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

  • Last Updated : 21 Dec, 2021

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

Javascript




<script>
// 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)
              {
                 document.write("Triplet found " + a.data +
                                     " " + b.data + " " + c.data+"<br>");
                 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;
      }
      document.write("No Triplet found<br>");
      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
</script>
  

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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