Peek Top Element of Stack in DSA Go - 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.

The question is: How does the time to peek change as the stack grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Peek returns the top element of the stack without removing it
func (s *Stack) Peek() (int, bool) {
    if len(s.items) == 0 {
        return 0, false // stack is empty
    }
    return s.items[len(s.items)-1], true
}
    

This code checks if the stack is empty and then returns the last item added.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Accessing the last element of the array (slice) directly.
• 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 constant as the stack grows.

Final Time Complexity

Time Complexity: O(1)

This means peeking the top element takes the same short time no matter how many items are in the stack.

Common Mistake

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

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

Interview Connect

Knowing that peek is a quick, constant-time operation helps you explain stack behavior clearly in interviews.

Self-Check

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

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