Peek Front Element of Queue in DSA Javascript

Mental Model

A queue is like a line where the first person in line is the next to be served. Peeking means looking at that first person without removing them.

Analogy: Imagine waiting in line at a coffee shop. You can see who is first in line without letting them leave the line. That is peeking the front of the queue.

front -> [1] -> 2 -> 3 -> null
          ↑peek

Dry Run Walkthrough

Input: queue: 1 -> 2 -> 3, peek front element

Goal: See the front element without removing it

Step 1: Access the front element of the queue

front -> [1] -> 2 -> 3 -> null
          ↑peek

Why: We want to see the first element without changing the queue

Result:

front -> 1 -> 2 -> 3 -> null
peeked element: 1

Annotated Code

DSA Javascript

class Queue {
  constructor() {
    this.items = [];
  }

  enqueue(element) {
    this.items.push(element);
  }

  dequeue() {
    if (this.isEmpty()) return null;
    return this.items.shift();
  }

  peek() {
    if (this.isEmpty()) return null;
    return this.items[0];
  }

  isEmpty() {
    return this.items.length === 0;
  }

  printQueue() {
    return this.items.join(' -> ') + ' -> null';
  }
}

// Driver code
const queue = new Queue();
queue.enqueue(1);
queue.enqueue(2);
queue.enqueue(3);
console.log(queue.printQueue());
const frontElement = queue.peek();
console.log('peeked element:', frontElement);

if (this.isEmpty()) return null;

check if queue is empty to avoid errors

return this.items[0];

return the first element without removing it

Output

1 -> 2 -> 3 -> null

peeked element: 1

Complexity Analysis

Time: O(1) because accessing the first element of an array is direct

Space: O(n) because the queue stores n elements

vs Alternative: Peeking is faster than dequeueing because it does not remove or shift elements

Edge Cases

empty queue

peek returns null indicating no elements

DSA Javascript

if (this.isEmpty()) return null;

When to Use This Pattern

When you need to see the next item to process in a queue without removing it, use the peek operation to check the front element safely.

Common Mistakes

Mistake: Removing the front element instead of just peeking
Fix: Use items[0] to access without removing, not shift()

Summary

Shows the first element in the queue without removing it

Use when you want to know the next item to process but keep the queue unchanged

The front element is always at index 0 and can be accessed directly

Practice

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

easy

A. Returns the front element without removing it

B. Removes the front element from the queue

C. Adds a new element to the back of the queue

D. Checks if the queue is empty

Solution

1. Step 1: Understand the peek operation

   The peek operation looks at the front element of the queue but does not remove it.

2. Step 2: Compare with other queue operations

   Removing is done by dequeue, adding by enqueue, and checking empty is a separate check.

3. Final Answer:

   Returns the front element without removing it -> Option A

4. Quick Check:

   Peek = front element without removal [OK]

Hint: Peek shows front without changing queue [OK]

Common Mistakes:

• Confusing peek with dequeue
• Thinking peek adds elements
• Assuming peek checks emptiness

2. Which of the following is the correct syntax to peek the front element of a queue named myQueue in JavaScript if it is implemented as an array?

easy

A. myQueue.peek()

B. myQueue[0]

C. myQueue.front()

D. myQueue.getFront()

Solution

1. Step 1: Understand array-based queue front access

   In JavaScript, arrays use index 0 for the first element, so myQueue[0] accesses the front.

2. Step 2: Check other options for validity

   Methods like peek(), front(), or getFront() are not standard array methods.

3. Final Answer:

   myQueue[0] -> Option B

4. Quick Check:

   Array front element = index 0 [OK]

Hint: Use index 0 to peek front in array queue [OK]

Common Mistakes:

• Using non-existent methods on arrays
• Using wrong index like myQueue[1]
• Confusing peek with dequeue method

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

const queue = [];
queue.push(10);
queue.push(20);
queue.push(30);
console.log(queue[0]);

medium

A. 10

B. 20

C. 30

D. undefined

Solution

1. Step 1: Understand queue insertion using push

   Elements 10, 20, 30 are added to the end of the array, so queue = [10, 20, 30].

2. Step 2: Access front element using index 0

   queue[0] returns the first element, which is 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Front element after pushes = 10 [OK]

Hint: Front element is first pushed value at index 0 [OK]

Common Mistakes:

• Thinking last pushed element is front
• Using queue.pop() instead of queue[0]
• Confusing index 0 with index 1

4. Identify the error in the following code that tries to peek the front element of a queue implemented as an array:

const queue = [5, 15, 25];
function peek() {
  return queue.pop();
}
console.log(peek());

medium

A. queue is not defined inside peek()

B. pop() returns undefined for non-empty arrays

C. peek() should use queue.push() instead

D. Using pop() removes the last element instead of peeking the front

Solution

1. Step 1: Understand pop() behavior on arrays

   pop() removes and returns the last element, not the front.

2. Step 2: Correct way to peek front element

   To peek front, access queue[0] without removing it.

3. Final Answer:

   Using pop() removes the last element instead of peeking the front -> Option D

4. Quick Check:

   pop() removes last, not front [OK]

Hint: pop() removes last; peek front uses index 0 [OK]

Common Mistakes:

• Using pop() to peek front
• Confusing pop() with shift()
• Trying to push in peek function

5. You have a queue implemented as an array: let q = [100, 200, 300]. You want to peek the front element safely, but the queue might be empty sometimes. Which code snippet correctly peeks the front element without causing an error?

hard

A. console.log(q.pop());

B. console.log(q[0]);

C. console.log(q.length > 0 ? q[0] : 'Queue is empty');

D. console.log(q.shift());

Solution

1. Step 1: Check for empty queue before peeking

   Accessing q[0] when empty returns undefined but may be unclear; better to check length.

2. Step 2: Use conditional to safely peek front

   Using q.length > 0 ? q[0] : 'Queue is empty' returns front if exists, else message.

3. Final Answer:

   console.log(q.length > 0 ? q[0] : 'Queue is empty'); -> Option C

4. Quick Check:

   Safe peek uses length check before access [OK]

Hint: Check length before peek to avoid errors [OK]

Common Mistakes:

• Peeking without checking empty queue
• Using pop() or shift() which remove elements
• Assuming q[0] always exists