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

Last Updated : 05 Sep, 2024

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.

Following code implements step 2 only. The solution can be easily modified for unsorted lists by adding the merge sort code discussed here.

JavaScript

// Javascript program to find a triplet from three linked lists with
// sum equal to a given number

/* Linked list Node*/
class Node {
    constructor(d) {
        this.data = d;
        this.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. */
function isSumSorted(la, lb, lc, givenNumber) {
    let a = la;

    // Traverse all nodes of la
    while (a != null) {
        let b = lb;
        let c = lc;

        // for every node in la pick 2 nodes from lb and lc
        while (b != null && c != null) {
            let 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;
}

/*  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. */
function push(head_ref, new_data) {
    /* 1 & 2: Allocate the Node &
                  Put in the data*/
    let new_node = new Node(new_data);

    /* 3. Make next of new Node as head */
    new_node.next = (head_ref);

    (head_ref) = new_node;

    return head_ref;

}

let headA = null;
headA = push(headA, 20)
headA = push(headA, 4)
headA = push(headA, 15)
headA = push(headA, 10)

// create a sorted linked list 'b' 2.4.9.10 
let headB = null;
headB = push(headB, 10)
headB = push(headB, 9)
headB = push(headB, 4)
headB = push(headB, 2)

// create another sorted 
// linked list 'c' 8.4.2.1 
let headC = null;
headC = push(headC, 1)
headC = push(headC, 2)
headC = push(headC, 4)
headC = push(headC, 8)

let givenNumber = 25

isSumSorted(headA, headB, headC, givenNumber)


// This code is contributed by avanitrachhadiya2155

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!