# C++ Program For Finding 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

## C++

 `// C++ program to find a triplet ``// from three linked lists with ``// sum equal to a given number ``#include ``using` `namespace` `std;`` ` `/* Link list node */``class` `Node ``{ ``    ``public``:``    ``int` `data; ``    ``Node* next; ``}; `` ` `/* A utility function to insert ``a node at the beginning of a ``linked list*/``void` `push (Node** head_ref, ``int` `new_data) ``{ ``    ``/* allocate node */``    ``Node* new_node = ``new` `Node();`` ` `    ``/* 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; ``} `` ` `/* 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. */``bool` `isSumSorted(Node *headA, Node *headB, ``                ``Node *headC, ``int` `givenNumber) ``{ ``    ``Node *a = headA; `` ` `    ``// Traverse through all nodes of a ``    ``while` `(a != NULL) ``    ``{ ``        ``Node *b = headB; ``        ``Node *c = headC; `` ` `        ``// For every node of list a, prick two nodes ``        ``// from lists b abd c ``        ``while` `(b != NULL && c != NULL) ``        ``{ ``            ``// If this a triplet with given sum, print ``            ``// it and return true ``            ``int` `sum = a->data + b->data + c->data; ``            ``if` `(sum == givenNumber) ``            ``{ ``            ``cout << ``"Triplet Found: "` `<< a->data << ``" "` `<< ``                                ``b->data << ``" "` `<< c->data; ``            ``return` `true``; ``            ``} `` ` `            ``// If sum of this triplet is smaller, look for ``            ``// greater values in b ``            ``else` `if` `(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 ``    ``} `` ` `    ``cout << ``"No such triplet"``; ``    ``return` `false``; ``} `` ` ` ` `/* Driver code*/``int` `main() ``{ ``    ``/* Start with the empty list */``    ``Node* headA = NULL; ``    ``Node* headB = NULL; ``    ``Node* headC = NULL; `` ` `    ``/*create a linked list 'a' 10->15->5->20 */``    ``push (&headA, 20); ``    ``push (&headA, 4); ``    ``push (&headA, 15); ``    ``push (&headA, 10); `` ` `    ``/*create a sorted linked list 'b' 2->4->9->10 */``    ``push (&headB, 10); ``    ``push (&headB, 9); ``    ``push (&headB, 4); ``    ``push (&headB, 2); `` ` `    ``/*create another sorted ``    ``linked list 'c' 8->4->2->1 */``    ``push (&headC, 1); ``    ``push (&headC, 2); ``    ``push (&headC, 4); ``    ``push (&headC, 8); `` ` `    ``int` `givenNumber = 25; `` ` `    ``isSumSorted (headA, headB, headC, givenNumber); `` ` `    ``return` `0; ``} `` ` `// This code is contributed by rathbhupendra`

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.

Please refer complete article on Find a triplet from three linked lists with sum equal to a given number for more details!

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