Count triplets in a sorted doubly linked list whose sum is equal to a given value x

Given a sorted doubly linked list of distinct nodes(no two nodes have the same data) and a value x. Count triplets in the list that sum up to a given value x.

Examples:

Method 1 (Naive Approach):
Using three nested loops generate all triplets and check whether elements in the triplet sum up to x or not.

C++

// C++ implementation to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
#include <bits/stdc++.h>

using namespace std;

// structure of node of doubly linked list
struct Node {
    int data;
    struct Node* next, *prev;
};

// function to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
int countTriplets(struct Node* head, int x)
{
    struct Node* ptr1, *ptr2, *ptr3;
    int count = 0;

    // generate all possible triplets
    for (ptr1 = head; ptr1 != NULL; ptr1 = ptr1->next)
        for (ptr2 = ptr1->next; ptr2 != NULL; ptr2 = ptr2->next)
            for (ptr3 = ptr2->next; ptr3 != NULL; ptr3 = ptr3->next)

                // if elements in the current triplet sum up to 'x'
                if ((ptr1->data + ptr2->data + ptr3->data) == x)

                    // increment count
                    count++;

    // required count of triplets
    return count;
}

// A utility function to insert a new node at the
// beginning of doubly linked list
void insert(struct Node** head, int data)
{
    // allocate node
    struct Node* temp = new Node();

    // put in the data
    temp->data = data;
    temp->next = temp->prev = NULL;

    if ((*head) == NULL)
        (*head) = temp;
    else {
        temp->next = *head;
        (*head)->prev = temp;
        (*head) = temp;
    }
}

// Driver program to test above
int main()
{
    // start with an empty doubly linked list
    struct Node* head = NULL;

    // insert values in sorted order
    insert(&head, 9);
    insert(&head, 8);
    insert(&head, 6);
    insert(&head, 5);
    insert(&head, 4);
    insert(&head, 2);
    insert(&head, 1);

    int x = 17;

    cout << "Count = "
         << countTriplets(head, x);
    return 0;
}

Java

// Java implementation to count triplets
// in a sorted doubly linked list
// whose sum is equal to a given value 'x'
import java.io.*;
import java.util.*;

// Represents node of a doubly linked list
class Node
{
    int data;
    Node prev, next;
    Node(int val)
    {
        data = val;
        prev = null;
        next = null;
    }
}

class GFG
{

    // function to count triplets in
    // a sorted doubly linked list
    // whose sum is equal to a given value 'x'
    static int countTriplets(Node head, int x)
    {
            Node ptr1, ptr2, ptr3;
            int count = 0;

            // generate all possible triplets
            for (ptr1 = head; ptr1 != null; ptr1 = ptr1.next)
                for (ptr2 = ptr1.next; ptr2 != null; ptr2 = ptr2.next)
                    for (ptr3 = ptr2.next; ptr3 != null; ptr3 = ptr3.next)

                        // if elements in the current triplet sum up to 'x'
                        if ((ptr1.data + ptr2.data + ptr3.data) == x)

                            // increment count
                            count++;

            // required count of triplets
            return count;
    }

    // A utility function to insert a new node at the
    // beginning of doubly linked list
    static Node insert(Node head, int val)
    {
            // allocate node
            Node temp = new Node(val);

            if (head == null)
                head = temp;

            else
            {
                temp.next = head;
                head.prev = temp;
                head = temp;
            }

            return head;
    }

    // Driver code
    public static void main(String args[])
    {
            // start with an empty doubly linked list
            Node head = null;

            // insert values in sorted order
            head = insert(head, 9);
            head = insert(head, 8);
            head = insert(head, 6);
            head = insert(head, 5);
            head = insert(head, 4);
            head = insert(head, 2);
            head = insert(head, 1);

            int x = 17;
            System.out.println("count = " + countTriplets(head, x));
    }
}

// This code is contributed by rachana soma

Python3

# Python3 implementation to count triplets
# in a sorted doubly linked list
# whose sum is equal to a given value 'x'

# structure of node of doubly linked list
class Node:
    def __init__(self):
        self.data = None
        self.prev = None
        self.next = None

# function to count triplets in a sorted doubly linked list
# whose sum is equal to a given value 'x'
def countTriplets( head, x):

    ptr1 = head
    ptr2 = None
    ptr3 = None
    count = 0

    # generate all possible triplets
    while (ptr1 != None ):
        ptr2 = ptr1.next
        while ( ptr2 != None ):
            ptr3 = ptr2.next
            while ( ptr3 != None ):

                # if elements in the current triplet sum up to 'x'
                if ((ptr1.data + ptr2.data + ptr3.data) == x):

                    # increment count
                    count = count + 1
                ptr3 = ptr3.next
            ptr2 = ptr2.next
        ptr1 = ptr1.next

    # required count of triplets
    return count

# A utility function to insert a new node at the
# beginning of doubly linked list
def insert(head, data):

    # allocate node
    temp = Node()

    # put in the data
    temp.data = data
    temp.next = temp.prev = None

    if ((head) == None):
        (head) = temp
    else :
        temp.next = head
        (head).prev = temp
        (head) = temp
    return head

