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# C Program For Finding 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

## C

 `// C program to find a triplet from three linked lists with``// sum equal to a given number``#include``#include``#include` `/* Link list node */``struct` `Node``{``    ``int` `data;``    ``struct` `Node* next;``};` `/* A utility function to insert a node at the beginning of a``   ``linked list*/``void` `push (``struct` `Node** head_ref, ``int` `new_data)``{``    ``/* allocate node */``    ``struct` `Node* new_node =``        ``(``struct` `Node*) ``malloc``(``sizeof``(``struct` `Node));` `    ``/* put in the data */``    ``new_node->data = new_data;` `    ``/* link the old list of 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(``struct` `Node *headA, ``struct` `Node *headB,``                 ``struct` `Node *headC, ``int` `givenNumber)``{``    ``struct` `Node *a = headA;` `    ``// Traverse through all nodes of a``    ``while` `(a != NULL)``    ``{``        ``struct` `Node *b = headB;``        ``struct` `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)``            ``{``               ``printf` `(``"Triplet Found: %d %d %d "``, 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``    ``}` `    ``printf` `(``"No such triplet"``);``    ``return` `false``;``}`  `/* Driver program to test above function*/``int` `main()``{``    ``/* Start with the empty list */``    ``struct` `Node* headA = NULL;``    ``struct` `Node* headB = NULL;``    ``struct` `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;``}`

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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