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 listdef push ( head_ref, new_data) : # allocate node new_node = Node(0) # put in the data new_node.data = new_data # link the old list of 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 = NoneheadB = NoneheadC = 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 = 25isSumSorted (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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