# Driver code

# start with an empty doubly linked list
head = None

# insert values in sorted order
head = insert(head, 9)
head = insert(head, 8)
head = insert(head, 6)
head = insert(head, 5)
head = insert(head, 4)
head = insert(head, 2)
head = insert(head, 1)

x = 17

print( "Count = ", countTriplets(head, x))

# This code is contributed by Arnab Kundu

C#

// C# implementation to count triplets
// in a sorted doubly linked list
// whose sum is equal to a given value 'x'
using System;

// Represents node of a doubly linked list
public class Node
{
    public int data;
    public Node prev, next;
    public Node(int val)
    {
        data = val;
        prev = null;
        next = null;
    }
}

class GFG
{

    // function to count triplets in
    // a sorted doubly linked list
    // whose sum is equal to a given value 'x'
    static int countTriplets(Node head, int x)
    {
        Node ptr1, ptr2, ptr3;
        int count = 0;

        // generate all possible triplets
        for (ptr1 = head; ptr1 != null; ptr1 = ptr1.next)
            for (ptr2 = ptr1.next; ptr2 != null; ptr2 = ptr2.next)
                for (ptr3 = ptr2.next; ptr3 != null; ptr3 = ptr3.next)

                    // if elements in the current triplet sum up to 'x'
                    if ((ptr1.data + ptr2.data + ptr3.data) == x)

                        // increment count
                        count++;

        // required count of triplets
        return count;
    }

    // A utility function to insert a new node at the
    // beginning of doubly linked list
    static Node insert(Node head, int val)
    {
        // allocate node
        Node temp = new Node(val);

        if (head == null)
            head = temp;

        else
        {
            temp.next = head;
            head.prev = temp;
            head = temp;
        }

        return head;
    }

    // Driver code
    public static void Main(String []args)
    {
            // start with an empty doubly linked list
            Node head = null;

            // insert values in sorted order
            head = insert(head, 9);
            head = insert(head, 8);
            head = insert(head, 6);
            head = insert(head, 5);
            head = insert(head, 4);
            head = insert(head, 2);
            head = insert(head, 1);

            int x = 17;
            Console.WriteLine("count = " + countTriplets(head, x));
    }
}

// This code is contributed by Arnab Kundu

Javascript

<script>
// javascript implementation to count triplets
// in a sorted doubly linked list
// whose sum is equal to a given value 'x'
// Represents node of a doubly linked list
class Node {
    constructor(val) {
        this.data = val;
        this.prev = null;
        this.next = null;
    }
}
    // function to count triplets in
    // a sorted doubly linked list
    // whose sum is equal to a given value 'x'
    function countTriplets( head , x) {
        var ptr1, ptr2, ptr3;
        var count = 0;

        // generate all possible triplets
        for (ptr1 = head; ptr1 != null; ptr1 = ptr1.next)
            for (ptr2 = ptr1.next; ptr2 != null; ptr2 = ptr2.next)
                for (ptr3 = ptr2.next; ptr3 != null; ptr3 = ptr3.next)

                    // if elements in the current triplet sum up to 'x'
                    if ((ptr1.data + ptr2.data + ptr3.data) == x)

                        // increment count
                        count++;

        // required count of triplets
        return count;
    }

    // A utility function to insert a new node at the
    // beginning of doubly linked list
    function insert( head , val) {
        // allocate node
         temp = new Node(val);

        if (head == null)
            head = temp;

        else {
            temp.next = head;
            head.prev = temp;
            head = temp;
        }

        return head;
    }

    // Driver code

        // start with an empty doubly linked list
         head = null;

        // insert values in sorted order
        head = insert(head, 9);
        head = insert(head, 8);
        head = insert(head, 6);
        head = insert(head, 5);
        head = insert(head, 4);
        head = insert(head, 2);
        head = insert(head, 1);

        var x = 17;
        document.write("count = " + countTriplets(head, x));

// This code is contributed by umadevi9616
</script>

Output

Count = 2

Output:

Count = 2

Time Complexity: O(n³) 
Auxiliary Space: O(1)

Method 2 (Hashing):
Create a hash table with (key, value) tuples represented as (node data, node pointer) tuples. Traverse the doubly linked list and store each node’s data and its pointer pair(tuple) in the hash table. Now, generate each possible pair of nodes. For each pair of nodes, calculate the p_sum(sum of data in the two nodes) and check whether (x-p_sum) exists in the hash table or not. If it exists, then also verify that the two nodes in the pair are not same to the node associated with (x-p_sum) in the hash table and finally increment count. Return (count / 3) as each triplet is counted 3 times in the above process.

