# Find 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.
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 tripel “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.

## C/C++

```// C/C++ program to find a triplet from three linked lists with
// sum equal to a given number
#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>

struct Node
{
int data;
struct Node* next;
};

/* A utility function to insert a node at the beginning of a
void push (struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node =
(struct Node*) malloc(sizeof(struct Node));

/* put in the data */
new_node->data = new_data;

/* link the old list off the new node */

/* move the head to point to the new node */
}

/* A function to chech 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. */
{

// Traverse through all nodes of a
while (a != NULL)
{

// For every node of list a, prick two nodes
// from lists b abd c
while (b != NULL && c != NULL)
{
// If this a triplet with given sum, print
// it and return true
int sum = a->data + b->data + c->data;
if (sum == givenNumber)
{
printf ("Triplet Found: %d %d %d ", a->data,
b->data, c->data);
return true;
}

// If sum of this triplet is smaller, look for
// greater values in b
else if (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
}

printf ("No such triplet");
return false;
}

/* Drier program to test above function*/
int main()
{

/*create a linked list 'a' 10->15->5->20 */

/*create a sorted linked list 'b' 2->4->9->10 */

/*create another sorted linked list 'c' 8->4->2->1 */

int givenNumber = 25;

return 0;
}
```

## Java

```// Java program to find a triplet from three linked lists with
// sum equal to a given number
{

class Node
{
int data;
Node next;
Node(int d) {data = d; next = null; }
}

/* A function to chech 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. */
int givenNumber)
{

// Traverse all nodes of la
while (a != null)
{

// 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 */

/* 4. Move the head to point to new Node */
}

/* Drier program to test above functions */
public static void main(String args[])
{

/* 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 */
```

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.

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