Bird
Raised Fist0
DSA Goprogramming~5 mins

Enqueue Using Linked List in DSA Go - Cheat Sheet & Quick Revision

Choose your learning style10 modes available
Recall & Review
beginner
What does 'enqueue' mean in the context of a queue data structure?
Enqueue means adding an element to the end (rear) of the queue.
Click to reveal answer
beginner
Why is a linked list a good choice for implementing a queue?
Because linked lists allow easy addition at the end without shifting elements, making enqueue operations efficient.
Click to reveal answer
intermediate
In enqueue using a linked list, what pointers are updated when adding a new node?
The current rear node's next pointer is set to the new node, and the rear pointer is updated to the new node.
Click to reveal answer
beginner
What happens if you enqueue an element when the queue is empty?
Both the front and rear pointers are set to the new node because it is the only element.
Click to reveal answer
beginner
Show the state of the queue after enqueueing 10, 20, and 30 using a linked list.
10 -> 20 -> 30 -> null
Click to reveal answer
What pointer is updated last when enqueueing a new node in a linked list queue?
AHead pointer
BFront pointer
CNext pointer of the new node
DRear pointer
If the queue is empty, what should the front pointer point to after enqueue?
ANew node
BNext node
COld rear node
DNull
Which operation is typically O(1) when using a linked list for a queue?
ASearching for an element
BEnqueue
CSorting the queue
DDequeuing all elements
What is the next pointer of the new node set to during enqueue?
ANull
BRear node
CFront node
DItself
Which pointer remains unchanged during enqueue if the queue is not empty?
ARear pointer
BNext pointer of rear
CFront pointer
DNew node pointer
Explain step-by-step how to enqueue an element in a queue implemented with a linked list.
Think about what happens when the queue is empty versus when it has elements.
You got /6 concepts.
    Describe the state of the queue after enqueueing elements 5, 15, and 25 using a linked list.
    Visualize the linked nodes and pointers after each enqueue.
    You got /5 concepts.

      Practice

      (1/5)
      1. What does the enqueue operation do in a queue implemented using a linked list?
      easy
      A. Deletes the last element from the linked list
      B. Adds a new element at the end of the linked list
      C. Removes the first element from the linked list
      D. Checks if the linked list is empty

      Solution

      1. Step 1: Understand enqueue operation

        Enqueue means adding an element to the queue.
      2. Step 2: Enqueue in linked list context

        In a linked list queue, enqueue adds the new node at the end (tail).
      3. Final Answer:

        Adds a new element at the end of the linked list -> Option B
      4. Quick Check:

        Enqueue = Add at end [OK]
      Hint: Enqueue always adds at the tail of the linked list [OK]
      Common Mistakes:
      • Confusing enqueue with dequeue operation
      • Adding element at the head instead of tail
      • Thinking enqueue removes elements
      • Assuming enqueue checks emptiness
      2. Which of the following is the correct syntax to create a new node and enqueue it in Go using a linked list?
      easy
      A. node := &Node{data: value, next: nil} queue.tail.next = node queue.tail = node
      B. node := Node{value, nil} queue.head = node
      C. node := &Node{value} queue.head.next = node
      D. node := Node{data: value} queue.tail = node.next

      Solution

      1. Step 1: Create new node correctly

        Use pointer to Node with data and next set to nil.
      2. Step 2: Link new node at tail and update tail

        Set current tail's next to new node, then update tail pointer.
      3. Final Answer:

        node := &Node{data: value, next: nil} queue.tail.next = node queue.tail = node -> Option A
      4. Quick Check:

        New node pointer + update tail = A [OK]
      Hint: Always use pointer and update tail's next and tail itself [OK]
      Common Mistakes:
      • Not using pointer for new node
      • Assigning new node to head instead of tail
      • Forgetting to update tail pointer
      • Incorrect struct initialization syntax
      3. What will be the state of the queue after executing the following Go code?
      type Node struct {
          data int
          next *Node
      }
      
      type Queue struct {
          head *Node
          tail *Node
      }
      
      func (q *Queue) enqueue(value int) {
          node := &Node{data: value}
          if q.tail == nil {
              q.head = node
              q.tail = node
          } else {
              q.tail.next = node
              q.tail = node
          }
      }
      
      func main() {
          q := &Queue{}
          q.enqueue(10)
          q.enqueue(20)
          q.enqueue(30)
          // Print queue
          current := q.head
          for current != nil {
              fmt.Print(current.data, " -> ")
              current = current.next
          }
          fmt.Print("null")
      }
      medium
      A. 30 -> 20 -> 10 -> null
      B. null
      C. 10 -> 30 -> 20 -> null
      D. 10 -> 20 -> 30 -> null

      Solution

      1. Step 1: Enqueue elements in order

        Elements 10, 20, 30 are added at the tail one by one.
      2. Step 2: Traverse from head to tail

        Traversal prints nodes in order: 10, then 20, then 30, then null.
      3. Final Answer:

        10 -> 20 -> 30 -> null -> Option D
      4. Quick Check:

        Enqueue order preserved in linked list [OK]
      Hint: Enqueue adds at tail, print from head to tail [OK]
      Common Mistakes:
      • Assuming new elements added at head
      • Printing tail to head order
      • Forgetting to print 'null' at end
      • Confusing enqueue with dequeue output
      4. Identify the error in the following Go enqueue function for a linked list queue and select the fix.
      func (q *Queue) enqueue(value int) {
          node := &Node{data: value}
          if q.head == nil {
              q.head = node
          } else {
              q.tail.next = node
          }
      }
      medium
      A. Missing update of q.tail to new node after adding
      B. Should check q.tail == nil instead of q.head == nil
      C. Node should be created without pointer
      D. Should assign q.head.next = node instead of q.tail.next

      Solution

      1. Step 1: Analyze enqueue logic

        Function sets head if empty, else links new node after tail.
      2. Step 2: Check tail pointer update

        Function does not update q.tail to new node, so tail remains old node.
      3. Final Answer:

        Missing update of q.tail to new node after adding -> Option A
      4. Quick Check:

        Update tail pointer after enqueue [OK]
      Hint: Always update tail pointer after adding new node [OK]
      Common Mistakes:
      • Not updating tail pointer after enqueue
      • Checking wrong pointer for empty queue
      • Creating node without pointer
      • Linking new node to head instead of tail
      5. You want to implement a queue using a linked list in Go that supports enqueue and dequeue efficiently. Which of the following changes will improve the enqueue operation's time complexity from O(n) to O(1)?
      hard
      A. Add new nodes at the head instead of the tail
      B. Use a doubly linked list instead of singly linked list
      C. Maintain a tail pointer to the last node and update it on enqueue
      D. Traverse the list to find the last node each time before enqueue

      Solution

      1. Step 1: Understand enqueue time complexity

        Without tail pointer, enqueue requires traversal to last node, O(n).
      2. Step 2: Using tail pointer

        Maintaining tail pointer allows direct access to last node, making enqueue O(1).
      3. Final Answer:

        Maintain a tail pointer to the last node and update it on enqueue -> Option C
      4. Quick Check:

        Tail pointer reduces enqueue to O(1) [OK]
      Hint: Keep tail pointer to avoid traversal on enqueue [OK]
      Common Mistakes:
      • Thinking doubly linked list alone improves enqueue
      • Adding at head breaks queue order
      • Traversing list each time causes O(n) complexity
      • Ignoring tail pointer update
      Recommended

      Ready to prove it?

      Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

      10 questions~5 mineasy hard