C++

// C++ implementation to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
#include <bits/stdc++.h>

using namespace std;

// structure of node of doubly linked list
struct Node {
    int data;
    struct Node* next, *prev;
};

// function to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
int countTriplets(struct Node* head, int x)
{
    struct Node* ptr, *ptr1, *ptr2;
    int count = 0;

    // unordered_map 'um' implemented as hash table
    unordered_map<int, Node*> um;

    // insert the <node data, node pointer> tuple in 'um'
    for (ptr = head; ptr != NULL; ptr = ptr->next)
        um[ptr->data] = ptr;

    // generate all possible pairs
    for (ptr1 = head; ptr1 != NULL; ptr1 = ptr1->next)
        for (ptr2 = ptr1->next; ptr2 != NULL; ptr2 = ptr2->next) {

            // p_sum - sum of elements in the current pair
            int p_sum = ptr1->data + ptr2->data;

            // if 'x-p_sum' is present in 'um' and either of the two nodes
            // are not equal to the 'um[x-p_sum]' node
            if (um.find(x - p_sum) != um.end() && um[x - p_sum] != ptr1
                && um[x - p_sum] != ptr2)

                // increment count
                count++;
        }

    // required count of triplets
    // division by 3 as each triplet is counted 3 times
    return (count / 3);
}

// A utility function to insert a new node at the
// beginning of doubly linked list
void insert(struct Node** head, int data)
{
    // allocate node
    struct Node* temp = new Node();

    // put in the data
    temp->data = data;
    temp->next = temp->prev = NULL;

    if ((*head) == NULL)
        (*head) = temp;
    else {
        temp->next = *head;
        (*head)->prev = temp;
        (*head) = temp;
    }
}

// Driver program to test above
int main()
{
    // start with an empty doubly linked list
    struct Node* head = NULL;

    // insert values in sorted order
    insert(&head, 9);
    insert(&head, 8);
    insert(&head, 6);
    insert(&head, 5);
    insert(&head, 4);
    insert(&head, 2);
    insert(&head, 1);

    int x = 17;

    cout << "Count = "
         << countTriplets(head, x);
    return 0;
}

Java

// Java implementation to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
import java.util.*;

class GFG{

// structure of node of doubly linked list
static class Node {
    int data;
    Node next, prev;
    Node(int val)
    {
        data = val;
        prev = null;
        next = null;
    }
};

// function to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
static int countTriplets(Node head, int x)
{
    Node ptr, ptr1, ptr2;
    int count = 0;

    // unordered_map 'um' implemented as hash table
    HashMap<Integer,Node> um = new HashMap<Integer,Node>();

    // insert the <node data, node pointer> tuple in 'um'
    for (ptr = head; ptr != null; ptr = ptr.next)
        um.put(ptr.data, ptr);

    // generate all possible pairs
    for (ptr1 = head; ptr1 != null; ptr1 = ptr1.next)
        for (ptr2 = ptr1.next; ptr2 != null; ptr2 = ptr2.next) {

            // p_sum - sum of elements in the current pair
            int p_sum = ptr1.data + ptr2.data;

            // if 'x-p_sum' is present in 'um' and either of the two nodes
            // are not equal to the 'um[x-p_sum]' node
            if (um.containsKey(x - p_sum) && um.get(x - p_sum) != ptr1
                && um.get(x - p_sum) != ptr2)

                // increment count
                count++;
        }

    // required count of triplets
    // division by 3 as each triplet is counted 3 times
    return (count / 3);
}

// A utility function to insert a new node at the
// beginning of doubly linked list
static Node insert(Node head, int val)
{
        // allocate node
        Node temp = new Node(val);

        if (head == null)
            head = temp;

        else
        {
            temp.next = head;
            head.prev = temp;
            head = temp;
        }

        return head;
}

// Driver program to test above
public static void main(String[] args)
{
    // start with an empty doubly linked list
    Node head = null;

    // insert values in sorted order
    head = insert(head, 9);
    head = insert(head, 8);
    head = insert(head, 6);
    head = insert(head, 5);
    head = insert(head, 4);
    head = insert(head, 2);
    head = insert(head, 1);

    int x = 17;

    System.out.print("Count = "
         + countTriplets(head, x));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to count triplets in a sorted doubly linked list
# whose sum is equal to a given value 'x'

# structure of node of doubly linked list
class Node:

    def __init__(self, data):

        self.data=data
        self.next=None
        self.prev=None


# function to count triplets in a sorted doubly linked list
# whose sum is equal to a given value 'x'
def countTriplets(head, x):

    ptr2=head
    count = 0;

    # unordered_map 'um' implemented as hash table
    um = dict()

    ptr = head
    # insert the <node data, node pointer> tuple in 'um'
    while ptr!=None:
        um[ptr.data] = ptr;
        ptr = ptr.next

    # generate all possible pairs
    ptr1=head

    while ptr1!=None:

        ptr2 = ptr1.next

        while ptr2!=None:

            # p_sum - sum of elements in the current pair
            p_sum = ptr1.data + ptr2.data;

            # if 'x-p_sum' is present in 'um' and either of the two nodes
            # are not equal to the 'um[x-p_sum]' node
            if ((x-p_sum) in um) and um[x - p_sum] != ptr1 and um[x - p_sum] != ptr2:

                # increment count
                count+=1
            ptr2 = ptr2.next
        ptr1 = ptr1.next


    # required count of triplets
    # division by 3 as each triplet is counted 3 times
    return (count // 3);


# A utility function to insert a new node at the
# beginning of doubly linked list
def insert(head, data):

    # allocate node
    temp = Node(data);

    if ((head) == None):
        (head) = temp;
    else:
        temp.next = head;
        (head).prev = temp;
        (head) = temp;
    return head


