Peek Top Element of Stack in DSA Java - 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.

How does the time needed change as the stack grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public class Stack {
    private int[] data;
    private int top = -1;

    public int peek() {
        if (top == -1) throw new RuntimeException("Stack is empty");
        return data[top];
    }
}
    

This code returns the top element of the stack without removing it.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Accessing the element at the top index in the array.
• How many times: Exactly once per peek call.

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 time stays constant regardless of stack size.

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 check many elements."

[OK] Correct: Peeking only looks at the top element directly by index, so it does not 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 an array? How would the time complexity of peek change?"

Practice

1. What does the peek() method do in a stack?

easy

A. Checks if the stack is empty

B. Removes and returns the top element

C. Adds a new element to the top

D. Returns the top element without removing it

Solution

1. Step 1: Understand the peek method purpose

   The peek method is designed to look at the top element of the stack without changing the stack.

2. Step 2: Compare with other stack methods

   Unlike pop, peek does not remove the element; it only returns 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 checks emptiness

2. Which of the following is the correct syntax to peek the top element of a Java Stack named stack?

easy

A. stack.peek();

B. stack.top();

C. stack.getTop();

D. stack.pop();

Solution

1. Step 1: Recall Java Stack method names

   The Java Stack class uses peek() to view the top element without removing it.

2. Step 2: Check other options for correctness

   top() and getTop() are not valid Stack methods; pop() removes the top element.

3. Final Answer:

   stack.peek(); -> Option A

4. Quick Check:

   Correct syntax for peek is stack.peek() [OK]

Hint: Use stack.peek() to view top element [OK]

Common Mistakes:

• Using pop() instead of peek()
• Using non-existent methods like top()
• Missing parentheses in method call

3. What will be the output of the following Java code?

Stack<Integer> stack = new Stack<>();
stack.push(10);
stack.push(20);
stack.push(30);
System.out.println(stack.peek());

medium

A. 10

B. 30

C. 20

D. StackEmptyException

Solution

1. Step 1: Trace the push operations

   Elements pushed in order: 10 (bottom), 20, then 30 (top).

2. Step 2: Understand peek output

   peek() returns the top element without removing it, which is 30.

3. Final Answer:

   30 -> Option B

4. Quick Check:

   Top element after pushes = 30 [OK]

Hint: Peek shows last pushed element [OK]

Common Mistakes:

• Confusing peek with pop output
• Assuming peek returns bottom element
• Expecting exception on non-empty stack

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

Stack<String> stack = new Stack<>();
System.out.println(stack.peek());

medium

A. Throws EmptyStackException at runtime

B. Prints empty string

C. Compilation error: peek() not found

D. No error, prints null

Solution

1. Step 1: Check stack initialization and usage

   The stack is created but no elements are pushed before peek is called.

2. Step 2: Understand peek behavior on empty stack

   Calling peek() on an empty stack throws EmptyStackException at runtime.

3. Final Answer:

   Throws EmptyStackException at runtime -> Option A

4. Quick Check:

   peek() on empty stack throws exception [OK]

Hint: Peek on empty stack throws exception [OK]

Common Mistakes:

• Assuming peek returns null if empty
• Expecting compile-time error
• Thinking it prints empty string

5. Given a stack with elements [5, 10, 15] (5 at bottom, 15 at top), which sequence of operations will peek the top element without removing it and then add 20 on top?

1. stack.peek();
2. stack.push(20);
3. stack.pop();
4. stack.peek();

Select the correct order of operations to achieve this.

hard

A. 3, 1, 2

B. 2, 1

C. 1, 2

D. 1, 3, 2

Solution

1. Step 1: Peek top element without removing

   Operation 1 (stack.peek()) returns 15 without removing it.

2. Step 2: Add 20 on top

   Operation 2 (stack.push(20)) adds 20 on top, so stack becomes [5, 10, 15, 20].

3. Final Answer:

   1, 2 -> Option C

4. Quick Check:

   Peek then push = 1, 2 [OK]

Hint: Peek first, then push new element [OK]

Common Mistakes:

• Popping before peek removes top
• Pushing before peek changes top
• Incorrect operation order