Count pairs from two linked lists whose product is equal to a given value
Last Updated :
06 Feb, 2023
Given two linked lists(can be sorted or unsorted) of size n1 and n2 of distinct elements. Given a value X. The problem is to count all pairs from both lists whose product is equal to the given value x.
Note:The pair must have an element from each linked list.
Examples:
Input : list1 = 3->1->5->7
list2 = 8->2->5->3
X = 10
Output : 1
The pair is: (5, 2)
Input : list1 = 4->3->5->7->11->2->1
list2 = 2->3->4->5->6->8-12
X = 9
Output : 1
The pair is: (3, 3)
A simple approach is using two loops pick elements from both the linked lists and check whether the product of the pair is equal to the given value X or not. Count all such pairs and print the result.
Below is the implementation of the above approach:
C++
// C++ program to count all pairs from both the
// linked lists whose product is equal to
// a given value
#include <iostream>
using namespace std;
/* A Linked list node */
struct Node {
int data;
struct Node* next;
};
// function to insert a node at the
// beginning of the linked list
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));
/* put in the data */
new_node->data = new_data;
/* link the old list to the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
// Function to count all pairs from both the linked lists
// whose product is equal to a given value
int countPairs(struct Node* head1, struct Node* head2, int x)
{
int count = 0;
struct Node *p1, *p2;
// Traverse the 1st linked list
for (p1 = head1; p1 != NULL; p1 = p1->next) {
// for each node of 1st list
// Traverse the 2nd list
for (p2 = head2; p2 != NULL; p2 = p2->next) {
// if sum of pair is equal to 'x'
// increment count
if ((p1->data * p2->data) == x)
count++;
}
}
// required count of pairs
return count;
}
// Driver Code
int main()
{
struct Node* head1 = NULL;
struct Node* head2 = NULL;
// create linked list1 3->1->5->7
push(&head1, 7);
push(&head1, 5);
push(&head1, 1);
push(&head1, 3);
// create linked list2 8->2->5->3
push(&head2, 3);
push(&head2, 5);
push(&head2, 2);
push(&head2, 8);
int x = 10;
cout << "Count = " << countPairs(head1, head2, x);
return 0;
}
// Java program to count all pairs from both the
// linked lists whose product is equal to
// a given value
class GFG
{
/* A Linked list node */
static class Node
{
int data;
Node next;
};
// function to insert a node at the
// beginning of the linked list
static Node push(Node head_ref, int new_data)
{
/* allocate node */
Node new_node = new Node();
/* put in the data */
new_node.data = new_data;
/* link the old list to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
return head_ref;
}
// Function to count all pairs from both the linked lists
// whose product is equal to a given value
static int countPairs(Node head1, Node head2, int x)
{
int count = 0;
Node p1, p2;
// Traverse the 1st linked list
for (p1 = head1; p1 != null; p1 = p1.next)
{
// for each node of 1st list
// Traverse the 2nd list
for (p2 = head2; p2 != null; p2 = p2.next)
{
// if sum of pair is equal to 'x'
// increment count
if ((p1.data * p2.data) == x)
{
count++;
}
}
}
// required count of pairs
return count;
}
// Driver Code
public static void main(String[] args)
{
Node head1 = null;
Node head2 = null;
// create linked list1 3.1.5.7
head1 = push(head1, 7);
head1 = push(head1, 5);
head1 = push(head1, 1);
head1 = push(head1, 3);
// create linked list2 8.2.5.3
head2 = push(head2, 3);
head2 = push(head2, 5);
head2 = push(head2, 2);
head2 = push(head2, 8);
int x = 10;
System.out.print("Count = " + countPairs(head1, head2, x));
}
}
// This code is contributed by Rajput-Ji
# Python3 program to count all pairs from both the
# linked lists whose product is equal to
# a given value
''' A Linked list node '''
class Node:
def __init__(self, data):
self.data = data
self.next = None
# function to insert a node at the
# beginning of the linked list
def push(head_ref, new_data):
''' allocate node '''
new_node = Node(new_data)
''' put in the data '''
new_node.data = new_data;