# Driver program to test above
if __name__=='__main__':

    # start with an empty doubly linked list
    head = None;

    # insert values in sorted order
    head = insert(head, 9);
    head = insert(head, 8);
    head = insert(head, 6);
    head = insert(head, 5);
    head = insert(head, 4);
    head = insert(head, 2);
    head = insert( head, 1);

    x = 17;

    print("Count = "+ str(countTriplets(head, x)))

# This code is contributed by rutvik_56

C#

// C# implementation to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
using System;
using System.Collections.Generic;

class GFG
{

// structure of node of doubly linked list
class Node {
    public int data;
    public Node next, prev;
    public Node(int val)
    {
        data = val;
        prev = null;
        next = null;
    }
};

// function to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
static int countTriplets(Node head, int x)
{
    Node ptr, ptr1, ptr2;
    int count = 0;

    // unordered_map 'um' implemented as hash table
    Dictionary<int,Node> um = new Dictionary<int,Node>();

    // insert the <node data, node pointer> tuple in 'um'
    for (ptr = head; ptr != null; ptr = ptr.next)
        if(um.ContainsKey(ptr.data))
            um[ptr.data] = ptr;
        else
            um.Add(ptr.data, ptr);

    // generate all possible pairs
    for (ptr1 = head; ptr1 != null; ptr1 = ptr1.next)
        for (ptr2 = ptr1.next; ptr2 != null; ptr2 = ptr2.next)
        {

            // p_sum - sum of elements in the current pair
            int p_sum = ptr1.data + ptr2.data;

            // if 'x-p_sum' is present in 'um' and either of the two nodes
            // are not equal to the 'um[x-p_sum]' node
            if (um.ContainsKey(x - p_sum) && um[x - p_sum] != ptr1
                && um[x - p_sum] != ptr2)

                // increment count
                count++;
        }

    // required count of triplets
    // division by 3 as each triplet is counted 3 times
    return (count / 3);
}

// A utility function to insert a new node at the
// beginning of doubly linked list
static Node insert(Node head, int val)
{
        // allocate node
        Node temp = new Node(val);

        if (head == null)
            head = temp;

        else
        {
            temp.next = head;
            head.prev = temp;
            head = temp;
        }

        return head;
}

// Driver code
public static void Main(String[] args)
{
    // start with an empty doubly linked list
    Node head = null;

    // insert values in sorted order
    head = insert(head, 9);
    head = insert(head, 8);
    head = insert(head, 6);
    head = insert(head, 5);
    head = insert(head, 4);
    head = insert(head, 2);
    head = insert(head, 1);

    int x = 17;

    Console.Write("Count = "
        + countTriplets(head, x));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript implementation to count
// triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'

// Structure of node of doubly linked list
class Node{
constructor(data)
{
    this.data = data;
    this.prev = null;
    this.next = null;
}
}

// Function to count triplets in a sorted
// doubly linked list whose sum is equal
// to a given value 'x'
function countTriplets(head, x)
{
    let ptr, ptr1, ptr2;
    let count = 0;

    // unordered_map 'um' implemented
    // as hash table
    let um = new Map();

    // Insert the <node data, node pointer>
    // tuple in 'um'
    for(ptr = head; ptr != null; ptr = ptr.next)
        um.set(ptr.data, ptr);

    // Generate all possible pairs
    for(ptr1 = head;
        ptr1 != null;
        ptr1 = ptr1.next)
        for(ptr2 = ptr1.next;
            ptr2 != null;
            ptr2 = ptr2.next)
        {

            // p_sum - sum of elements in
            // the current pair
            let p_sum = ptr1.data + ptr2.data;

            // If 'x-p_sum' is present in 'um'
            // and either of the two nodes are
            // not equal to the 'um[x-p_sum]' node
            if (um.has(x - p_sum) &&
                um.get(x - p_sum) != ptr1 &&
                um.get(x - p_sum) != ptr2)

                // Increment count
                count++;
        }

    // Required count of triplets
    // division by 3 as each triplet
    // is counted 3 times
    return (count / 3);
}

// A utility function to insert a new
// node at the beginning of doubly linked list
function insert(head, val)
{

    // Allocate node
    let temp = new Node(val);

    if (head == null)
        head = temp;

    else
    {
        temp.next = head;
        head.prev = temp;
        head = temp;
    }
    return head;
}

// Driver code

// Start with an empty doubly linked list
let head = null;

// Insert values in sorted order
head = insert(head, 9);
head = insert(head, 8);
head = insert(head, 6);
head = insert(head, 5);
head = insert(head, 4);
head = insert(head, 2);
head = insert(head, 1);

let x = 17;

document.write("Count = " +
               countTriplets(head, x));

// This code is contributed by patel2127

</script>

Output

Count = 2

Output:

Count = 2

Time Complexity: O(n²) 
Auxiliary Space: O(n)

Method 3 Efficient Approach(Use of two pointers):
Traverse the doubly linked list from left to right. For each current node during the traversal, initialize two pointers first = pointer to the node next to the current node and last = pointer to the last node of the list. Now, count pairs in the list from first to last pointer that sum up to value (x – current node’s data) (algorithm described in this post). Add this count to the total_count of triplets. Pointer to the last node can be found only once in the beginning.

