Why Doubly Linked List Over Singly Linked List in DSA Go - See It Work

Why Doubly Linked List Over Singly Linked List

📖 Scenario: Imagine you are managing a playlist of songs on a music player. You want to move forward and backward through the songs easily.A singly linked list lets you move forward only, but a doubly linked list lets you move both forward and backward.

🎯 Goal: Build a simple doubly linked list in Go to see how it allows moving both forward and backward through the list, unlike a singly linked list.

📋 What You'll Learn

Create a doubly linked list node struct with data, next, and prev pointers

Create a doubly linked list with three nodes containing song names

Traverse the list forward and print the song names

Traverse the list backward and print the song names

💡 Why This Matters

🌍 Real World

Music players, web browsers, and undo-redo features use doubly linked lists to move forward and backward through items.

💼 Career

Understanding doubly linked lists helps in software development roles that involve data navigation, editing, and memory management.

1

Create Doubly Linked List Nodes

Create a struct called Node with fields: data of type string, next and prev both of type *Node. Then create three nodes named node1, node2, and node3 with data values "Song1", "Song2", and "Song3" respectively.

DSA Go

// Define Node struct and create three nodes node1, node2, node3 with data "Song1", "Song2", "Song3"
// Your code here

Hint

Define the struct with three fields. Create nodes using &Node{data: "Song1"} syntax.

2

Link Nodes Forward and Backward

Link the nodes to form a doubly linked list by setting node1.next = node2, node2.prev = node1, node2.next = node3, and node3.prev = node2.

DSA Go

package main

import "fmt"

type Node struct {
	data string
	next *Node
	prev *Node
}

func main() {
	node1 := &Node{data: "Song1"}
	node2 := &Node{data: "Song2"}
	node3 := &Node{data: "Song3"}

	// Link nodes forward and backward
	// Your code here
}

Hint

Set the next pointer of one node to the next node, and the prev pointer of the next node back to the previous node.

3

Traverse Forward and Backward

Create two variables current and last. Use a for loop to traverse forward from node1 using current = current.next and print each current.data. Then use another for loop to traverse backward from last using last = last.prev and print each last.data.

DSA Go

package main

import "fmt"

type Node struct {
	data string
	next *Node
	prev *Node
}

func main() {
	node1 := &Node{data: "Song1"}
	node2 := &Node{data: "Song2"}
	node3 := &Node{data: "Song3"}

	node1.next = node2
	node2.prev = node1
	node2.next = node3
	node3.prev = node2

	// Traverse forward and print data
	// Traverse backward and print data
	// Your code here
}

Hint

Use a loop to move forward until current is nil, printing data. Then use another loop to move backward from last.

4

Print Forward and Backward Traversal

Run the program to print the song names in forward order first, then backward order.

DSA Go

package main

import "fmt"

type Node struct {
	data string
	next *Node
	prev *Node
}

func main() {
	node1 := &Node{data: "Song1"}
	node2 := &Node{data: "Song2"}
	node3 := &Node{data: "Song3"}

	node1.next = node2
	node2.prev = node1
	node2.next = node3
	node3.prev = node2

	current := node1
	var last *Node

	for current != nil {
		fmt.Println(current.data)
		last = current
		current = current.next
	}

	for last != nil {
		fmt.Println(last.data)
		last = last.prev
	}

	// Program output should show forward and backward traversal
}

Hint

Run the program and check the printed song names forward then backward.

Practice

1. Why might you choose a doubly linked list over a singly linked list in a program?

easy

A. Because it uses less memory per node

B. Because it only stores one pointer per node

C. Because it is simpler to implement

D. Because it allows traversal in both directions easily

Solution

1. Step 1: Understand traversal capabilities

   A doubly linked list stores two pointers per node: one to the next node and one to the previous node, allowing movement forward and backward.

2. Step 2: Compare with singly linked list

   A singly linked list stores only one pointer to the next node, so it can only be traversed forward easily.

3. Final Answer:

   Because it allows traversal in both directions easily -> Option D

4. Quick Check:

   Traversal both ways = B [OK]

Hint: Doubly means two directions for traversal [OK]

Common Mistakes:

• Thinking doubly linked lists use less memory
• Assuming singly linked lists allow backward traversal
• Confusing simplicity of implementation

2. Which of the following is the correct way to declare a node struct for a doubly linked list in Go?

easy

A. type Node struct { data int; prev *Node; next *Node }

B. type Node struct { data int; next *Node }

C. type Node struct { data int; prev int; next int }

D. type Node struct { data int; next int }

Solution

1. Step 1: Identify required pointers for doubly linked list node

   A doubly linked list node must have pointers to both previous and next nodes, so it needs two pointers of type *Node.

