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++
#include <bits/stdc++.h> using namespace std; // CPP program to find a triplet from three linked lists // with sum equal to a given number /* Linked list Node*/ class Node { public : int data; Node* next; Node( int d) { this ->data = d; this ->next = NULL; } }; Node* head; // head of list /* 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*& la, Node* lb, Node* lc, int givenNumber) { Node* a = la; // Traverse all nodes of la while (a != NULL) { Node* b = lb; Node* c = lc; // 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) { cout << "Triplet Found: " << a->data << " " << b->data << " " << c->data << endl; 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; } cout << "No Triplet found" << endl; 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(Node*& list, 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 = list; /* 4. Move the head to point to new Node */ list = new_node; } int main() { Node* list1 = new Node(20); Node* list2 = new Node(10); Node* list3 = new Node(1); /* Create Linked List llist1 100->15->5->20 */ // list1.push(20); push(list1, 5); push(list1, 15); push(list1, 100); /*create a sorted linked list 'b' 2->4->9->10 */ // list2.push(10); push(list2, 9); push(list2, 4); push(list2, 2); /*create another sorted linked list 'c' 8->4->2->1 */ // list3.push(1); push(list3, 2); push(list3, 4); push(list3, 8); int givenNumber = 25; isSumSorted(list1, list2, list3, givenNumber); return 0; } // This code is contributed by akashish__ |
Java
// Java program to find a triplet from three linked lists // with sum equal to a given number public class LinkedList { /* Linked list Node*/ public class Node { int data; Node next; public Node( int d) { this .data = d; next = null ; } } public Node head; // head of list /* 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. */ public boolean 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) { System.out.println( "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; } System.out.println( "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. */ public 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; } public static void main(String[] args) { LinkedList list1 = new LinkedList(); LinkedList list2 = new LinkedList(); LinkedList list3 = new LinkedList(); /* Create Linked List llist1 100->15->5->20 */ list1.push( 20 ); list1.push( 5 ); list1.push( 15 ); list1.push( 100 ); /*create a sorted linked list 'b' 2->4->9->10 */ list2.push( 10 ); list2.push( 9 ); list2.push( 4 ); list2.push( 2 ); /*create another sorted linked list 'c' 8->4->2->1 */ list3.push( 1 ); list3.push( 2 ); list3.push( 4 ); list3.push( 8 ); int givenNumber = 25 ; list1.isSumSorted(list1, list2, list3, givenNumber); } } // This code is contributed by lokesh (lokeshmvs21). |
Python3
# Python program to find a triplet # from three linked lists with # sum equal to a given number # Node class class Node: def __init__( self , data): self .data = data self . next = None # Linked List class class LinkedList: def __init__( self ): self .head = None # function to add new node at the beginning def push( self , new_data): new_node = Node(new_data) new_node. next = self .head self .head = 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. def isSumSorted( self , la, lb, lc, givenNumber): a = la.head # Traverse all nodes of la while a is not None : b = lb.head c = lc.head # for every node in la pick # 2 nodes from lb and lc while b is not None and c is not None : sum = a.data + b.data + c.data if sum = = givenNumber: print ( "Triplet found" , a.data, b.data, c.data) return True # If sum is smaller then # look for greater value of b elif sum < givenNumber: b = b. next else : c = c. next a = a. next print ( "No Triplet found" ) return False # Driver code if __name__ = = '__main__' : llist1 = LinkedList() llist2 = LinkedList() llist3 = 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 ) givenNumber = 25 llist1.isSumSorted(llist1, llist2, llist3, givenNumber) # This code is contributed by akashish__ |
C#
// C# program to find a triplet // from three linked lists with // sum equal to a given number using 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 |
Javascript
// Javascript code for the above approach class Node { constructor(data) { this .data = data; this .next = null ; } } class LinkedList { constructor() { this .head = null ; } // function to add new node at the beginning push(newData) { const newNode = new Node(newData); newNode.next = this .head; this .head = newNode; } // 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. isSumSorted(la, lb, lc, givenNumber) { let a = la.head; // Traverse all nodes of la while (a !== null ) { let b = lb.head; let c = lc.head; // for every node in la pick // 2 nodes from lb and lc while (b !== null && c !== null ) { const sum = a.data + b.data + c.data; if (sum === givenNumber) { console.log( "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.log( "No Triplet found" ); return false ; } } // Driver code if ( typeof exports !== "undefined" ) { exports.LinkedList = LinkedList; } // usage const llist1 = new LinkedList(); const llist2 = new LinkedList(); const 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); const givenNumber = 25; llist1.isSumSorted(llist1, llist2, llist3, givenNumber); // This code is contributed by lokeshpotta20. |
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.
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