Given a sorted doubly linked list of distinct nodes(no two nodes have the same data) and a value x. The task is to count the triplets in the list that product up to a given value x.
Examples:
Input: list = 1->2->4->5->6->8->9, x = 8
Output: 1
Triplet is (1, 2, 4)Input: list = 1->2->4->5->6->8->9, x = 120
Output: 1
Triplet is (4, 5, 6)
Naive Approach: Using three nested loops generate all triplets and check whether elements in the triplet product up to x or not.
Below is the implementation of the above approach:
// C++ implementation to count triplets // in a sorted doubly linked list // whose product 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 product 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 product 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 = 8;
cout << "Count = "
<< countTriplets(head, x);
return 0;
} |
// Java implementation to count triplets // in a sorted doubly linked list // whose sum is equal to a given value 'x' import java.util.*;
// Represents node of a doubly linked list 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 = 8 ;
System.out.println( "count = " + countTriplets(head, x));
}
} // This code is contributed by 29AjayKumar |
# Python3 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:
data = None
prev = None
next_ = None
def __init__( self , val):
self .data = val
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, ptr2, ptr3 = Node( 0 ), Node( 0 ), Node( 0 )
count = 0
# generate all possible triplets
ptr1 = head
while ptr1 is not None :
ptr2 = ptr1.next_
while ptr2 is not None :
ptr3 = ptr2.next_
while ptr3 is not None :
# if elements in the current
# triplet sum up to 'x'
if ptr1.data * ptr2.data * ptr3.data = = x:
# increment 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, val):
# allocate node
temp = Node(val)
if head is 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 = Node( 0 )
# 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 = 8
print ( "count =" , countTriplets(head, x))
# This code is contributed by # sanjeev2552 |
// 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 = 8;
Console.WriteLine( "count = " + countTriplets(head, x));
}
} // This code is contributed by Arnab Kundu |
<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 = 8;
document.write( "count = " + countTriplets(head, x));
// This code contributed by umadevi9616 </script> |
Count = 1
Complexity Analysis:
- Time Complexity: O(n^3)
- 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_product(product of data in the two nodes) and check whether (x/p_product) exists in the hash table or not.
If it exists, then also verify that the two nodes in the pair are not same as to the node associated with (x/p_product) in the hash table and finally increment count. Return (count / 3) as each triplet is counted 3 times in the above process.
Below is the implementation of the above approach:
// C++ implementation to count triplets // in a sorted doubly linked list // whose product 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 product 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_product = product of elements in the current pair
int p_product = (ptr1->data * ptr2->data);
// if 'x/p_product' is present in 'um' and
// either of the two nodes
// are not equal to the 'um[x/p_product]' node
if (um.find(x / p_product) != um.end() && um[x / p_product] != ptr1
&& um[x / p_product] != 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 functions 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 = 8;
cout << "Count = "
<< countTriplets(head, x);
return 0;
} |
// Java implementation to count triplets // in a sorted doubly linked list whose // product is equal to a given value 'x' import java.io.*;
import java.util.*;
// Structure of node of doubly linked list class Node
{ int data;
Node next, prev;
} class GFG{
// Function to count triplets in a sorted // doubly linked list whose product 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
Map<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_product = product of elements
// in the current pair
int p_product = (ptr1.data * ptr2.data);
// If 'x/p_product' is present in 'um' and
// either of the two nodes are not equal
// to the 'um[x/p_product]' node
if (um.containsKey(x / p_product) &&
um.get(x / p_product) != ptr1 &&
um.get(x / p_product) != 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 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 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 = 8 ;
System.out.println( "Count = " +
countTriplets(head, x));
} } // This code is contributed by avanitrachhadiya2155 |
# Python3 implementation to # count triplets in a sorted # doubly linked list whose # product is equal to a given # value 'x' from math import ceil
# 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 triplets # in a sorted doubly linked # list whose product is equal # to a given value 'x' def countTriplets(head, x):
ptr, ptr1, ptr2 = None , None , None
count = 0
# unordered_map 'um' implemented
# as hash table
um = {}
# insert the tuple in 'um'
ptr = head
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_product = product of
# elements in the current
# pair
p_product = (ptr1.data *
ptr2.data)
# if 'x/p_product' is present
# in 'um' and either of the
# two nodes are not equal to
# the 'um[x/p_product]' node
if ((x / p_product) in um and
(x / p_product) ! = ptr1 and
um[x / p_product] ! = 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)
# 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 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 = 8
print ( "Count =" ,
countTriplets(head, x))
