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 <bits/stdc++.h> 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!