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

Challenge - 5 Problems

❓ Predict Output

intermediate

Find the middle element output of a linked list

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

DSA C++

struct Node {
    int data;
    Node* next;
    Node(int val) : data(val), next(nullptr) {}
};

Node* findMiddle(Node* head) {
    Node* slow = head;
    Node* fast = head;
    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
    }
    return slow;
}

int main() {
    Node* head = new Node(10);
    head->next = new Node(20);
    head->next->next = new Node(30);
    head->next->next->next = new Node(40);
    head->next->next->next->next = new Node(50);

    Node* mid = findMiddle(head);
    if (mid) {
        std::cout << mid->data << std::endl;
    }
    return 0;
}

A 30

B 20

C 40

D 50

💡 Hint

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

The fast pointer moves twice as fast as the slow pointer. When fast reaches the end, slow is at the middle.

For 5 nodes, the middle is the 3rd node with value 30.

❓ Predict Output

intermediate

Output when linked list has even number of nodes

What will be printed by this code when the linked list has 6 nodes?

DSA C++

struct Node {
    int data;
    Node* next;
    Node(int val) : data(val), next(nullptr) {}
};

Node* findMiddle(Node* head) {
    Node* slow = head;
    Node* fast = head;
    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
    }
    return slow;
}

int main() {
    Node* head = new Node(1);
    head->next = new Node(2);
    head->next->next = new Node(3);
    head->next->next->next = new Node(4);
    head->next->next->next->next = new Node(5);
    head->next->next->next->next->next = new Node(6);

    Node* mid = findMiddle(head);
    if (mid) {
        std::cout << mid->data << std::endl;
    }
    return 0;
}

A 3

B 4

C 2

D 5

💡 Hint

For even nodes, the function returns the second middle node.

When the list has even nodes, the slow pointer ends at the second middle node.

Here, nodes are 1 to 6, so middle is node 4.

🔧 Debug

advanced

Identify the error in this middle element finder

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

DSA C++

struct Node {
    int data;
    Node* next;
    Node(int val) : data(val), next(nullptr) {}
};

Node* findMiddle(Node* head) {
    Node* slow = head;
    Node* fast = head->next;
    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
    }
    return slow;
}

int main() {
    Node* head = new Node(1);
    head->next = new Node(2);
    head->next->next = new Node(3);

    Node* mid = findMiddle(head);
    if (mid) {
        std::cout << mid->data << std::endl;
    }
    return 0;
}

A Outputs 3

B Outputs 1

C Outputs 2

D Segmentation fault (runtime error)

💡 Hint

Check where fast pointer starts and how it moves.

Starting fast at head->next causes the slow pointer to stop one node earlier than usual.

For 3 nodes, the middle should be 2, which is correctly output.

❓ Predict Output

advanced

Output of middle element after deleting nodes

Given the linked list and operations below, what is the middle element printed at the end?

DSA C++

struct Node {
    int data;
    Node* next;
    Node(int val) : data(val), next(nullptr) {}
};

Node* findMiddle(Node* head) {
    Node* slow = head;
    Node* fast = head;
    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
    }
    return slow;
}

void deleteNode(Node*& head, int val) {
    if (!head) return;
    if (head->data == val) {
        Node* temp = head;
        head = head->next;
        delete temp;
        return;
    }
    Node* curr = head;
    while (curr->next && curr->next->data != val) {
        curr = curr->next;
    }
    if (curr->next) {
        Node* temp = curr->next;
        curr->next = curr->next->next;
        delete temp;
    }
}

int main() {
    Node* head = new Node(5);
    head->next = new Node(10);
    head->next->next = new Node(15);
    head->next->next->next = new Node(20);
    head->next->next->next->next = new Node(25);

    deleteNode(head, 15);
    deleteNode(head, 5);

    Node* mid = findMiddle(head);
    if (mid) {
        std::cout << mid->data << std::endl;
    }
    return 0;
}

A 20

B 25

C 10

D 15

💡 Hint

After deletions, list is 10 -> 20 -> 25. Find middle of this list.

After deleting nodes with values 15 and 5, the list is 10 -> 20 -> 25.

The middle node is 20.

🧠 Conceptual

expert

Why does the two-pointer method find the middle node?

Why does moving one pointer twice as fast as the other find the middle node of a linked list?

A Because the slow pointer moves twice as fast as the fast pointer.

B Because the slow pointer skips every other node, landing on the middle.

C Because the fast pointer counts the nodes and tells the slow pointer where to stop.

D Because the fast pointer reaches the end in half the steps, so the slow pointer is at the middle.

💡 Hint

Think about relative speeds and positions of pointers.

The fast pointer moves two steps for every one step of the slow pointer.

When fast reaches the end, slow has moved half the distance, so it is at the middle.

