What is the brute force approach and complexity for Delete the Middle Node of a Linked List?

Count n, walk n/2 − 1 steps, skip the next node with time complexity O(N) and space complexity O(1).

What is the optimal approach and complexity for Delete the Middle Node of a Linked List?

Start fast at head->next->next so slow stops one node before the middle with optimal time complexity O(N) and space complexity O(1).

Delete the Middle Node of a Linked List

Q: How do you identify the pattern for Delete the Middle Node of a Linked List?

Look for key signals in the prompt: "middle node (index ⌊n/2⌋)" indicates Halfway along a list of unknown length → slow/fast.; "delete / remove" indicates In a singly linked list you remove a node through the node **before** it → stop slow one node early.; "return the head" indicates The head only changes when the list has 1 node, so handle that case first..

Medium·LeetCode #2095 Pro OnlySlight Variation

LeetCode

Problem Statement

You are given the head of a singly linked list with n nodes. Remove the middle node and return the head of the changed list.

The middle node is the one at index ⌊n / 2⌋, counting from 0. So for an even length it is the second of the two middles. For example: n = 1 → index 0, n = 2 → index 1, n = 4 → index 2, n = 5 → index 2.

Input

head, the first node of a singly linked list.

Output

The head of the list after the middle node has been removed (null if nothing is left).

Example 1

Input: head = [1, 3, 4, 7, 1, 2, 6]

Output: [1, 3, 4, 1, 2, 6]

n = 7, so the middle is index 3, the node with value 7. It is removed.

Example 2

Input: head = [1, 2, 3, 4]

Output: [1, 2, 4]

n = 4, so the middle is index 2, the node with value 3.

Example 3

Input: head = [2, 1]

Output: [2]

n = 2, so the middle is index 1, the node with value 1.

Constraints

• The list has between 1 and 10^5 nodes.
• 1 ≤ Node.val ≤ 10^5

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“middle node (index ⌊n/2⌋)”

→ Halfway along a list of unknown length → slow/fast.

“delete / remove”

→ In a singly linked list you remove a node through the node **before** it → stop slow one node early.

“return the head”

→ The head only changes when the list has 1 node, so handle that case first.

Approach Complexity Ladder

Brute Force — count, then walk to the node before the middleO(N)O(1)

Count n, walk n/2 − 1 steps, skip the next node

Optimal — slow/fast with a head startO(N)O(1)

Start fast at head->next->next so slow stops one node before the middle

Brute Force Intuition

Count the nodes to get n. The middle is at index n / 2, so the node before it is at n / 2 − 1. Walk there from the head and set its next to skip the middle.

Recognition rule for this sub-pattern:

When fast reaches the end, slow is in the middle. Use this to split a list in half.

Remember this · Linked List — Fast & Slow

Trigger

Cycle detection, middle of list, "n-th from end", two lists meeting, repeating sequences.

Think

If one pointer moves faster or starts earlier, where will they meet?

Key question

What speed or gap makes the meeting point tell me the answer?

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