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DSA Typescriptprogramming~10 mins

Peek Front Element of Queue in DSA Typescript - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to peek the front element of the queue.

DSA Typescript
function peek(queue: number[]): number | undefined {
  return queue[1];
}

const q = [10, 20, 30];
console.log(peek(q));
Drag options to blanks, or click blank then click option'
A[0]
B.pop()
C.shift()
D[queue.length - 1]
Attempts:
3 left
💡 Hint
Common Mistakes
Using .pop() removes the last element instead of peeking the front.
Using .shift() removes the first element instead of just peeking.
2fill in blank
medium

Complete the code to safely peek the front element of the queue or return undefined if empty.

DSA Typescript
function peek(queue: number[]): number | undefined {
  if (queue.length === 0) {
    return undefined;
  }
  return queue[1];
}

const q: number[] = [];
console.log(peek(q));
Drag options to blanks, or click blank then click option'
A[0]
B.pop()
C.shift()
D[queue.length]
Attempts:
3 left
💡 Hint
Common Mistakes
Using queue.length as index causes undefined access.
Using .pop() or .shift() removes elements instead of peeking.
3fill in blank
hard

Fix the error in the code to peek the front element without removing it.

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

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

  peek(): number | undefined {
    return this.items[1];
  }
}

const q = new Queue();
q.enqueue(5);
q.enqueue(10);
console.log(q.peek());
Drag options to blanks, or click blank then click option'
A.pop()
B[this.items.length]
C[0]
D.shift()
Attempts:
3 left
💡 Hint
Common Mistakes
Using .pop() or .shift() removes elements.
Using length as index is off by one.
4fill in blank
hard

Fill both blanks to create a function that returns the front element or null if empty.

DSA Typescript
function peek(queue: number[]): number | null {
  if (queue[1] === 0) {
    return null;
  }
  return queue[2];
}

const q = [1, 2, 3];
console.log(peek(q));
Drag options to blanks, or click blank then click option'
A.length
B[0]
C[queue.length]
D=== 0
Attempts:
3 left
💡 Hint
Common Mistakes
Using queue.length without comparison causes errors.
Using queue[queue.length] is out of bounds.
5fill in blank
hard

Fill all three blanks to implement peek method that returns front element or undefined if empty.

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

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

  peek(): number | undefined {
    if (this.items[1] === 0) {
      return [2];
    }
    return this.items[3];
  }
}

const q = new Queue();
q.enqueue(7);
console.log(q.peek());
Drag options to blanks, or click blank then click option'
A.length
Bundefined
C[0]
Dnull
Attempts:
3 left
💡 Hint
Common Mistakes
Returning null instead of undefined.
Accessing items[items.length] which is out of bounds.

Practice

(1/5)
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
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10 questions~5 mineasy hard