Peek Front Element of Queue in DSA Typescript

Mental Model

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

Analogy: Imagine waiting in line at a coffee shop. Peeking is like glancing at the person at the front of the line to see who will be served next, without letting them leave the line.

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

Dry Run Walkthrough

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

Goal: To see the front element of the queue without removing it

Step 1: Check if the queue is empty

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

Why: We must ensure there is an element to peek at

Step 2: Access the front element value

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

Why: The front element is the next to be served, so we look at it

Step 3: Return the front element value without removing it

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

Why: Peeking means seeing the front without changing the queue

Result:

front -> [1] -> 2 -> 3 -> null
peeked value: 1

Annotated Code

DSA Typescript

class Queue {
  private items: number[] = [];

  enqueue(value: number): void {
    this.items.push(value);
  }

  dequeue(): number | undefined {
    return this.items.shift();
  }

  peek(): number | undefined {
    if (this.items.length === 0) {
      return undefined; // queue is empty
    }
    return this.items[0]; // front element
  }

  print(): void {
    console.log(this.items.join(' -> ') + ' -> null');
  }
}

// Driver code
const queue = new Queue();
queue.enqueue(1);
queue.enqueue(2);
queue.enqueue(3);
queue.print();
const front = queue.peek();
console.log('peeked value:', front);

if (this.items.length === 0) {

check if queue is empty to avoid errors

return this.items[0];

return the front element without removing it

Output

1 -> 2 -> 3 -> null

peeked value: 1

Complexity Analysis

Time: O(1) because accessing the front element is direct without traversal

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

vs Alternative: Compared to dequeue which removes the element, peek only reads it, so peek is faster and does not change the queue

Edge Cases

empty queue

peek returns undefined indicating no elements to view

DSA Typescript

if (this.items.length === 0) {

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 viewing it during peek
Fix: Return the front element without calling dequeue or modifying the queue

Summary

Peek shows the front element of a queue without removing it.

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

The key is to access the first element directly without altering the queue.

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