Find Middle Element of Linked List in DSA C - Build from Scratch

Find Middle Element of Linked List

📖 Scenario: You are working with a simple linked list that stores numbers. You want to find the middle number in the list, just like finding the middle book on a shelf.

🎯 Goal: Build a program that creates a linked list with 5 nodes, then finds and prints the middle element's value.

📋 What You'll Learn

Create a linked list with exactly 5 nodes with values 10, 20, 30, 40, 50

Use a pointer to traverse the list

Find the middle element using the slow and fast pointer technique

Print the middle element's value

💡 Why This Matters

🌍 Real World

Finding the middle element is useful in many applications like splitting data, balancing trees, or optimizing searches.

💼 Career

Understanding linked lists and pointer techniques is essential for software developers working with low-level data structures and memory management.

1

Create the linked list with 5 nodes

Create a struct called Node with an int data and a Node* called next. Then create 5 nodes named node1 to node5 with data values 10, 20, 30, 40, and 50 respectively. Link them so node1.next = &node2, node2.next = &node3, and so on, ending with node5.next = NULL. Finally, create a pointer head pointing to &node1.

DSA C

// Define struct Node and create 5 linked nodes with values 10, 20, 30, 40, 50
// Link nodes and set head pointer

Hint

Remember to define the struct first, then create nodes with the given values, link them, and set the head pointer.

2

Create pointers for slow and fast traversal

Inside main, create two pointers called slow and fast and set both to point to head.

DSA C

#include 
#include 

typedef struct Node {
    int data;
    struct Node* next;
} Node;

int main() {
    Node node1 = {10, NULL};
    Node node2 = {20, NULL};
    Node node3 = {30, NULL};
    Node node4 = {40, NULL};
    Node node5 = {50, NULL};

    node1.next = &node2;
    node2.next = &node3;
    node3.next = &node4;
    node4.next = &node5;
    node5.next = NULL;

    Node* head = &node1;

    // Create pointers slow and fast and set to head
    // Your code here

    return 0;
}

Hint

Both slow and fast pointers start at the head of the list.

3

Find the middle element using slow and fast pointers

Use a while loop to move fast two steps and slow one step until fast is NULL or fast->next is NULL. This will make slow point to the middle node.

DSA C

#include 
#include 

typedef struct Node {
    int data;
    struct Node* next;
} Node;

int main() {
    Node node1 = {10, NULL};
    Node node2 = {20, NULL};
    Node node3 = {30, NULL};
    Node node4 = {40, NULL};
    Node node5 = {50, NULL};

    node1.next = &node2;
    node2.next = &node3;
    node3.next = &node4;
    node4.next = &node5;
    node5.next = NULL;

    Node* head = &node1;

    Node* slow = head;
    Node* fast = head;

    // Use while loop to move fast two steps and slow one step
    // Your code here

    return 0;
}

Hint

Move slow by one node and fast by two nodes each loop until fast reaches the end.

4

Print the middle element's value

Use printf to print the data of the node pointed to by slow in the format: "Middle element: %d\n".

DSA C

#include 
#include 

typedef struct Node {
    int data;
    struct Node* next;
} Node;

int main() {
    Node node1 = {10, NULL};
    Node node2 = {20, NULL};
    Node node3 = {30, NULL};
    Node node4 = {40, NULL};
    Node node5 = {50, NULL};

    node1.next = &node2;
    node2.next = &node3;
    node3.next = &node4;
    node4.next = &node5;
    node5.next = NULL;

    Node* head = &node1;

    Node* slow = head;
    Node* fast = head;

    while (fast != NULL && fast->next != NULL) {
        slow = slow->next;
        fast = fast->next->next;
    }

    // Print the middle element's data
    // Your code here

    return 0;
}

Hint

Use printf("Middle element: %d\n", slow->data); to print the middle node's value.

Practice

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

easy

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

B. Reverse the list and pick the first element

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

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

Solution

1. Step 1: Understand two-pointer technique

   One pointer moves one node at a time (slow), the other moves two nodes at a time (fast).