''' link the old list to the new node '''
new_node.next = (head_ref);
''' move the head to point to the new node '''
(head_ref) = new_node;
return head_ref
# Function to count all pairs from both the linked lists
# whose product is equal to a given value
def countPairs(head1, head2, x):
count = 0;
p1 = head1
# Traverse the 1st linked list
while p1 != None:
p2 = head2
# for each node of 1st list
# Traverse the 2nd list
while p2 != None:
# if sum of pair is equal to 'x'
# increment count
if ((p1.data * p2.data) == x):
count += 1
p2 = p2.next
p1 = p1.next
# required count of pairs
return count;
# Driver Code
if __name__=='__main__':
head1 = None;
head2 = None;
# create linked list1 3.1.5.7
head1 = push(head1, 7);
head1 = push(head1, 5);
head1 = push(head1, 1);
head1 = push(head1, 3);
# create linked list2 8.2.5.3
head2 = push(head2, 3);
head2 = push(head2, 5);
head2 = push(head2, 2);
head2 = push(head2, 8);
x = 10;
print("Count = " + str(countPairs(head1, head2, x)))
# This code is contributed by rutvik_56
// C# program to count all pairs from both the
// linked lists whose product is equal to
// a given value
using System;
class GFG
{
/* A Linked list node */
class Node
{
public int data;
public Node next;
};
// function to insert a node at the
// beginning of the linked list
static Node push(Node head_ref, int new_data)
{
/* allocate node */
Node new_node = new Node();
/* put in the data */
new_node.data = new_data;
/* link the old list to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
return head_ref;
}
// Function to count all pairs from both the linked lists
// whose product is equal to a given value
static int countPairs(Node head1, Node head2, int x)
{
int count = 0;
Node p1, p2;
// Traverse the 1st linked list
for (p1 = head1; p1 != null; p1 = p1.next)
{
// for each node of 1st list
// Traverse the 2nd list
for (p2 = head2; p2 != null; p2 = p2.next)
{
// if sum of pair is equal to 'x'
// increment count
if ((p1.data * p2.data) == x)
{
count++;
}
}
}
// required count of pairs
return count;
}
// Driver Code
public static void Main(String[] args)
{
Node head1 = null;
Node head2 = null;
// create linked list1 3->1->5->7
head1 = push(head1, 7);
head1 = push(head1, 5);
head1 = push(head1, 1);
head1 = push(head1, 3);
// create linked list2 8->2->5->3
head2 = push(head2, 3);
head2 = push(head2, 5);
head2 = push(head2, 2);
head2 = push(head2, 8);
int x = 10;
Console.Write("Count = " + countPairs(head1, head2, x));
}
}
// This code is contributed by Rajput-Ji
<script>
// javascript program to count all pairs from both the
// linked lists whose product is equal to
// a given value
/* A Linked list node */
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
// function to insert a node at the
// beginning of the linked list
function push(head_ref , new_data) {
/* allocate node */
var new_node = new Node();
/* put in the data */
new_node.data = new_data;
/* link the old list to the new node */
new_node.next = head_ref;
/* move the head to point to the new node */
head_ref = new_node;
return head_ref;
}
// Function to count all pairs from both the linked lists
// whose product is equal to a given value
function countPairs(head1, head2 , x) {
var count = 0;
var p1, p2;
// Traverse the 1st linked list
for (p1 = head1; p1 != null; p1 = p1.next) {
// for each node of 1st list
// Traverse the 2nd list
for (p2 = head2; p2 != null; p2 = p2.next) {
// if sum of pair is equal to 'x'
// increment count
if ((p1.data * p2.data) == x) {
count++;
}
}
}
// required count of pairs
return count;
}
// Driver Code
var head1 = null;
var head2 = null;
// create linked list1 3.1.5.7
head1 = push(head1, 7);
head1 = push(head1, 5);
head1 = push(head1, 1);
head1 = push(head1, 3);
// create linked list2 8.2.5.3
head2 = push(head2, 3);
head2 = push(head2, 5);
head2 = push(head2, 2);
head2 = push(head2, 8);
var x = 10;
document.write("Count = " + countPairs(head1, head2, x));
// This code contributed by umadevi9616
</script>
Time complexity: O(N^2)
Auxiliary Space: O(1)
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