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
- 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.
- Reverse the second list. In this example the second list becomes 6 -> 5 -> 4 -> NULL.
- 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.
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