avoid runtime errors?

A Check if the stack is empty

B Clear the stack

C Pop the top element

D Push a new element

Always check if the stack is empty before peeking to avoid errors.

Explain how to safely peek the top element of a stack implemented as a slice in Go.

Think about slice length and indexing.

Describe the difference between peek and pop operations on a stack.

Focus on whether the element stays or is removed.

Practice

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

easy

A. Returns the top element without removing it

B. Removes the top element from the stack

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

D. Checks if the stack is empty

Solution

1. Step 1: Understand stack operations

   A stack stores elements in a last-in, first-out order. Common operations include push (add), pop (remove), and peek (view top).

2. Step 2: Define peek operation

   Peek returns the top element without removing it, allowing you to see the current top without changing the stack.

3. Final Answer:

   Returns the top element without removing it -> Option A

4. Quick Check:

   Peek = View top element without removal [OK]

Hint: Peek shows top without removing it [OK]

Common Mistakes:

• Confusing peek with pop operation
• Thinking peek adds elements
• Assuming peek checks emptiness

2. Which of the following is the correct Go syntax to peek the top element of a stack implemented as a slice named stack?

easy

A. top := stack[len(stack)-1]

B. top := stack[-1]

C. top := stack[0]

D. top := stack[len(stack)]

Solution

1. Step 1: Understand Go slice indexing

   Go slices are zero-indexed. The last element is at index length-1.

2. Step 2: Identify correct top element access

   To peek top, access stack[len(stack)-1]. Using len(stack) is out of range, 0 is bottom, and negative index is invalid.

3. Final Answer:

   top := stack[len(stack)-1] -> Option A

4. Quick Check:

   Last element index = length-1 [OK]

Hint: Use len(stack)-1 to get top index [OK]

Common Mistakes:

• Using len(stack) as index (out of range)
• Using negative index (not supported in Go)
• Accessing stack[0] which is bottom

3. What will be the output of the following Go code snippet?

stack := []int{10, 20, 30}
top := stack[len(stack)-1]
fmt.Println(top)

medium

A. 10

B. Runtime error

C. 20

D. 30

Solution

1. Step 1: Analyze stack contents

   The stack slice contains elements [10, 20, 30], where 30 is the last element (top).

2. Step 2: Evaluate peek operation

   top := stack[len(stack)-1] accesses stack[2], which is 30, then prints it.

3. Final Answer:

   30 -> Option D

4. Quick Check:

   Top element = 30 [OK]

Hint: Last element is top, printed value is 30 [OK]

Common Mistakes:

• Printing bottom element instead of top
• Indexing out of range causing error
• Confusing output with pop operation

4. Identify the error in the following Go code that tries to peek the top element of a stack:

stack := []int{}
top := stack[len(stack)-1]
fmt.Println(top)

medium

A. Syntax error due to missing import

B. Index out of range runtime error

C. top will be zero by default

D. Code runs fine and prints 0

Solution

1. Step 1: Check stack initialization

   The stack is initialized as an empty slice with length 0.

2. Step 2: Analyze peek operation

   Accessing stack[len(stack)-1] means stack[-1], which is invalid and causes runtime panic: index out of range.

3. Final Answer:

   Index out of range runtime error -> Option B

4. Quick Check:

   Empty slice peek causes runtime error [OK]

Hint: Empty stack peek causes index error [OK]

Common Mistakes:

• Assuming peek returns zero on empty stack
• Ignoring runtime panic possibility
• Confusing syntax error with runtime error

5. Given a stack implemented as a slice stack := []int{5, 10, 15}, write the correct Go code snippet to safely peek the top element and handle the case when the stack is empty.

hard

A. top := stack[0]
fmt.Println(top)

B. top := stack[len(stack)]
fmt.Println(top)

C. if len(stack) > 0 {
top := stack[len(stack)-1]
fmt.Println(top)
} else {
fmt.Println("Stack is empty")
}

D. if stack != nil {
top := stack[len(stack)-1]
fmt.Println(top)
}

Solution

1. Step 1: Check stack length before peeking

   To avoid runtime error, verify stack length > 0 before accessing top element.

2. Step 2: Access top element safely and handle empty case

   If stack is not empty, peek top with stack[len(stack)-1]. Otherwise, print "Stack is empty" message.

3. Final Answer:

   if len(stack) > 0 { top := stack[len(stack)-1] fmt.Println(top) } else { fmt.Println("Stack is empty") } -> Option C

4. Quick Check:

   Check length before peek to avoid errors [OK]

Hint: Check length > 0 before peek [OK]

Common Mistakes:

• Not checking length before peek
• Using len(stack) as index
• Assuming nil check is enough