Sudo Placement[1.4] | K Sum

Given head of a linked list of integers and an integer k, your task is to modify the linked list as follows:

• Consider nodes in groups of size k. In every group, replace value of first node with group sum.
• Also, delete the elements of group except the first node.
• During traversal, if the remaining nodes in linked list are less than k then also do the above considering remaining nodes.

Examples: 

Input: N = 6, K = 2 
1->2->3->4->5->6 
Output: 3 7 11
We have denoted grouping of k elements by (). The elements inside () are summed. 
1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null 
(1 -> 2) -> 3 -> 4 -> 5 -> 6 -> null 
(3) -> 3 -> 4 -> 5 -> 6 -> null 
3 -> (3 -> 4) -> 5 -> 6 -> null 
3 -> (7) -> 5 -> 6 -> null 
3 -> 7 -> (5 -> 6) -> null 
3 -> 7 -> (11) -> null 
3 -> 7 -> 11 -> null 

Approach: 

Insert the given nodes in the Linked list. The approach of insertion has been discussed in this post. Once the nodes are inserted, iterate from the beginning of the list. Mark the first node as temp node. Iterate for the next k-1 nodes and sum up the nodes in sum variable. If the number of nodes is less than K, then replace the temp node's data with sum and point temp to NULL. 

If there is a group with K nodes, replace the temp's data with sum and move temp to the node which is just after K nodes. Continue the above operation till temp points to NULL. Once temp reaches the end, it means all groups of size K has been traversed and the node has been replaced with the sum of nodes of size K. Once the operation of replacements have been done, the linked list thus formed can be printed. The method of printing the linked list has been discussed in this post. 

Below is the implementation of the above approach:  

3 7 11