2. Step 2: Identify middle when fast reaches end

   When fast pointer reaches the end, 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 D

4. Quick Check:

   Two-pointer technique = slow one step, fast two steps [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 C code snippets correctly initializes pointers to find the middle of a linked list named head?

easy

A. Node *slow = head; Node *fast = NULL;

B. Node *slow = head; Node *fast = head;

C. Node *slow = head->next; Node *fast = head->next;

D. Node *slow = NULL; Node *fast = head;

Solution

1. Step 1: Initialize both pointers at head

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

2. Step 2: Check pointer initialization correctness

   Only Node *slow = head; Node *fast = head; initializes both pointers correctly at head; others start one pointer at NULL or next node incorrectly.

3. Final Answer:

   Node *slow = head; Node *fast = head; -> Option B

4. Quick Check:

   Both pointers start at head [OK]

Hint: Start both pointers at head node [OK]

Common Mistakes:

• Starting one pointer at NULL
• Starting pointers at next node instead of head
• Not initializing pointers before loop

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

typedef struct Node {
  int data;
  struct Node *next;
} Node;

int main() {
  Node n3 = {30, NULL};
  Node n2 = {20, &n3};
  Node n1 = {10, &n2};
  Node *head = &n1;

  Node *slow = head, *fast = head;
  while (fast != NULL && fast->next != NULL) {
    slow = slow->next;
    fast = fast->next->next;
  }
  printf("%d", slow->data);
  return 0;
}

medium

A. 30

B. 10

C. 20

D. Runtime error

Solution

1. Step 1: Trace pointer movements

   Initially, slow and fast point to 10. After first loop, slow moves to 20, fast moves to 30. Next loop fast->next is NULL, loop ends.

2. Step 2: Identify slow pointer position

   Slow pointer is at node with data 20, which is the middle element.

3. Final Answer:

   20 -> Option C

4. Quick Check:

   Middle element = 20 [OK]

Hint: Slow pointer stops at middle when fast reaches end [OK]

Common Mistakes:

• Assuming middle is first node
• Confusing fast pointer position with slow
• Ignoring loop exit condition

4. Identify the error in the following code snippet intended to find the middle element of a linked list and select the correct fix.

Node *slow = head;
Node *fast = head;
while (fast != NULL) {
  slow = slow->next;
  fast = fast->next->next;
}
return slow->data;

medium

A. Change loop condition to while (fast != NULL && fast->next != NULL)

B. Initialize slow to head->next instead of head

C. Move slow pointer two steps instead of one

D. No error, code is correct

Solution

1. Step 1: Analyze loop condition

   The loop checks only fast != NULL, but fast->next could be NULL causing fast->next->next to crash.

2. Step 2: Correct loop condition

   Adding fast->next != NULL prevents dereferencing NULL pointer, avoiding runtime error.

3. Final Answer:

   Change loop condition to while (fast != NULL && fast->next != NULL) -> Option A

4. Quick Check:

   Check both fast and fast->next not NULL [OK]

Hint: Check both fast and fast->next in loop condition [OK]

Common Mistakes:

• Not checking fast->next before accessing
• Moving slow pointer incorrectly
• Assuming code runs without null checks

5. Given a linked list with an even number of nodes, which node will the two-pointer method return as the middle element? Consider the list: 5 -> 10 -> 15 -> 20 -> NULL

hard

A. 15 (the second middle node)

B. 10 (the first middle node)

C. 5 (the head node)

D. 20 (the last node)

Solution

1. Step 1: Trace pointers on even-length list

   Start slow and fast at 5. After first iteration, slow=10, fast=15. Next iteration, slow=15, fast=NULL (end).

2. Step 2: Identify middle node returned

   When fast reaches NULL, slow points to 15, which is the second middle node in even-length list.

3. Final Answer:

   15 (the second middle node) -> Option A

4. Quick Check:

   Two-pointer method returns second middle in even list [OK]

Hint: Two-pointer method returns second middle in even-length list [OK]

Common Mistakes:

• Assuming first middle is returned
• Confusing head or last node as middle
• Not tracing pointer steps carefully