C++

// C++ implementation to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
#include <bits/stdc++.h>

using namespace std;

// structure of node of doubly linked list
struct Node {
    int data;
    struct Node* next, *prev;
};

// function to count pairs whose sum equal to given 'value'
int countPairs(struct Node* first, struct Node* second, int value)
{
    int count = 0;

    // The loop terminates when either of two pointers
    // become NULL, or they cross each other (second->next
    // == first), or they become same (first == second)
    while (first != NULL && second != NULL &&
           first != second && second->next != first) {

        // pair found
        if ((first->data + second->data) == value) {

            // increment count
            count++;

            // move first in forward direction
            first = first->next;

            // move second in backward direction
            second = second->prev;
        }

        // if sum is greater than 'value'
        // move second in backward direction
        else if ((first->data + second->data) > value)
            second = second->prev;

        // else move first in forward direction
        else
            first = first->next;
    }

    // required count of pairs
    return count;
}

// function to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
int countTriplets(struct Node* head, int x)
{
    // if list is empty
    if (head == NULL)
        return 0;

    struct Node* current, *first, *last;
    int count = 0;

    // get pointer to the last node of
    // the doubly linked list
    last = head;
    while (last->next != NULL)
        last = last->next;

    // traversing the doubly linked list
    for (current = head; current != NULL; current = current->next) {

        // for each current node
        first = current->next;

        // count pairs with sum(x - current->data) in the range
        // first to last and add it to the 'count' of triplets
        count += countPairs(first, last, x - current->data);
    }

    // required count of triplets
    return count;
}

// A utility function to insert a new node at the
// beginning of doubly linked list
void insert(struct Node** head, int data)
{
    // allocate node
    struct Node* temp = new Node();

    // put in the data
    temp->data = data;
    temp->next = temp->prev = NULL;

    if ((*head) == NULL)
        (*head) = temp;
    else {
        temp->next = *head;
        (*head)->prev = temp;
        (*head) = temp;
    }
}

// Driver program to test above
int main()
{
    // start with an empty doubly linked list
    struct Node* head = NULL;

    // insert values in sorted order
    insert(&head, 9);
    insert(&head, 8);
    insert(&head, 6);
    insert(&head, 5);
    insert(&head, 4);
    insert(&head, 2);
    insert(&head, 1);

    int x = 17;

    cout << "Count = "
         << countTriplets(head, x);
    return 0;
}

Java

// Java implementation to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
import java.util.*;

class GFG{

// structure of node of doubly linked list
static class Node {
    int data;
    Node next, prev;
};

// function to count pairs whose sum equal to given 'value'
static int countPairs(Node first, Node second, int value)
{
    int count = 0;

    // The loop terminates when either of two pointers
    // become null, or they cross each other (second.next
    // == first), or they become same (first == second)
    while (first != null && second != null &&
           first != second && second.next != first) {

        // pair found
        if ((first.data + second.data) == value) {

            // increment count
            count++;

            // move first in forward direction
            first = first.next;

            // move second in backward direction
            second = second.prev;
        }

        // if sum is greater than 'value'
        // move second in backward direction
        else if ((first.data + second.data) > value)
            second = second.prev;

        // else move first in forward direction
        else
            first = first.next;
    }

    // required count of pairs
    return count;
}

// function to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
static int countTriplets(Node head, int x)
{
    // if list is empty
    if (head == null)
        return 0;

    Node current, first, last;
    int count = 0;

    // get pointer to the last node of
    // the doubly linked list
    last = head;
    while (last.next != null)
        last = last.next;

    // traversing the doubly linked list
    for (current = head; current != null; current = current.next) {

        // for each current node
        first = current.next;

        // count pairs with sum(x - current.data) in the range
        // first to last and add it to the 'count' of triplets
        count += countPairs(first, last, x - current.data);
    }

    // required count of triplets
    return count;
}

// A utility function to insert a new node at the
// beginning of doubly linked list
static Node insert(Node head, int data)
{
    // allocate node
    Node temp = new Node();

    // put in the data
    temp.data = data;
    temp.next = temp.prev = null;

    if ((head) == null)
        (head) = temp;
    else {
        temp.next = head;
        (head).prev = temp;
        (head) = temp;
    }
    return head;
}

// Driver program to test above
public static void main(String[] args)
{
    // start with an empty doubly linked list
    Node head = null;

    // insert values in sorted order
    head = insert(head, 9);
    head = insert(head, 8);
    head = insert(head, 6);
    head = insert(head, 5);
    head = insert(head, 4);
    head = insert(head, 2);
    head = insert(head, 1);

    int x = 17;

    System.out.print("Count = "
         + countTriplets(head, x));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation to count triplets
# in a sorted doubly linked list whose sum
# is equal to a given value 'x'

# Structure of node of doubly linked list
class Node:

    def __init__(self, x):

        self.data = x
        self.next = None
        self.prev = None

# Function to count pairs whose sum
# equal to given 'value'
def countPairs(first, second, value):

    count = 0

    # The loop terminates when either of two pointers
    # become None, or they cross each other (second.next
    # == first), or they become same (first == second)
    while (first != None and second != None and
           first != second and second.next != first):

