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C++ Program For Finding A Triplet From Three Linked Lists With Sum Equal To A Given Number

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  • 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++ program to find a triplet 
// from three linked lists with 
// sum equal to a given number 
#include <bits/stdc++.h>
using namespace std;
/* Link list node */
class Node 
    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


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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