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.
// Java program to find a triplet from three linked lists with // sum equal to a given number class LinkedList
{ Node head; // head of list
/* Linked list Node*/
class Node
{
int data;
Node next;
Node( int d) {data = d; 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. */
boolean isSumSorted(LinkedList la, LinkedList lb, LinkedList lc,
int givenNumber)
{
Node a = la.head;
// Traverse all nodes of la
while (a != null )
{
Node b = lb.head;
Node c = lc.head;
// for every node in la pick 2 nodes from lb and lc
while (b != null && c!= null )
{
int sum = a.data + b.data + c.data;
if (sum == givenNumber)
{
System.out.println( "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;
}
System.out.println( "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. */
void push( int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
/* Driver program to test above functions */
public static void main(String args[])
{
LinkedList llist1 = new LinkedList();
LinkedList llist2 = new LinkedList();
LinkedList llist3 = new LinkedList();
/* Create Linked List llist1 100->15->5->20 */
llist1.push( 20 );
llist1.push( 5 );
llist1.push( 15 );
llist1.push( 100 );
/*create a sorted linked list 'b' 2->4->9->10 */
llist2.push( 10 );
llist2.push( 9 );
llist2.push( 4 );
llist2.push( 2 );
/*create another sorted linked list 'c' 8->4->2->1 */
llist3.push( 1 );
llist3.push( 2 );
llist3.push( 4 );
llist3.push( 8 );
int givenNumber = 25 ;
llist1.isSumSorted(llist1,llist2,llist3,givenNumber);
}
} /* This code is contributed by Rajat Mishra */
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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!