        # Pair found
        if ((first.data + second.data) == value):

            # Increment count
            count += 1

            # Move first in forward direction
            first = first.next

            # Move second in backward direction
            second = second.prev

        # If sum is greater than 'value'
        # move second in backward direction
        elif ((first.data + second.data) > value):
            second = second.prev

        # Else move first in forward direction
        else:
            first = first.next

    # Required count of pairs
    return count

# Function to count triplets in a sorted
# doubly linked list whose sum is equal
# to a given value 'x'
def countTriplets(head, x):

    # If list is empty
    if (head == None):
        return 0

    current, first, last = head, None, None
    count = 0

    # Get pointer to the last node of
    # the doubly linked list
    last = head

    while (last.next != None):
        last = last.next

    # Traversing the doubly linked list
    while current != None:

        # For each current node
        first = current.next

        # count pairs with sum(x - current.data) in
        # the range first to last and add it to the
        # 'count' of triplets
        count, current = count + countPairs(
            first, last, x - current.data), current.next

    # Required count of triplets
    return count

# A utility function to insert a new node
# at the beginning of doubly linked list
def insert(head, data):

    # Allocate node
    temp = Node(data)

    # Put in the data
    # temp.next = temp.prev = None
    if (head == None):
        head = temp
    else:
        temp.next = head
        head.prev = temp
        head = temp

    return head

# Driver code
if __name__ == '__main__':

    # Start with an empty doubly linked list
    head = None

    # Insert values in sorted order
    head = insert(head, 9)
    head = insert(head, 8)
    head = insert(head, 6)
    head = insert(head, 5)
    head = insert(head, 4)
    head = insert(head, 2)
    head = insert(head, 1)

    x = 17

    print("Count = ", countTriplets(head, x))

# This code is contributed by mohit kumar 29

C#

// C# implementation to count triplets
// in a sorted doubly linked list
// whose sum is equal to a given value 'x'
using System;

class GFG
{

// structure of node of doubly linked list
class Node
{
    public int data;
    public Node next, prev;
};

// function to count pairs whose sum equal to given 'value'
static int countPairs(Node first, Node second, int value)
{
    int count = 0;

    // The loop terminates when either of two pointers
    // become null, or they cross each other (second.next
    // == first), or they become same (first == second)
    while (first != null && second != null &&
        first != second && second.next != first) {

        // pair found
        if ((first.data + second.data) == value) {

            // increment count
            count++;

            // move first in forward direction
            first = first.next;

            // move second in backward direction
            second = second.prev;
        }

        // if sum is greater than 'value'
        // move second in backward direction
        else if ((first.data + second.data) > value)
            second = second.prev;

        // else move first in forward direction
        else
            first = first.next;
    }

    // required count of pairs
    return count;
}

// function to count triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'
static int countTriplets(Node head, int x)
{
    // if list is empty
    if (head == null)
        return 0;

    Node current, first, last;
    int count = 0;

    // get pointer to the last node of
    // the doubly linked list
    last = head;
    while (last.next != null)
        last = last.next;

    // traversing the doubly linked list
    for (current = head; current != null; current = current.next) {

        // for each current node
        first = current.next;

        // count pairs with sum(x - current.data) in the range
        // first to last and add it to the 'count' of triplets
        count += countPairs(first, last, x - current.data);
    }

    // required count of triplets
    return count;
}

// A utility function to insert a new node at the
// beginning of doubly linked list
static Node insert(Node head, int data)
{
    // allocate node
    Node temp = new Node();

    // put in the data
    temp.data = data;
    temp.next = temp.prev = null;

    if ((head) == null)
        (head) = temp;
    else {
        temp.next = head;
        (head).prev = temp;
        (head) = temp;
    }
    return head;
}

// Driver program to test above
public static void Main(String[] args)
{
    // start with an empty doubly linked list
    Node head = null;

    // insert values in sorted order
    head = insert(head, 9);
    head = insert(head, 8);
    head = insert(head, 6);
    head = insert(head, 5);
    head = insert(head, 4);
    head = insert(head, 2);
    head = insert(head, 1);

    int x = 17;

    Console.Write("Count = "
        + countTriplets(head, x));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript implementation to count
// triplets in a sorted doubly linked list
// whose sum is equal to a given value 'x'

// Structure of node of doubly linked list
class Node
{
    constructor(data)
    {
        this.data = data;
        this.next = this.prev = null;
    }
}

// Function to count pairs whose sum
// equal to given 'value'
function countPairs(first, second, value)
{
    let count = 0;

    // The loop terminates when either of two pointers
    // become null, or they cross each other (second.next
    // == first), or they become same (first == second)
    while (first != null && second != null &&
           first != second && second.next != first)
    {

        // Pair found
        if ((first.data + second.data) == value)
        {

            // Increment count
            count++;

            // Move first in forward direction
            first = first.next;

            // Move second in backward direction
            second = second.prev;
        }

        // If sum is greater than 'value'
        // move second in backward direction
        else if ((first.data + second.data) > value)
            second = second.prev;

        // Else move first in forward direction
        else
            first = first.next;
    }

    // Required count of pairs
    return count;
}

// Function to count triplets in a sorted
// doubly linked list whose sum is equal
// to a given value 'x'
function countTriplets(head, x)
{

