# Python Program for Find a triplet from three linked lists with sum equal to a given number

• Last Updated : 21 Dec, 2021

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

## Python

 `# Python program to find a triplet ``# from three linked lists with ``# sum equal to a given number `` ` `# Link list node ``class` `Node: ``    ``def` `__init__(``self``, new_data): ``        ``self``.data ``=` `new_data ``        ``self``.``next` `=` `None`` ` `# A utility function to insert ``# a node at the beginning of a ``# linked list``def` `push ( head_ref, new_data) :`` ` `    ``# allocate node ``    ``new_node ``=` `Node(``0``)`` ` `    ``# put in the data ``    ``new_node.data ``=` `new_data `` ` `    ``# link the old list off the new node ``    ``new_node.``next` `=` `(head_ref) `` ` `    ``# move the head to point to the new node ``    ``(head_ref) ``=` `new_node``     ` `    ``return` `head_ref;`` ` `# A function to check if there are three elements in a, b ``# and c whose sum is equal to givenNumber. The function ``# assumes that the list b is sorted in ascending order ``# and c is sorted in descending order. ``def` `isSumSorted(headA, headB,headC, givenNumber) :`` ` `    ``a ``=` `headA `` ` `    ``# Traverse through all nodes of a ``    ``while` `(a !``=` `None``) :``     ` `        ``b ``=` `headB ``        ``c ``=` `headC `` ` `        ``# For every node of list a, prick two nodes ``        ``# from lists b abd c ``        ``while` `(b !``=` `None` `and` `c !``=` `None``) :``         ` `            ``# If this a triplet with given sum, print ``            ``# it and return true ``            ``sum` `=` `a.data ``+` `b.data ``+` `c.data ``            ``if` `(``sum` `=``=` `givenNumber) :``             ` `                ``print` `"Triplet Found: "` `, a.data , ``" "` `, b.data , ``" "` `, c.data, ``                ``return` `True``             ` `            ``# If sum of this triplet is smaller, look for ``            ``# greater values in b ``            ``elif` `(``sum` `< givenNumber): ``                ``b ``=` `b.``next``            ``else` `:``# If sum is greater, look for smaller values in c ``                ``c ``=` `c.``next``         ` `        ``a ``=` `a.``next` `# Move ahead in list a ``     ` `    ``print``(``"No such triplet"``) ``    ``return` `False`` ` `# Driver code`` ` `# Start with the empty list ``headA ``=` `None``headB ``=` `None``headC ``=` `None`` ` `# create a linked list 'a' 10.15.5.20 ``headA ``=` `push (headA, ``20``) ``headA ``=` `push (headA, ``4``) ``headA ``=` `push (headA, ``15``) ``headA ``=` `push (headA, ``10``) `` ` `# create a sorted linked list 'b' 2.4.9.10 ``headB ``=` `push (headB, ``10``) ``headB ``=` `push (headB, ``9``) ``headB ``=` `push (headB, ``4``) ``headB ``=` `push (headB, ``2``) `` ` `# create another sorted ``# linked list 'c' 8.4.2.1 ``headC ``=` `push (headC, ``1``) ``headC ``=` `push (headC, ``2``) ``headC ``=` `push (headC, ``4``) ``headC ``=` `push (headC, ``8``) `` ` `givenNumber ``=` `25`` ` `isSumSorted (headA, headB, headC, givenNumber) `` ` `# This code is contributed by Arnab Kundu`

Output:

`Triplet Found: 15 2 8`

Time complexity: 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.

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