Open In App

# 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 ``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 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(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!