Problem of the day
You are given a Singly Linked List of integers and a positive integer 'K'. Modify the linked list by reversing every alternate 'K' nodes of the linked list.
A singly linked list is a type of linked list that is unidirectional, that is, it can be traversed in only one direction from head to the last node (tail).
Note:
If the number of nodes in the list or in the last group is less than 'K', just reverse the remaining nodes.
Example:
Linked list: 5 6 7 8 9 10 11 12
K: 3
Output: 7 6 5 8 9 10 12 11
We reverse the first 'K' (3) nodes and then skip the next 'K'(3) nodes. Now, since the number of nodes remaining in the list (2) is less than 'K', we just reverse the remaining nodes (11 and 12).
Note:
You need to reverse the first 'K' nodes and then skip the 'K' nodes and so on. 5 6 7 10 9 8 11 12 is not the correct answer for the given linked list.
The first line of input contains an integer 'T', the number of test cases.
The first line of every test case contains the elements of the singly linked list separated by a single space and terminated by -1. Hence, -1 would never be a list element.
The second line of every test case contains the positive integer ‘K’.
Output Format:
For every test case, return the modified linked list. The elements of the modified list should be single-space separated, terminated by -1.
Note:
You do not need to print anything; it has already been taken care of. Just implement the given function.
1 <= T <= 10
1 <= N <= 5 * 10^4
1 <= K <= N
-10^3 <= data <= 10^3 and data != -1
Time Limit: 1 sec
1
1 2 3 4 5 6 7 -1
2
2 1 3 4 6 5 7 -1
In the given linked list, we have to reverse the first two nodes, then skip two nodes, and so on… until all the nodes are processed in the same way.
The modified linked list after the above process is 2 1 3 4 6 5 7 -1
2
5 10 -6 4 7 -1
3
10 20 30 40 50 -1
1
-6 10 5 4 7 -1
10 20 30 40 50 -1
Can we use recursion to solve this problem?
The idea is very simple. We will process 2 * ‘K’ nodes at a time. Firstly, we will reverse the first ‘K’ nodes of the linked list and then we will skip the next ‘K’ nodes. We will do this recursively for the remaining linked list.
Algorithm:
O(N), where ‘N’ is the number of nodes in the linked list.
Since each node of the linked list is traversed exactly once, the final time complexity is O(N).
O(N / (2 * K)), where ‘N’ is the number of nodes in the linked list and ‘K’ is the given positive integer.
In each recursive step, we are processing (2 * K) nodes of the linked list. Thus, O(N / (2 * K)) is the total recursion stack space used by the algorithm.