    // If list is empty
    if (head == null)
        return 0;

    let current, first, last;
    let count = 0;

    // Get pointer to the last node of
    // the doubly linked list
    last = head;
    while (last.next != null)
        last = last.next;

    // Traversing the doubly linked list
    for(current = head;
        current != null;
        current = current.next)
    {

        // For each current node
        first = current.next;

        // Count pairs with sum(x - current.data)
        // in the range first to last and add it
        // to the 'count' of triplets
        count += countPairs(first, last,
                            x - current.data);
    }

    // Required count of triplets
    return count;
}

// A utility function to insert a new node at the
// beginning of doubly linked list
function insert(head, data)
{

    // Allocate node
    let temp = new Node();

    // Put in the data
    temp.data = data;
    temp.next = temp.prev = null;

    if ((head) == null)
        (head) = temp;
    else
    {
        temp.next = head;
        (head).prev = temp;
        (head) = temp;
    }
    return head;
}

// Driver code

// Start with an empty doubly linked list
let head = null;

// Insert values in sorted order
head = insert(head, 9);
head = insert(head, 8);
head = insert(head, 6);
head = insert(head, 5);
head = insert(head, 4);
head = insert(head, 2);
head = insert(head, 1);

let x = 17;

document.write("Count = " +
    countTriplets(head, x));

// This code is contributed by unknown2108

</script>

Output

Count = 2

Output:

Count = 2

Time Complexity: O(n²) 
Auxiliary Space: O(1)

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Another Solution  

find tripple sum in Doubly linkedList.

C++

#include <iostream>
#include <vector>

using namespace std;

struct Node
{
    int data;
    Node *next;
    Node *prev;
    Node(int x)
    {
        data = x;
        next = nullptr;
        prev = nullptr;
    }
};

class Solution
{
  public:
    vector<vector<int>> trippleSumInLinkedList(Node *head, int sumv)
    {
        vector<vector<int>> res;
        Node *s, *m, *e;
        s = head;
        m = head;
        e = head;
        while (e->next != nullptr)
            e = e->next;
        while (s->next->next != nullptr)
        {
            int currSum = sumv - s->data;
            m = s->next;
            Node *ev = e;
            while (m != nullptr && ev != nullptr && m != ev)
            {
                int newSum = m->data + ev->data;
                if (newSum == currSum)
                {
                    res.push_back({s->data, m->data, ev->data});
                    m = m->next;
                }
                else if (newSum > currSum)
                    ev = ev->prev;
                else
                    m = m->next;
            }
            s = s->next;
        }
        return res;
    }
};

int main()
{
    Node *head = new Node(1);
    Node *node2 = new Node(2);
    Node *node3 = new Node(4);
    Node *node4 = new Node(5);
    Node *node5 = new Node(6);
    Node *node6 = new Node(8);
    Node *node7 = new Node(9);

    head->next = node2;
    node2->prev = head;
    node2->next = node3;

    node3->prev = node2;
    node3->next = node4;

    node4->next = node5;
    node4->prev = node3;

    node5->prev = node4;
    node5->next = node6;

    node6->prev = node5;
    node6->next = node7;

    node7->prev = node6;

    Solution sol;
    vector<vector<int>> res = sol.trippleSumInLinkedList(head, 15);

    for (int i = 0; i < res.size(); i++)
        cout << res[i][0] << ", " << res[i][1] << ", " << res[i][2] << endl;
    return 0;
}

Java

import java.util.ArrayList;

class Node {
    public int data;
    public Node next;
    public Node prev;

    public Node(int x) {
        data = x;
        next = null;
        prev = null;
    }
}

public class Solution {
    public ArrayList<ArrayList<Integer>> trippleSumInLinkedList(Node head, int sumv) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        Node s, m, e;
        s = head;
        m = head;
        e = head;
        while (e.next != null)
            e = e.next;
        while (s.next.next != null) {
            int currSum = sumv - s.data;
            m = s.next;
            Node ev = e;
            while (m != null && ev != null && m != ev) {
                int newSum = m.data + ev.data;
                if (newSum == currSum) {
                    ArrayList<Integer> triple = new ArrayList<Integer>();
                    triple.add(s.data);
                    triple.add(m.data);
                    triple.add(ev.data);
                    res.add(triple);
                    m = m.next;
                }
                else if (newSum > currSum)
                    ev = ev.prev;
                else
                    m = m.next;
            }
            s = s.next;
        }
        return res;
    }

    public static void main(String[] args) {
        Node head = new Node(1);
        Node node2 = new Node(2);
        Node node3 = new Node(4);
        Node node4 = new Node(5);
        Node node5 = new Node(6);
        Node node6 = new Node(8);
        Node node7 = new Node(9);

        head.next = node2;
        node2.prev = head;
        node2.next = node3;

        node3.prev = node2;
        node3.next = node4;

        node4.next = node5;
        node4.prev = node3;

        node5.prev = node4;
        node5.next = node6;

        node6.prev = node5;
        node6.next = node7;

        node7.prev = node6;

