LeetCode Challenge Day 47 — 3217. Delete Nodes From Linked List Present in Array

Hey folks 👋

This is Day 47 of my LeetCode streak 🚀
Today’s problem is 3217. Delete Nodes From Linked List Present in Array — a medium-level linked list problem that focuses on efficiently removing nodes based on an array of values.

📌 Problem Statement

You are given an array of integers nums and the head of a linked list.
Return the head of the modified linked list after removing all nodes whose values exist in nums.

Example 1:

Input: nums = [1,2,3], head = [1,2,3,4,5]
Output: [4,5]

Example 2:

Input: nums = [1], head = [1,2,1,2,1,2]
Output: [2,2,2]

Example 3:

Input: nums = [5], head = [1,2,3,4]
Output: [1,2,3,4]

💡 Intuition

The key insight is that repeatedly checking if a node’s value exists in nums could be slow if we use an array lookup.
To make this efficient, we can store all elements of nums in a Set, allowing for O(1) lookups when determining whether a node should be deleted.

🔑 Approach

1. Convert the array nums into a Set for constant-time value lookup.
2. Create a dummy node that points to the head — this simplifies removal of the head node when necessary.
3. Traverse the linked list using a pointer current.
   • If current.next.val exists in the set, skip the node (current.next = current.next.next).
   • Otherwise, move to the next node.
4. Continue until all nodes are processed.
5. Return dummy.next as the new head of the modified list.

⏱️ Complexity Analysis

• Time complexity:
  (O(n + m)) — where n is the number of linked list nodes and m is the number of elements in nums.

• Space complexity:
  (O(m)) — for storing nums in a Set.

🧑‍💻 Code (JavaScript)

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */

/**
 * @param {number[]} nums
 * @param {ListNode} head
 * @return {ListNode}
 */
var modifiedList = function(nums, head) {
    const removeSet = new Set(nums);
    const dummy = new ListNode(0, head);
    let current = dummy;

    while (current.next) {
        if (removeSet.has(current.next.val)) {
            current.next = current.next.next;
        } else {
            current = current.next;
        }
    }

    return dummy.next;
};

🧪 Example Walkthrough

Input:
nums = [1,2,3], head = [1,2,3,4,5]

Steps:

1. Convert nums → 3
2. Start with dummy → 0 → 1 → 2 → 3 → 4 → 5
3. Remove 1, 2, and 3 since they exist in the Set
4. Remaining list: 4 → 5

✅ Output = [4,5]

🎯 Reflection

This problem highlights how hash-based lookups and a dummy node pattern can simplify linked list manipulation. It’s an elegant one-pass solution that’s both clean and efficient — a must-know technique for any linked list problem!

That’s it for Day 47 of my LeetCode journey 💪 See you tomorrow for Day 48 — let’s keep building consistency and clarity in problem-solving! 🔥

Happy Coding 👨‍💻