Reverse DLL nodes in groups

Moderate
0/80
Average time to solve is 10m
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Problem statement

You are given a Doubly Linked List of integers and a positive integer 'K' representing the group size. Modify the linked list by reversing every group of 'K' nodes in the linked list.

A doubly linked list is a type of linked list that is bidirectional, that is, it can be traversed in both directions, forward and backward.
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: 8 9 10 11 12
K: 3 

Output: 10 9 8 12 11

We reverse the first 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).
Detailed explanation ( Input/output format, Notes, Images )
Input Format 
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, print 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.
Constraints: 
1 <= T <= 10    
1 <= N <= 5 * 10^4
1 <= K <=  10^5
-10^3 <= data <= 10^3 and data != -1

Time Limit: 1 sec
Sample Input 1:
1 
1 2 3 4 5 6 7 -1
2
Sample Output 1:
2 1 4 3 6 5 7 -1
Explanation of sample input 1:
In the given linked list, we have to reverse the first two nodes, then reverse the next 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 4 3 6 5 7 -1
Sample Input 2:
2
5 10 -6 4 7 -1 
3 
10 20 30 40 50 -1
1
Sample Output 2:
-6 10 5 7 4 -1
10 20 30 40 50 -1
Hint

Can we use recursion to solve this problem?

Approaches (2)
Recursion

The idea is very simple. We will process ‘K’ nodes at a time. Firstly, we will reverse the first ‘K’ nodes of the doubly linked list and then we will do this recursively for the remaining linked list.

Algorithm:

  • If the node does not exist, simply return ‘NULL’.
  • ‘HEAD’ is pointing to the first node of the linked list.
  • If there are less than ‘K’ nodes, just reverse them and return the reversed linked list. Else reverse the first ‘K’ nodes of the linked list.
  • After reversing the ‘K’ nodes, the head points to the Kth node and the Kth node in the original linked list will become the new head.
  • To connect the reversed part with the remaining part of the linked list, update the next of the head to (K + 1)th node and the previous of the (K + 1)th node to the head node.
  • Now, recursively modify the rest of the linked list and link the two sub-linked lists.
  • Return the new head of the linked list.
Time Complexity

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). 

Space Complexity

O(N / 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 ‘K’ nodes of the linked list. Thus, O(N / K) is the total recursion stack space used by the algorithm.

Code Solution
(100% EXP penalty)
Reverse DLL nodes in groups
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