Why Peek Front Element of Queue in DSA Typescript?

The Big Idea

Peek lets you see the next in line without breaking the flow--curious how?

The Scenario

Imagine you are waiting in line at a coffee shop. You want to know who is next to be served without removing anyone from the line.

The Problem

If you try to check the next person manually by asking everyone to step aside, it wastes time and confuses the line. You might accidentally skip someone or lose the order.

The Solution

Using the peek operation on a queue lets you see the front person without disturbing the line. It keeps the order intact and saves time.

Before vs After

✗ Before

let line = ['Alice', 'Bob', 'Charlie'];
// To see next without removing, you might do:
let next = line[0]; // but this breaks abstraction

✓ After

class Queue {
  private items: string[] = [];
  peek(): string | undefined {
    return this.items[0];
  }
}

What It Enables

It allows quick access to the next item to process without changing the queue.

Real Life Example

Checking the next customer to serve in a bank queue without making anyone leave the line.

Key Takeaways

Peek shows the front element without removing it.

It keeps the queue order safe and unchanged.

Useful for previewing next item to process.

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 an 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 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 removal [OK]

Common Mistakes:

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

2. Which of the following is the correct TypeScript syntax to peek the front element of a queue implemented as an array named queue?

easy

A. const front = queue.pop();

B. const front = queue.shift();

C. const front = queue.push();

D. const front = queue[0];

Solution

1. Step 1: Understand array methods for queue

   In a queue using an array, the front element is at index 0.

2. Step 2: Identify correct syntax for peek

   Accessing queue[0] returns the front without removing it. pop() removes from end, shift() removes from front, push() adds to end.

3. Final Answer:

   const front = queue[0]; -> Option D

4. Quick Check:

   Peek = queue[0] [OK]

Hint: Peek front is queue[0], not pop or shift [OK]

Common Mistakes:

• Using pop() which removes last element
• Using shift() which removes front element
• Using push() which adds element

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

const queue: number[] = [10, 20, 30];
const front = queue[0];
console.log(front);
console.log(queue);

medium

A. undefined
[10, 20, 30]

B. 10
[20, 30]

C. 10
[10, 20, 30]

D. 30
[10, 20, 30]

Solution

1. Step 1: Peek front element from queue array

   Accessing queue[0] returns 10, the first element.

2. Step 2: Check if queue is modified

   Since we only accessed queue[0], the array remains unchanged as [10, 20, 30].

3. Final Answer:

   10 [10, 20, 30] -> Option C

4. Quick Check:

   Peek returns 10, queue unchanged [OK]

Hint: Peek does not remove, queue stays same [OK]

Common Mistakes:

• Assuming peek removes element
• Confusing front with last element
• Expecting undefined for valid index

4. Identify the error in this TypeScript code that tries to peek the front element of a queue:

class Queue {
  items: number[] = [];
  peek() {
    return this.items.pop();
  }
}

medium

A. pop() removes last element, should use items[0]

B. pop() is correct for peek operation

C. items array should be private

D. peek method should return void

Solution

1. Step 1: Understand pop() behavior

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

2. Step 2: Correct way to peek front

   To peek front, return this.items[0] without removing it.

3. Final Answer:

   pop() removes last element, should use items[0] -> Option A

4. Quick Check:

   pop() ≠ peek front [OK]

Hint: pop() removes last, peek front is items[0] [OK]

Common Mistakes:

• Using pop() for peek
• Not returning front element correctly
• Ignoring array indexing

5. You have a queue implemented as a linked list with nodes having value and next. How do you peek the front element safely when the queue might be empty?

hard

A. Return head.value directly without checks

B. Check if head is null; if not, return head.value, else return null

C. Traverse to the last node and return its value

D. Remove the head node and return its value

Solution

1. Step 1: Understand linked list queue front

   The front element is at the head node of the linked list.

2. Step 2: Handle empty queue case

   If head is null, queue is empty, so return null or undefined safely.

3. Step 3: Return front value safely

   If head exists, return head.value without removing the node.

4. Final Answer:

   Check if head is null; if not, return head.value, else return null -> Option B

5. Quick Check:

   Peek front safely = check head then return value [OK]

Hint: Check head null before peek to avoid errors [OK]

Common Mistakes:

• Not checking for empty queue
• Returning value without null check
• Removing node instead of peeking