Decoding Interview Problems: Mastering Linked Lists in JavaScript

Today, we will tackle common interview challenges regarding linked lists, focusing on honing your problem-solving skills. Like figuring out the best way to organize a messy drawer, we'll explore ways to streamline and refine our coding techniques when working with linked lists. By the end, you'll have a better grasp of crafting efficient algorithms and writing clean, effective JavaScript code.

In this lesson, we will use the same linked list implementation as in the previous one, but with one additional method to display the list:

show() {
    let currentNode = this.head;
    while (currentNode) {
        console.log(currentNode.value);
        currentNode = currentNode.next;
    }
}

Our journey begins with a linked list with several duplicate values, resembling a situation where an email campaign mistakenly sends multiple invites to the same guests. Our goal is equivalent to ensuring each guest receives only one invitation, which means we must eliminate these duplicate nodes from our linked list.

Much as you would mark songs you have already checked, we can use a Set to keep track of the node values we've seen. When a duplicate value surfaces, we bypass the node altogether, simplifying the process to a manageable single pass through the list or an $O(n)$ operation — much as you would realize you've already heard this song and skip it. This method efficiently preserves our list's uniqueness, much like that perfect, non-repetitive playlist you'd enjoy on a long drive.

Here’s a step-by-step breakdown of how you'd implement such a solution:

function removeDuplicates(list) {
    if (list.head === null || list.head.next === null) return head;

    let currentNode = list.head;
    // Imagine the set as a guest list where we mark off each attendee.
    const seen = new Set([currentNode.value]);

    while (currentNode.next !== null) {
        // Upon encountering a guest who's already marked, we avoid re-inviting them.
        if (seen.has(currentNode.next.value)) {
            currentNode.next = currentNode.next.next;
        } else {
            // A new guest is marked as 'invited'.
            seen.add(currentNode.next.value);
            currentNode = currentNode.next;
        }
    }
}

We kick things off with an empty 'guest list' — our Set. Moving through the linked list, we add each unique value to the set. Whenever we encounter a value already in the set, we skip over the node containing it. This method ensures we only 'invite' each unique value once.