Find a triplet from three linked lists with sum equal to a given number

Given three linked lists, say a, b and c, find one node from each list such that the sum of the values of the nodes is equal to a given number. 
For example, if the three linked lists are 12->6->29, 23->5->8, and 90->20->59, and the given number is 101, the output should be triple "6 5 90".
In the following solutions, size of all three linked lists is assumed same for simplicity of analysis. The following solutions work for linked lists of different sizes also.

A simple method to solve this problem is to run three nested loops. The outermost loop picks an element from list a, the middle loop picks an element from b and the innermost loop picks from c. The innermost loop also checks whether the sum of values of current nodes of a, b and c is equal to given number. The time complexity of this method will be O(n^3).
Sorting can be used to reduce the time complexity to O(n*n). Following are the detailed steps. 
1) Sort list b in ascending order, and list c in descending order. 
2) After the b and c are sorted, one by one pick an element from list a and find the pair by traversing both b and c. See isSumSorted() in the following code. The idea is similar to Quadratic algorithm of 3 sum problem.

Following code implements step 2 only. The solution can be easily modified for unsorted lists by adding the merge sort code discussed here. 

Output: 

Triplet Found: 15 2 8

Time complexity: O(n²)

The linked lists b and c can be sorted in O(nLogn) time using Merge Sort (See this). The step 2 takes O(n*n) time. So the overall time complexity is O(nlogn) + O(nlogn) + O(n*n) = O(n*n). 
In this approach, the linked lists b and c are sorted first, so their original order will be lost. If we want to retain the original order of b and c, we can create copy of b and c. 

Auxiliary Space: O(1)

Here constant extra space is used, but if we wish to maintain the original order of lists b and c our time complexity will become O(n) as extra space will be used to store the sorted list elements.

This article is compiled by Abhinav Priyadarshi and reviewed by GeeksforGeeks team.