        Solution sol = new Solution();
        ArrayList<ArrayList<Integer>> res = sol.trippleSumInLinkedList(head, 15);

        for (int i = 0; i < res.size(); i++)
            System.out.println(res.get(i).get(0) + ", " + res.get(i).get(1) + ", " + res.get(i).get(2));
    }
}

Python3

class Solution:
    def trippleSumInLinkedList(self, head,sumv):
        res = []
        s,m,e = head,head,head
        while e.next != None:
            e = e.next
        while s.next.next != None:
            currSum = sumv-s.data
            m = s.next
            ev = e
            while m and ev and m != ev:
                newSum = m.data + ev.data
                if newSum == currSum:
                    res.append([s.data,m.data,ev.data])
                    m = m.next
                elif newSum > currSum:
                    ev = ev.prev
                else:
                    m= m.next
            s = s.next
        return res
sol = Solution()

head = Node(1)
node2 = Node(2)
node3 = Node(4)
node4 = Node(5)
node5 = Node(6)
node6 = Node(8)
node7 = Node(9)

head.next = node2
node2.prev = head
node2.next= node3

node3.prev = node2
node3.next = node4

node4.next= node5
node4.prev = node3

node5.prev = node4
node5.next = node6

node6.prev = node5
node6.next = node7

node7.prev= node6


print('solution: ',sol.trippleSumInLinkedList(head,15))

C#

using System;
using System.Collections.Generic;

class Node {
    public int data;
    public Node next;
    public Node prev;

    public Node(int x)
    {
        data = x;
        next = null;
        prev = null;
    }
}

class Solution {
    public List<List<int> >
    trippleSumInLinkedList(Node head, int sumv)
    {
        List<List<int> > res = new List<List<int> >();
        Node s, m, e;
        s = head;
        m = head;
        e = head;
        while (e.next != null)
            e = e.next;
        while (s.next.next != null) {
            int currSum = sumv - s.data;
            m = s.next;
            Node ev = e;
            while (m != null && ev != null && m != ev) {
                int newSum = m.data + ev.data;
                if (newSum == currSum) {
                    res.Add(new List<int>{ s.data, m.data,
                                           ev.data });
                    m = m.next;
                }
                else if (newSum > currSum)
                    ev = ev.prev;
                else
                    m = m.next;
            }
            s = s.next;
        }
        return res;
    }
}

class Program {
    static void Main(string[] args)
    {
        Node head = new Node(1);
        Node node2 = new Node(2);
        Node node3 = new Node(4);
        Node node4 = new Node(5);
        Node node5 = new Node(6);
        Node node6 = new Node(8);
        Node node7 = new Node(9);

        head.next = node2;
        node2.prev = head;
        node2.next = node3;

        node3.prev = node2;
        node3.next = node4;

        node4.next = node5;
        node4.prev = node3;

        node5.prev = node4;
        node5.next = node6;

        node6.prev = node5;
        node6.next = node7;

        node7.prev = node6;

        Solution sol = new Solution();
        List<List<int> > res
            = sol.trippleSumInLinkedList(head, 15);
        foreach(List<int> triple in res)
        {
            Console.WriteLine("{0}, {1}, {2}", triple[0],
                              triple[1], triple[2]);
        }
        Console.ReadKey();
    }
}
// This code is contributed by sarojmcy2e

Javascript

class Node {
    constructor(x) {
        this.data = x;
        this.next = null;
        this.prev = null;
    }
}

class Solution {
    trippleSumInLinkedList(head, sumv) {
        let res = [];
        let s, m, e;
        s = head;
        m = head;
        e = head;
        while (e.next != null)
            e = e.next; // move e to the end of the list
        while (s.next.next != null) { // iterate through all nodes except last two
            let currSum = sumv - s.data; // calculate current sum
            m = s.next; // set m to next node of s
            let ev = e; // set ev to last node of list
            while (m != null && ev != null && m != ev) { // iterate through all nodes between m and ev
                let newSum = m.data + ev.data; // calculate new sum
                if (newSum == currSum) { // if new sum equals current sum
                    res.push([s.data, m.data, ev.data]); // add to result array
                    m = m.next; // move m to next node
                } else if (newSum > currSum)
                    ev = ev.prev; // move ev to previous node
                else
                    m = m.next; // move m to next node
            }
            s = s.next; // move s to next node
        }
        return res; // return result array
    }
}

let head = new Node(1);
let node2 = new Node(2);
let node3 = new Node(4);
let node4 = new Node(5);
let node5 = new Node(6);
let node6 = new Node(8);
let node7 = new Node(9);

head.next = node2;
node2.prev = head;
node2.next = node3;

node3.prev = node2;
node3.next = node4;

node4.next = node5;
node4.prev = node3;

node5.prev = node4;
node5.next = node6;

node6.prev = node5;
node6.next = node7;

node7.prev = node6;

let sol = new Solution();
let res = sol.trippleSumInLinkedList(head, 15);

for (let i in res)
    console.log(res[i][0] + ", " + res[i][1] + ", " + res[i][2]);

Output

1, 5, 9
1, 6, 8
2, 4, 9
2, 5, 8
4, 5, 6

Time complexity : O(n^2), where n is the length of the linked list.  
Space complexity :O(1),because it only uses a constant amount of extra memory to store pointers to the nodes being examined