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Python Program for Find 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

Python




# Python program to find a triplet 
# from three linked lists with 
# sum equal to a given number 
  
# Link list node 
class Node: 
    def __init__(self, new_data): 
        self.data = new_data 
        self.next = None
  
# A utility function to insert 
# a node at the beginning of a 
# linked list
def push ( head_ref, new_data) :
  
    # allocate node 
    new_node = Node(0)
  
    # put in the data 
    new_node.data = new_data 
  
    # link the old list off the new node 
    new_node.next = (head_ref) 
  
    # move the head to point to the new node 
    (head_ref) = new_node
      
    return head_ref;
  
# 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(headA, headB,headC, givenNumber) :
  
    a = headA 
  
    # Traverse through all nodes of a 
    while (a != None) :
      
        b = headB 
        c = headC 
  
        # For every node of list a, prick two nodes 
        # from lists b abd c 
        while (b != None and c != None) :
          
            # If this a triplet with given sum, print 
            # it and return true 
            sum = a.data + b.data + c.data 
            if (sum == givenNumber) :
              
                print "Triplet Found: " , a.data , " " , b.data , " " , c.data, 
                return True
              
            # If sum of this triplet is smaller, look for 
            # greater values in b 
            elif (sum < givenNumber): 
                b = b.next
            else :# If sum is greater, look for smaller values in c 
                c = c.next
          
        a = a.next # Move ahead in list a 
      
    print("No such triplet"
    return False
  
# Driver code
  
# Start with the empty list 
headA = None
headB = None
headC = None
  
# create a linked list 'a' 10.15.5.20 
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 
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 
headC = push (headC, 1
headC = push (headC, 2
headC = push (headC, 4
headC = push (headC, 8
  
givenNumber = 25
  
isSumSorted (headA, headB, headC, givenNumber) 
  
# This code is contributed by Arnab Kundu

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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