Given a sorted doubly linked list of positive distinct elements, the task is to find pairs in the doubly linked list whose product is equal to given value x, without using any extra space.
Examples:
Input : List = 1 <=> 2 <=> 4 <=> 5 <=> 6 <=> 8 <=> 9 x = 8 Output: (1, 8), (2, 4) Input : List = 1 <=> 2 <=> 3 <=> 4 <=> 5 <=> 6 <=> 7 x = 6 Output: (1, 6), (2, 3)
A Simple Approach for this problem is to traverse the linked list using two nested loops and find all pairs and check for the pairs with product equals to x. Time complexity for this problem will be O(n^2), n is total number of nodes in doubly linked list.
An Efficient Solution for this problem is same as this article. Here is the algorithm :
- Initialize two pointer variables to find the candidate elements in the sorted doubly linked list. Initialize first with start of doubly linked list i.e; first=head and initialize second with last node of doubly linked list i.e; second=last_node.
- We initialize first and second pointers as first and last nodes. Here we don’t have random access, so to find second pointer, we traverse the list to initialize second.
- If current product of first and second is less than x, then we move first in forward direction. If current product of first and second element is greater than x, then we move second in backward direction.
- Loop termination conditions are also different from arrays. The loop terminates when either of two pointers become NULL, or they cross each other (second->next = first), or they become same (first == second)
Below is the implementation of the above approach:
// C++ program to find a pair with // given product x in sorted Doubly // Linked List #include <bits/stdc++.h> using namespace std;
// Doubly Linked List Node struct Node {
int data;
struct Node *next, *prev;
}; // Function to find pair whose product // equal to given value x void pairProduct( struct Node* head, int x)
{ // Set two pointers,
// first to the beginning of DLL
// and second to the end of DLL.
struct Node* first = head;
struct Node* second = head;
while (second->next != NULL)
second = second->next;
// To track if we find a pair or not
bool found = false ;
// 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) == x) {
found = true ;
cout << "(" << first->data << ", "
<< second->data << ")" << endl;
// move first in forward direction
first = first->next;
// move second in backward direction
second = second->prev;
}
else {
if ((first->data * second->data) < x)
first = first->next;
else
second = second->prev;
}
}
// if pair is not present
if (found == false )
cout << "No pair found" ;
} // A utility function to insert a new node at the // beginning of doubly linked list void insert( struct Node** head, int data)
{ struct Node* temp = new Node;
temp->data = data;
temp->next = temp->prev = NULL;
if (!(*head))
(*head) = temp;
else {
temp->next = *head;
(*head)->prev = temp;
(*head) = temp;
}
} // Driver Code int main()
{ // Create Doubly Linked List
struct Node* head = NULL;
insert(&head, 9);
insert(&head, 8);
insert(&head, 6);
insert(&head, 5);
insert(&head, 4);
insert(&head, 2);
insert(&head, 1);
int x = 8;
pairProduct(head, x);
return 0;
} |
// Java program to find a pair with // given product x in sorted Doubly // Linked List class GFG {
// Doubly Linked List Node
static class Node {
int data;
Node next, prev;
};
// Function to find pair whose product
// equal to given value x
static void pairProduct(Node head, int x)
{
// Set two pointers,
// first to the beginning of DLL
// and second to the end of DLL.
Node first = head;
Node second = head;
while (second.next != null )
second = second.next;
// To track if we find a pair or not
boolean found = false ;
// 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) == x) {
found = true ;
System.out.println( "(" + first.data + ", "
+ second.data + ")" );
// move first in forward direction
first = first.next;
// move second in backward direction
second = second.prev;
}
else {
if ((first.data * second.data) < x)
first = first.next;
else
second = second.prev;
}
}
// if pair is not present
if (found == false )
System.out.println( "No pair found" );
}
// A utility function to insert a new node at the
// beginning of doubly linked list
static Node insert(Node head, int data)
{
Node temp = new Node();
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[])
{
// Create Doubly Linked List
Node head = null ;
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 ;
pairProduct(head, x);
}
} // This code is contributed by Arnab Kundu |
# Python3 program to find a pair with # given product x in sorted Doubly # Linked List # Node of the doubly linked list class Node:
def __init__( self , data):
self .data = data
self .prev = None
self . next = None
# Function to find pair whose product # equal to given value x def pairProduct(head, x):
# Set two pointers,
# first to the beginning of DLL
# and second to the end of DLL.
first = head
second = head
while (second. next ! = None ):
second = second. next
# To track if we find a pair or not
found = False
# 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) = = x) :
found = True
print ( "(" , first.data, ", " , second.data, ")" )
# move first in forward direction
first = first. next
# move second in backward direction
second = second.prev
else :
if ((first.data * second.data) < x):
first = first. next
else :
second = second.prev
# if pair is not present
if (found = = False ):
print ( "No pair found" )
# A utility function to insert a new node at the # beginning of doubly linked list def insert(head, data):
temp = Node( 0 )
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__" :
# Create Doubly Linked List
head = None
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
pairProduct(head, x)
# This code is contributed by Arnab Kundu |
// C# program to find a pair with // given product x in sorted Doubly // Linked List using System;
class GFG {
// Doubly Linked List Node
public class Node {
public int data;
public Node next, prev;
};
// Function to find pair whose product
// equal to given value x
static void pairProduct(Node head, int x)
{
// Set two pointers,
// first to the beginning of DLL
// and second to the end of DLL.
Node first = head;
Node second = head;
while (second.next != null )
second = second.next;
// To track if we find a pair or not
bool found = false ;
// 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) == x) {
found = true ;
Console.WriteLine( "(" + first.data + ", "
+ second.data + ")" );
// move first in forward direction
first = first.next;
// move second in backward direction
second = second.prev;
}
else {
if ((first.data * second.data) < x)
first = first.next;
else
second = second.prev;
}
}
// if pair is not present
if (found == false )
Console.WriteLine( "No pair found" );
}
// A utility function to insert a new node at the
// beginning of doubly linked list
static Node insert(Node head, int data)
{
Node temp = new Node();
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)
{
// Create Doubly Linked List
Node head = null ;
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;
pairProduct(head, x);
}
} // This code contributed by Rajput-Ji |
<script> // javascript program to find a pair with // given product x in sorted Doubly // Linked List // Doubly Linked List Node
class Node { constructor() {
this .data = 0;
this .prev = null ;
this .next = null ;
}
} // Function to find pair whose product
// equal to given value x
function pairProduct( head , x) {
// Set two pointers,
// first to the beginning of DLL
// and second to the end of DLL.
first = head;
second = head;
while (second.next != null )
second = second.next;
// To track if we find a pair or not
var found = false ;
// 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) == x) {
found = true ;
document.write( "(" + first.data + ", " + second.data + ")<br/>" );
// move first in forward direction
first = first.next;
// move second in backward direction
second = second.prev;
} else {
if ((first.data * second.data) < x)
first = first.next;
else
second = second.prev;
}
}
// if pair is not present
if (found == false )
document.write( "No pair found" );
}
// A utility function to insert a new node at the
// beginning of doubly linked list
function insert( head , data) {
temp = new Node();
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
// Create Doubly Linked List
head = null ;
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;
pairProduct(head, x);
// This code contributed by aashish1995 </script> |
(1, 8) (2, 4)
Time complexity: O(n)
Auxiliary Space: O(1)