Zip Linked List From Two Ends Problem Statement

Given a linked list, zip it from its two ends in place, using constant extra space. The nodes in the resulting zipped linked list should go in this order: first, last, second, second last, and so on.

Follow up:

Implement functions to zip two linked lists and to unzip such that unzip(zip(L1, L2)) returns L1 and L2.

Example One

{
"head": [1, 2, 3, 4, 5, 6]
}

Output:

[1, 6, 2, 5, 3, 4]

Example Two

{
"head": [1, 2, 3, 4, 5]
}

Output:

[1, 5, 2, 4, 3]

Notes

• The function has one parameter: head of the given linked list.
• Return the head of zipped linked list.

Constraints:

• 0 <= number of nodes <= 100000
• -2 * 10⁹ <= node value <= 2 * 10⁹

We provided one solution.

Zip Linked List From Two Ends Solution: Optimal

1. Split the given list into halves so that, for example, 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL becomes 1 -> 2 -> 3 -> NULL and 4 -> 5 -> 6 -> NULL.
2. Reverse the second list. In this example the second list becomes 6 -> 5 -> 4 -> NULL.
3. Now merge the two lists by alternating the nodes from first and second lists. Build the resulting list by updating the next pointers of the nodes.

Time Complexity

O(n).

Auxiliary Space Used

O(1).

Space Complexity

O(n).

Code For Zip Linked List From Two Ends Solution: Optimal

/*
Asymptotic complexity in terms of length of given linked list `n`:
* Time: O(n).
* Auxiliary space: O(1).
* Total space: O(n).
*/

// Reverse singly linked list in O(len) time and O(1) space.
LinkedListNode *reverse_linked_list(LinkedListNode *cur)
{
    LinkedListNode *prev = NULL;
    LinkedListNode *next;
    while (cur)
    {
        next = cur->next;
        cur->next = prev;
        prev = cur;
        cur = next;
    }
    return prev;
}

LinkedListNode *zip_given_linked_list(LinkedListNode *head)
{
    if (head == NULL)
    {
        return NULL;
    }
    /*
    Using slow-fast technique find the middle element.
    If head: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL,
    then slow should stop at 3.
    */
    LinkedListNode *slow = head;
    LinkedListNode *fast = head->next;
    while (fast && fast->next)
    {
        slow = slow->next;
        fast = fast->next->next;
    }
    /*
    Separate linked lists from the middle.
    list1: 1 -> 2 -> 3 -> NULL
    list2: 4 -> 5 -> 6 -> NULL
    */
    LinkedListNode *list1 = head;
    LinkedListNode *list2 = slow->next;
    /*
    Till now:
    1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL
    With list1 pointing to 1, list2 pointing to 4 and slow pointing to 3.

    Now break main linked list into two parts.
    So do 3->next = NULL.
    */
    slow->next = NULL;
    /*
    Reverse list2 so that from
    list2: 4 -> 5 -> 6 -> NULL
    it becomes
    list2: 6 -> 5 -> 4 -> NULL
    */
    list2 = reverse_linked_list(list2);
    /*
    For readability we declare two new pointers instead of reusing
    already declared "slow" and "fast", for example.
    */
    LinkedListNode *next1;
    LinkedListNode *next2;
    /*
    Merge list1 and list2.
    list1: 1 -> 2 -> 3 -> NULL
    list2: 6 -> 5 -> 4 -> NULL
    merged: 1 -> 6 -> 2 -> 5 -> 3 -> 4 -> NULL
    */
    while (list2)
    {
        next1 = list1->next;
        next2 = list2->next;
        list1->next = list2;
        list2->next = next1;
        list1 = next1;
        list2 = next2;
    }
    return head;
}

We hope that these solutions to zipping-unzipping two linked lists problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart’s FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

• Back-end Engineering Interview Course
• Front-end Engineering Interview Course
• Full Stack Developer Interview Course

‍To learn more, register for the FREE webinar.