2. Step 2: Check each option's struct fields

   type Node struct { data int; prev *Node; next *Node } correctly declares both prev *Node and next *Node. Others miss one or use wrong types.

3. Final Answer:

   type Node struct { data int; prev *Node; next *Node } -> Option A

4. Quick Check:

   Two pointers prev and next = A [OK]

Hint: Doubly linked node needs prev and next pointers [OK]

Common Mistakes:

• Using int instead of pointer for prev/next
• Missing prev pointer in struct
• Confusing singly linked list node with doubly linked

3. What will be the output of the following Go code snippet that uses a doubly linked list?

type Node struct {
    data int
    prev *Node
    next *Node
}

func printForward(head *Node) {
    for current := head; current != nil; current = current.next {
        fmt.Print(current.data, " ")
    }
}

func printBackward(tail *Node) {
    for current := tail; current != nil; current = current.prev {
        fmt.Print(current.data, " ")
    }
}

func main() {
    n1 := &Node{data: 1}
    n2 := &Node{data: 2}
    n3 := &Node{data: 3}

    n1.next = n2
    n2.prev = n1
    n2.next = n3
    n3.prev = n2

    printForward(n1)
    fmt.Println()
    printBackward(n3)
}

medium

A. 1 2 3 \n 1 2 3

B. 3 2 1 \n 1 2 3

C. 1 2 3 \n 3 2 1

D. 3 2 1 \n 3 2 1

Solution

1. Step 1: Trace printForward function

   Starting at n1 (data=1), it prints 1, then moves to n2 (2), then n3 (3), then nil. Output: "1 2 3 "

2. Step 2: Trace printBackward function

   Starting at n3 (3), it prints 3, then moves to n2 (2), then n1 (1), then nil. Output: "3 2 1 "

3. Final Answer:

   1 2 3 \n 3 2 1 -> Option C

4. Quick Check:

   Forward then backward print = A [OK]

Hint: Forward prints from head, backward from tail [OK]

Common Mistakes:

• Confusing prev and next pointers in loops
• Printing backward list in forward order
• Missing nil check causing runtime error

4. The following Go code tries to insert a new node after a given node in a doubly linked list. What is the bug?

func insertAfter(node *Node, newNode *Node) {
    newNode.next = node.next
    node.next = newNode
    newNode.prev = node
}

medium

A. It does not set node.next correctly

B. It does not update the prev pointer of the node after newNode

C. It does not set newNode.prev correctly

D. It does not set newNode.next correctly

Solution

1. Step 1: Analyze pointer updates in insertAfter

   The code sets newNode.next to node.next and newNode.prev to node, and updates node.next to newNode.

2. Step 2: Identify missing pointer update

   If node.next was not nil, the previous pointer of node.next (the node after newNode) should be updated to newNode. This step is missing.

3. Final Answer:

   It does not update the prev pointer of the node after newNode -> Option B

4. Quick Check:

   Missing prev update of next node = C [OK]

Hint: Always update prev pointer of next node after insertion [OK]

Common Mistakes:

• Forgetting to update next node's prev pointer
• Confusing prev and next pointers
• Assuming node.next is always nil

5. You want to implement a feature that requires frequent backward traversal and deletion of nodes from the middle of a list. Why is a doubly linked list better than a singly linked list for this task?

hard

A. Because doubly linked lists allow backward traversal and easy access to previous nodes for deletion

B. Because singly linked lists use less memory and are faster for deletion

C. Because doubly linked lists do not need pointers to previous nodes

D. Because singly linked lists allow backward traversal without extra pointers

Solution

1. Step 1: Understand backward traversal requirement

   Backward traversal means moving from a node to its previous node, which singly linked lists cannot do easily because they only have next pointers.

2. Step 2: Understand deletion from middle nodes

   To delete a node, you need to update the previous node's next pointer. Doubly linked lists have direct access to previous nodes, making deletion easier and efficient.

3. Final Answer:

   Because doubly linked lists allow backward traversal and easy access to previous nodes for deletion -> Option A

4. Quick Check:

   Backward traversal and deletion ease = D [OK]

Hint: Backward traversal needs prev pointers for easy deletion [OK]

Common Mistakes:

• Thinking singly linked lists can traverse backward easily
• Ignoring need to update previous node on deletion
• Assuming doubly linked lists use less memory