Find Middle Element of Linked List in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find the middle element output

What is the output of the following Go code that finds the middle element of a singly linked list?

DSA Go

package main
import "fmt"
type Node struct {
    value int
    next  *Node
}
func findMiddle(head *Node) int {
    slow, fast := head, head
    for fast != nil && fast.next != nil {
        slow = slow.next
        fast = fast.next.next
    }
    return slow.value
}
func main() {
    head := &Node{1, &Node{2, &Node{3, &Node{4, &Node{5, nil}}}}}
    fmt.Println(findMiddle(head))
}

A 5

B 3

C 4

D 2

💡 Hint

Use two pointers: one moves one step, the other moves two steps.

The slow pointer moves one step at a time, fast pointer moves two steps. When fast reaches the end, slow is at the middle. For list 1->2->3->4->5, middle is 3.

❓ Predict Output

intermediate

Middle element with even number of nodes

What is the output of this Go code that finds the middle element of a linked list with an even number of nodes?

DSA Go

package main
import "fmt"
type Node struct {
    value int
    next  *Node
}
func findMiddle(head *Node) int {
    slow, fast := head, head
    for fast != nil && fast.next != nil {
        slow = slow.next
        fast = fast.next.next
    }
    return slow.value
}
func main() {
    head := &Node{10, &Node{20, &Node{30, &Node{40, nil}}}}
    fmt.Println(findMiddle(head))
}

A 30

B 40

C 10

D 20

💡 Hint

For even length, the middle is the second of the two middle nodes.

With even nodes 10->20->30->40, slow pointer ends at 30, which is considered the middle in this approach.

🔧 Debug

advanced

Identify the error in middle element function

What error does this Go code produce when trying to find the middle element of a linked list?

DSA Go

package main
import "fmt"
type Node struct {
    value int
    next  *Node
}
func findMiddle(head *Node) int {
    slow, fast := head, head
    for fast.next != nil && fast.next.next != nil {
        slow = slow.next
        fast = fast.next.next
    }
    return slow.value
}
func main() {
    head := &Node{1, &Node{2, &Node{3, nil}}}
    fmt.Println(findMiddle(head))
}

A compile error: missing return statement

B 3

C 1

D runtime error: invalid memory address or nil pointer dereference

💡 Hint

Check the loop condition for nil pointer safety.

The loop condition accesses fast.next.next without checking if fast != nil, causing a nil pointer dereference at runtime.

🧠 Conceptual

advanced

Why use two pointers to find middle?

Why does using two pointers (slow and fast) help find the middle element of a linked list efficiently?

A Because fast pointer moves twice as fast, slow pointer reaches middle when fast reaches end

B Because slow pointer skips nodes to reach middle faster

C Because fast pointer counts nodes and slow pointer stores values

D Because slow pointer moves randomly to find middle

💡 Hint

Think about relative speeds of pointers.

Fast pointer moves two nodes at a time, slow moves one. When fast reaches end, slow is at middle. This avoids counting nodes first.

❓ Predict Output

expert

Output of middle element function with single node

What is the output of this Go code that finds the middle element of a linked list with only one node?

DSA Go

package main
import "fmt"
type Node struct {
    value int
    next  *Node
}
func findMiddle(head *Node) int {
    slow, fast := head, head
    for fast != nil && fast.next != nil {
        slow = slow.next
        fast = fast.next.next
    }
    return slow.value
}
func main() {
    head := &Node{42, nil}
    fmt.Println(findMiddle(head))
}

A runtime error: invalid memory address or nil pointer dereference

B 0

C 42

D compile error: undefined variable

💡 Hint

Check how the loop behaves with one node.

With one node, fast and slow start at head. Loop condition fails immediately, so slow.value is returned as 42.

Practice

1. What is the main idea behind finding the middle element of a singly linked list using two pointers?

easy

A. Use one slow pointer moving one step and one fast pointer moving two steps until fast reaches the end

B. Count all nodes first, then access the middle by index

C. Use two pointers both moving one step at a time

D. Reverse the list and pick the first element

Solution

1. Step 1: Understand two-pointer technique

   The slow pointer moves one node at a time, while the fast pointer moves two nodes at a time.

2. Step 2: Identify middle when fast pointer reaches end

   When the fast pointer reaches the end of the list, the slow pointer will be at the middle node.

