# 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: ## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` ` `  `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  `

## 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 `

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++ implementation to count triplets in a sorted doubly linked list ` `// whose sum is equal to a given value 'x' ` `#include ` ` `  `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 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; ` `} `

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, initailze 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++ implementation to count triplets in a sorted doubly linked list ` `// whose sum is equal to a given value 'x' ` `#include ` ` `  `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; ` `} `

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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : rachana soma, andrew1234

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