Traversal Forward and Backward in DSA Go - Time & Space Complexity

Time Complexity: Traversal Forward and Backward

O(n)

Understanding Time Complexity

We want to understand how long it takes to move through a list from start to end and then back again.

How does the time needed change as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

package main

import "fmt"

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

func traverseForwardBackward(head *Node) {
    current := head
    // Traverse forward
    for current != nil {
        fmt.Print(current.value, " -> ")
        current = current.next
    }
    fmt.Print("nil\n")

    // Traverse backward
    // Move to tail first
    tail := head
    for tail != nil && tail.next != nil {
        tail = tail.next
    }
    for tail != nil {
        fmt.Print(tail.value, " -> ")
        tail = tail.prev
    }
    fmt.Print("nil\n")
}

This code moves through a doubly linked list first from the start to the end, then from the end back to the start.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two loops that each visit every node once.
• How many times: Each loop runs once over all nodes, so together they visit nodes twice.

How Execution Grows With Input

As the list grows, the time to go forward and backward grows roughly twice as fast.

Input Size (n)	Approx. Operations
10	About 20 steps (10 forward + 10 backward)
100	About 200 steps
1000	About 2000 steps

Pattern observation: The total steps grow linearly with the number of nodes, doubling because of forward and backward passes.

Final Time Complexity

Time Complexity: O(n)

This means the time to traverse forward and backward grows directly in proportion to the list size.

Common Mistake

[X] Wrong: "Because we traverse twice, the time complexity is O(n²)."

[OK] Correct: Each traversal is separate and linear, so total steps add up to 2n, which is still O(n) because constants don't change growth rate.

Interview Connect

Understanding how traversing data structures forward and backward affects time helps you explain your code clearly and shows you know how to reason about efficiency.

Self-Check

"What if we only traversed forward without going backward? How would the time complexity change?"

Practice

1. What does "Traversal Forward and Backward" mean in the context of a doubly linked list?

easy

A. Visiting nodes from the start to the end and then from the end back to the start

B. Only visiting nodes from the start to the end

C. Only visiting nodes from the end to the start

D. Skipping nodes randomly during traversal

Solution

1. Step 1: Understand forward traversal

   Forward traversal means visiting nodes starting from the first node and moving to the last node.

2. Step 2: Understand backward traversal

   Backward traversal means visiting nodes starting from the last node and moving back to the first node.

3. Final Answer:

   Visiting nodes from the start to the end and then from the end back to the start -> Option A

4. Quick Check:

   Traversal forward and backward = visiting nodes both ways [OK]

Hint: Think of walking forward then turning back [OK]

Common Mistakes:

• Confusing forward traversal with backward traversal
• Thinking traversal only goes one way
• Assuming traversal skips nodes

2. Which of the following Go code snippets correctly traverses a doubly linked list forward?

easy

A. for node := head; node != nil; node = node.prev { fmt.Println(node.value) }

B. for node := head; node != nil; node = node.next { fmt.Println(node.value) }

C. for node := tail; node != nil; node = node.next { fmt.Println(node.value) }

D. for node := tail; node != nil; node = node.prev { fmt.Println(node.value) }

Solution

1. Step 1: Identify forward traversal start point

   Forward traversal starts at the head (first node).

2. Step 2: Identify correct pointer for forward traversal

   Moving forward uses the next pointer, so node = node.next is correct.

3. Final Answer:

   for node := head; node != nil; node = node.next { fmt.Println(node.value) } -> Option B

4. Quick Check:

   Forward traversal uses head and next pointer [OK]

Hint: Forward traversal starts at head and uses next [OK]

Common Mistakes:

• Starting traversal at tail for forward direction
• Using prev pointer for forward traversal
• Confusing next and prev pointers

3. What is the output of the following Go code snippet that traverses a doubly linked list backward?

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

func main() {
  n1 := &Node{value: 1}
  n2 := &Node{value: 2}
  n3 := &Node{value: 3}
  n1.next = n2
  n2.prev = n1
  n2.next = n3
  n3.prev = n2

  for node := n3; node != nil; node = node.prev {
    fmt.Print(node.value, " ")
  }
}

medium

A. 3 1 2

B. 1 2 3

C. 3 2 1

D. 2 3 1

Solution

1. Step 1: Understand backward traversal start

   The loop starts at n3 (value 3) and moves backward using prev pointer.

2. Step 2: Trace the loop values

   First prints 3, then moves to n2 (2), prints 2, then moves to n1 (1), prints 1, then stops at nil.

3. Final Answer:

   3 2 1 -> Option C

4. Quick Check:

   Backward traversal prints 3 2 1 [OK]

Hint: Backward traversal prints values from tail to head [OK]

Common Mistakes:

• Confusing order of printing
• Starting traversal at head for backward
• Using next pointer instead of prev

4. Identify the error in the following Go code that attempts to traverse a doubly linked list backward:

for node := head; node != nil; node = node.prev {
  fmt.Println(node.value)
}

medium

A. Starting traversal at head but moving backward with prev pointer

B. Using fmt.Println instead of fmt.Print

C. Missing initialization of node variable

D. Loop condition should be node == nil

Solution

1. Step 1: Check traversal start and direction

   The code starts at head (first node) but moves backward using prev pointer, which is incorrect because head.prev is nil.

2. Step 2: Identify correct traversal logic

   Backward traversal should start at tail, not head, when moving with prev pointer.

3. Final Answer:

   Starting traversal at head but moving backward with prev pointer -> Option A

4. Quick Check:

   Backward traversal must start at tail [OK]

Hint: Backward traversal starts at tail, not head [OK]

Common Mistakes:

• Starting backward traversal at head
• Confusing loop condition
• Ignoring pointer directions

5. Given a doubly linked list, how would you print all elements first forward and then backward without repeating the last element twice in Go?

hard

A. Traverse backward printing all nodes, then traverse forward starting from head to tail

B. Traverse forward printing all nodes, then traverse backward starting from tail to head

C. Traverse forward printing all nodes, then traverse backward starting from head.prev to tail

D. Traverse forward printing all nodes, then traverse backward starting from tail.prev to head

Solution

1. Step 1: Traverse forward and print all nodes

   Start from head and print each node until tail inclusive.

2. Step 2: Traverse backward starting from tail.prev to avoid duplicate tail

   Start backward traversal from tail.prev to head to avoid printing tail twice.

3. Final Answer:

   Traverse forward printing all nodes, then traverse backward starting from tail.prev to head -> Option D

4. Quick Check:

   Skip last node in backward traversal to avoid duplicate print [OK]

Hint: Skip tail in backward traversal to avoid duplicate print [OK]

Common Mistakes:

• Printing tail twice by starting backward at tail
• Starting backward traversal incorrectly
• Confusing traversal directions