Java Program For Finding A Triplet From Three Linked Lists With Sum Equal To A Given Number
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.Â
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.Â
1) Sort list b in ascending order, and list c in descending order.Â
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.Â
Java
class LinkedList
{
Node head;
class Node
{
int data;
Node next;
Node( int d) {data = d; next = null ; }
}
boolean isSumSorted(LinkedList la, LinkedList lb, LinkedList lc,
int givenNumber)
{
Node a = la.head;
while (a != null )
{
Node b = lb.head;
Node c = lc.head;
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 ;
}
else if (sum < givenNumber)
b = b.next;
else
c = c.next;
}
a = a.next;
}
System.out.println( "No Triplet found" );
return false ;
}
void push( int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
public static void main(String args[])
{
LinkedList llist1 = new LinkedList();
LinkedList llist2 = new LinkedList();
LinkedList llist3 = new LinkedList();
llist1.push( 20 );
llist1.push( 5 );
llist1.push( 15 );
llist1.push( 100 );
llist2.push( 10 );
llist2.push( 9 );
llist2.push( 4 );
llist2.push( 2 );
llist3.push( 1 );
llist3.push( 2 );
llist3.push( 4 );
llist3.push( 8 );
int givenNumber = 25 ;
llist1.isSumSorted(llist1,llist2,llist3,givenNumber);
}
}
|
Output:Â
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).Â
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.Â
Please refer complete article on Find a triplet from three linked lists with sum equal to a given number for more details!
Last Updated :
21 Dec, 2021
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