Practice

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

easy

A. Use two pointers, one moving twice as fast as the other; when the fast pointer reaches the end, the slow pointer is at the middle.

B. Count the total nodes first, then traverse again to the middle node.

C. Reverse the linked list and pick the first element.

D. Use a stack to store nodes and pop half of them.

Solution

1. Step 1: Understand the two-pointer technique

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

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

   When the fast pointer reaches the end (null), the slow pointer will be at the middle node.

3. Final Answer:

   Use two pointers, one moving twice as fast as the other; when the fast pointer reaches the end, the slow pointer is at the middle. -> Option A

4. Quick Check:

   Two-pointer approach = Fast and slow pointers [OK]

Hint: Move one pointer twice as fast; slow pointer lands in middle [OK]

Common Mistakes:

• Trying to reverse the list unnecessarily
• Counting nodes twice instead of one pass
• Using extra space like stack without need

2. Which of the following C++ code snippets correctly initializes the slow and fast pointers to find the middle of a linked list named head?

easy

A. ListNode* slow = head->next; ListNode* fast = head;

B. ListNode* slow = nullptr; ListNode* fast = head;

C. ListNode* slow = head; ListNode* fast = nullptr;

D. ListNode* slow = head; ListNode* fast = head;

Solution

1. Step 1: Initialize both pointers at the head

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

2. Step 2: Avoid null initialization for slow or fast

   Initializing slow or fast to nullptr will cause errors or skip nodes.

3. Final Answer:

   ListNode* slow = head; ListNode* fast = head; -> Option D

4. Quick Check:

   Both pointers start at head [OK]

Hint: Start both pointers at head node [OK]

Common Mistakes:

• Initializing pointers to nullptr causing runtime errors
• Starting slow pointer at head->next skipping first node
• Starting fast pointer at nullptr causing immediate stop

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

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

Node* findMiddle(Node* head) {
  Node* slow = head;
  Node* fast = head;
  while (fast != nullptr && fast->next != nullptr) {
    slow = slow->next;
    fast = fast->next->next;
  }
  return slow;
}

int main() {
  Node n1 = {1, nullptr};
  Node n2 = {2, nullptr};
  Node n3 = {3, nullptr};
  Node n4 = {4, nullptr};
  Node n5 = {5, nullptr};
  n1.next = &n2;
  n2.next = &n3;
  n3.next = &n4;
  n4.next = &n5;
  Node* mid = findMiddle(&n1);
  std::cout << mid->data << std::endl;
  return 0;
}

medium

A. 3

B. 2

C. 4

D. 5

Solution

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

   Initially, slow=1, fast=1; after first loop, slow=2, fast=3; second loop, slow=3, fast=5; next loop fast->next is nullptr, stop.

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

   Slow pointer is at node with data 3, which is the middle element.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Middle element = 3 [OK]

Hint: Fast moves two steps, slow moves one; slow ends at middle [OK]

Common Mistakes:

• Assuming middle is 2 or 4 instead of 3
• Not checking fast->next before moving fast
• Confusing zero-based indexing with node values

4. Identify the error in the following code snippet that attempts to find the middle element of a linked list and fix it:

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

medium

A. Return fast pointer instead of slow.

B. The while condition should check both fast and fast->next to avoid null pointer dereference.

C. Move slow pointer two steps instead of one.

D. Initialize slow to nullptr instead of head.

Solution

1. Step 1: Analyze the while loop condition

   The loop only checks if fast != nullptr, but fast->next might be nullptr causing a crash when accessing fast->next->next.

2. Step 2: Correct the condition to check fast->next as well

   Change the loop condition to while (fast != nullptr && fast->next != nullptr) to safely move fast pointer two steps.

3. Final Answer:

   The while condition should check both fast and fast->next to avoid null pointer dereference. -> Option B

4. Quick Check:

   Check fast and fast->next before moving fast [OK]

Hint: Always check fast and fast->next in loop condition [OK]

Common Mistakes:

• Not checking fast->next causing runtime errors
• Moving slow pointer two steps incorrectly
• Returning wrong pointer at the end

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

hard

A. Initialize slow pointer to head->next instead of head.

B. Move fast pointer one step at a time instead of two.

C. Initialize fast pointer to head->next instead of head.

D. Use a stack to store nodes and pop half.

Solution

1. Step 1: Understand default behavior for even length lists

   Normally, with both pointers starting at head, slow ends at the second middle node for even length lists.

2. Step 2: Modify fast pointer start to head->next

   Starting fast at head->next causes slow to stop at the first middle node when fast reaches the end.

3. Final Answer:

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

4. Quick Check:

   Fast starts at head->next for first middle [OK]

Hint: Start fast at second node to get first middle in even list [OK]

Common Mistakes:

• Changing slow pointer start instead of fast
• Moving fast pointer one step breaks logic
• Using extra space unnecessarily