# This code is contributed by Mohit Kumar 29 |
// C# implementation to count triplets // in a sorted doubly linked list whose // product is equal to a given value 'x' using System;
using System.Collections.Generic;
// Structure of node of doubly linked list class Node
{ public int data;
public Node next, prev;
} class GFG
{ // Function to count triplets in a sorted // doubly linked list whose product 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)
{
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_product = product of elements
// in the current pair
int p_product = (ptr1.data * ptr2.data);
// If 'x/p_product' is present in 'um' and
// either of the two nodes are not equal
// to the 'um[x/p_product]' node
if (um.ContainsKey(x / p_product) &&
um[x / p_product] != ptr1 &&
um[x / p_product] != 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 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 code static public void Main ()
{ // 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 = 8;
Console.WriteLine( "Count = " + countTriplets(head, x));
} } // This code is contributed by rag2127 |
<script> // JavaScript implementation to count triplets // in a sorted doubly linked list whose // product is equal to a given value 'x' class Node { constructor(data)
{
this .data=data;
this .next= this .prev= null ;
}
} // Function to count triplets in a sorted // doubly linked list whose product 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_product = product of elements
// in the current pair
let p_product = (ptr1.data * ptr2.data);
// If 'x/p_product' is present in 'um' and
// either of the two nodes are not equal
// to the 'um[x/p_product]' node
if (um.has(x / p_product) &&
um.get(x / p_product) != ptr1 &&
um.get(x / p_product) != 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,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 = 8; document.write( "Count = " +
countTriplets(head, x));
// This code is contributed by patel2127 </script> |
Count = 1
Complexity Analysis:
- Time Complexity: O(n^2)
- Auxiliary Space: O(n)
Method-3 (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 the last pointer that product up to the 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.
Below is the implementation of the above approach:
// C++ implementation to count triplets // in a sorted doubly linked list // whose product 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 product 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 product 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 product 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 product(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 = 8;
cout << "Count = "
<< countTriplets(head, x);
return 0;
} |
// Java implementation to count triplets // in a sorted doubly linked list // whose product 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 product // 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 product 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 product 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 product(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 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 = 8 ;
System.out.println( "Count = " + countTriplets(head, x));
} } // This code is contributed by Arnab Kundu |
# Python3 implementation to count triplets # in a sorted doubly linked list whose # product 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 pairs whose product # 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 product 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 product is # equal to a given value 'x' def countTriplets(head, x):
# If list is empty
if (head = = None ):
return 0
count = 0
# Get pointer to the last node of
# the doubly linked list
last = head
while (last. next ! = None ):
last = last. next
current = head
# Traversing the doubly linked list
while current ! = None :
# For each current node
first = current. next
# Count pairs with product(x / current.data)
# in the range first to last and
# add it to the 'count' of triplets
count + = countPairs(first, last,
x / / current.data)
current = 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.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 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 = 8
print ( "Count = " + str (countTriplets(head, x)))
# This code is contributed by rutvik_56 |
// C# implementation to count triplets // in a sorted doubly linked list // whose product 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 product // 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 product 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 product 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 product(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 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 = 8;
Console.WriteLine( "Count = " + countTriplets(head, x));
} } // This code is contributed by Rajput-Ji |
<script> // JavaScript implementation to count triplets // in a sorted doubly linked list // whose product is equal to a given value 'x' // structure of node of doubly linked list class Node { constructor()
{
let data;
let next, prev;
}
} // function to count pairs whose product // 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 product 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 product 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 product(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 = 8; document.write( "Count = " + countTriplets(head, x));
// This code is contributed by unknown2108 </script> |
Count = 1
Complexity Analysis:
- Time Complexity: O(n^2)
- Auxiliary Space: O(1)