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 numberusing System; public class LinkedList{ public Node head; // head of list /* Linked list Node*/ public class Node { public int data; public Node next; public Node(int d) { data = d; next = null; } } /* 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(LinkedList la, LinkedList lb, LinkedList lc, int givenNumber){ Node a = la.head; // Traverse all nodes of la while (a != null) { Node b = lb.head; Node c = lc.head; // for every node in la pick // 2 nodes from lb and lc while (b != null && c!=null) { int sum = a.data + b.data + c.data; if (sum == givenNumber) { Console.WriteLine("Triplet found " + a.data + " " + b.data + " " + c.data); return true; } // If sum is smaller then // look for greater value of b else if (sum < givenNumber) b = b.next; else c = c.next; } a = a.next; } Console.WriteLine("No Triplet found"); return false;} /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void push(int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; } /* Driver code*/ public static void Main(String []args) { LinkedList llist1 = new LinkedList(); LinkedList llist2 = new LinkedList(); LinkedList llist3 = new LinkedList(); /* Create Linked List llist1 100->15->5->20 */ llist1.push(20); llist1.push(5); llist1.push(15); llist1.push(100); /*create a sorted linked list 'b' 2->4->9->10 */ llist2.push(10); llist2.push(9); llist2.push(4); llist2.push(2); /*create another sorted linked list 'c' 8->4->2->1 */ llist3.push(1); llist3.push(2); llist3.push(4); llist3.push(8); int givenNumber = 25; llist1.isSumSorted(llist1,llist2,llist3,givenNumber); }} // This code contributed by Rajput-Ji |
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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