Count pairs from two linked lists whose product is equal to a given value

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:

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// 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;
}

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Output:

Count = 1

Time complexity : O(N^2)



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Improved By : VishalBachchas