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Python Program for Find a triplet from three linked lists with sum equal to a given number

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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 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): = new_data = 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_data
    # link the old list of the new node = (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 = + +
            if (sum == givenNumber) :
                print "Triplet Found: " , , " " , , " " ,,
                return True
            # If sum of this triplet is smaller, look for
            # greater values in b
            elif (sum < givenNumber):
                b =
            else :# If sum is greater, look for smaller values in c
                c =
        a = # 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'
headA = push (headA, 20)
headA = push (headA, 4)
headA = push (headA, 15)
headA = push (headA, 10)
# create a sorted linked list 'b'
headB = push (headB, 10)
headB = push (headB, 9)
headB = push (headB, 4)
headB = push (headB, 2)
# create another sorted
# linked list 'c'
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


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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Last Updated : 03 Nov, 2022
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