# 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)
```

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

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++ program to count all pairs from both the ` `// linked lists whose product is equal to ` `// a given value ` ` `  `#include ` `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; ` `} `

Output:

```Count = 1
```

Time complexity : O(N^2)

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Improved By : VishalBachchas

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