Javascript Program For Finding A Triplet From Three Linked Lists With Sum Equal To A Given Number
Last Updated :
23 Jul, 2025
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
OutputTriplet 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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