3. Final Answer:

   Use one slow pointer moving one step and one fast pointer moving two steps until fast reaches the end -> Option A

4. Quick Check:

   Two-pointer technique = Use one slow pointer moving one step and one fast pointer moving two steps until fast reaches the end [OK]

Hint: Move fast pointer twice as fast as slow pointer [OK]

Common Mistakes:

• Moving both pointers at the same speed
• Counting nodes first instead of using pointers
• Reversing the list unnecessarily

2. Which of the following Go code snippets correctly initializes two pointers to find the middle of a linked list named head?

easy

A. slow, fast := nil, head

B. slow := head; fast := nil

C. slow, fast := head, head

D. slow, fast := head.next, head.next.next

Solution

1. Step 1: Initialize both pointers at head

   Both slow and fast pointers should start at the head node to traverse the list properly.

2. Step 2: Check pointer initialization correctness

   slow, fast := head, head correctly sets slow and fast to head, allowing traversal from the start.

3. Final Answer:

   slow, fast := head, head -> Option C

4. Quick Check:

   Both pointers start at head = slow, fast := head, head [OK]

Hint: Start both pointers at head node [OK]

Common Mistakes:

• Setting fast pointer to nil initially
• Starting pointers at different nodes
• Skipping nodes at initialization

3. What will be the output of the following Go code snippet that finds the middle element of a linked list?

type Node struct {
    val  int
    next *Node
}

func findMiddle(head *Node) int {
    slow, fast := head, head
    for fast != nil && fast.next != nil {
        slow = slow.next
        fast = fast.next.next
    }
    return slow.val
}

// List: 10 -> 20 -> 30 -> 40 -> 50 -> nil
head := &Node{10, &Node{20, &Node{30, &Node{40, &Node{50, nil}}}}}
fmt.Println(findMiddle(head))

medium

A. 30

B. 20

C. 40

D. 50

Solution

1. Step 1: Trace slow and fast pointers through the list

   Initially, slow=10, fast=10. After first loop: slow=20, fast=30. Second loop: slow=30, fast=50. Third loop ends as fast.next is nil.

2. Step 2: Identify slow pointer value at loop end

   Slow pointer is at node with value 30, which is the middle element.

3. Final Answer:

   30 -> Option A

4. Quick Check:

   Middle element = 30 [OK]

Hint: Trace slow pointer moves until fast reaches end [OK]

Common Mistakes:

• Returning fast pointer value instead of slow
• Stopping loop too early
• Confusing node values

4. The following Go code is intended to find the middle element of a linked list but has a bug. What is the bug?

func findMiddle(head *Node) int {
    slow, fast := head, head
    for fast.next != nil && fast.next.next != nil {
        slow = slow.next
        fast = fast.next.next
    }
    return slow.val
}

medium

A. The slow pointer should move two steps instead of one

B. The pointers should be initialized to nil

C. The function should return fast.val instead of slow.val

D. The loop condition should check fast != nil before fast.next

Solution

1. Step 1: Analyze loop condition safety

   The loop checks fast.next and fast.next.next but does not check if fast itself is nil, risking a runtime error if fast is nil.

2. Step 2: Correct loop condition to avoid nil pointer dereference

   The condition should first check fast != nil, then fast.next != nil to safely access fast.next.next.

3. Final Answer:

   The loop condition should check fast != nil before fast.next -> Option D

4. Quick Check:

   Check fast != nil first to avoid errors [OK]

Hint: Always check pointer != nil before accessing next [OK]

Common Mistakes:

• Ignoring nil checks causing runtime panic
• Moving slow pointer incorrectly
• Returning wrong pointer value

5. Given a singly linked list with an even number of nodes, how can you modify the two-pointer approach to return the second middle element instead of the first middle element?

hard

A. Reverse the list and pick the middle element

B. Initialize fast pointer to head.next instead of head

C. Return fast pointer value instead of slow pointer value

D. Move slow pointer two steps and fast pointer one step

Solution

1. Step 1: Understand middle element in even-length list

   For even nodes, two middle elements exist; the second middle is the one after the first middle.

2. Step 2: Modify fast pointer start to shift middle selection

   Starting fast pointer at head.next causes slow pointer to move to the second middle when fast reaches the end.

3. Final Answer:

   Initialize fast pointer to head.next instead of head -> Option B

4. Quick Check:

   Start fast at head.next for second middle [OK]

Hint: Start fast pointer one step ahead for second middle [OK]

Common Mistakes:

• Moving pointers incorrectly
• Returning fast pointer value
• Reversing list unnecessarily