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

2.1

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 sum is equal to the given value x.

Note: The pair has an element from each linked list.

Examples:

Input : list1 = 3->1->5->7
        list2 = 8->2->5->3
        x = 10
Output : 2
The pairs are:
(5, 5) and (7, 3)

Input : list1 = 4->3->5->7->11->2->1
        list2 = 2->3->4->5->6->8-12
        x = 9         
Output : 5

Method 1 (Naive Approach): Using two loops pick elements from both the linked lists and check whether the sum of the pair is equal to x or not.

C/C++

// C++ implementation to count pairs from both linked 
// lists  whose sum is equal to a given value
#include <bits/stdc++.h>
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 sum 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 program to test above
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

// Java implementation to count pairs from both linked 
// lists  whose sum is equal to a given value

// Note : here we use java.util.LinkedList for 
// linked list implementation

import java.util.Arrays;
import java.util.Iterator;
import java.util.LinkedList;

class GFG 
{
	// method to count all pairs from both the linked lists
	// whose sum is equal to a given value
	static int countPairs(LinkedList<Integer> head1, LinkedList<Integer> head2, int x)
	{
	    int count = 0;
	     
	    // traverse the 1st linked list
	    Iterator<Integer> itr1 = head1.iterator();
	    while(itr1.hasNext())
	    {
	        // for each node of 1st list
	        // traverse the 2nd list
	    	Iterator<Integer> itr2 = head2.iterator();
	     
	    	while(itr2.hasNext())
	    	{
	    		// if sum of pair is equal to 'x'
	            // increment count
	            if ((itr1.next() + itr2.next()) == x)
	                count++; 
	    	}
	    }
	                       
	    // required count of pairs     
	    return count;
	}
	
	// Driver method
	public static void main(String[] args) 
	{
		Integer arr1[] = {3, 1, 5, 7};
		Integer arr2[] = {8, 2, 5, 3};
		
		// create linked list1 3->1->5->7
		LinkedList<Integer> head1 = new LinkedList<>(Arrays.asList(arr1));
		
		// create linked list2 8->2->5->3
	    LinkedList<Integer> head2 = new LinkedList<>(Arrays.asList(arr2));
	   
	    int x = 10;
	     
	    System.out.println("Count = " + countPairs(head1, head2, x));
	}	
}


Output:

Count = 2

Time Complexity: O(n1*n2)
Auxiliary Space: O(1)

 

Method 2 (Sorting): Sort the 1st linked list in ascending order and the 2nd linked list in descending order using merge sort technique. Now traverse both the lists from left to right in the following way:

Algorithm:

countPairs(list1, list2, x)
  Initialize count = 0
  while list != NULL and list2 != NULL
     if (list1->data + list2->data) == x
        list1 = list1->next    
        list2 = list2->next
        count++
    else if (list1->data + list2->data) > x
        list2 = list2->next
    else
        list1 = list1->next

  return count     

For simplicity, the implementation given below assumes that list1 is sorted in ascending order and list2 is sorted in descending order.

C/C++

// C++ implementation to count pairs from both linked 
// lists  whose sum is equal to a given value
#include <bits/stdc++.h>
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 sum is equal to a given value
int countPairs(struct Node* head1, struct Node* head2,
                                              int x)
{
    int count = 0;
    
    // sort head1 in ascending order and
    // head2 in descending order
    // sort (head1), sort (head2)
    // For simplicity both lists are considered to be 
    // sorted in the respective orders
    
    // traverse both the lists from left to right
    while (head1 != NULL && head2 != NULL)
    {
        // if this sum is equal to 'x', then move both 
        // the lists to next nodes and increment 'count'
        if ((head1->data + head2->data) == x)
        {
            head1 = head1->next;
            head2 = head2->next;
            count++;    
        }    
        
        // if this sum is greater than x, then
        // move head2 to next node
        else if ((head1->data + head2->data) > x)
            head2 = head2->next;
            
        // else move head1 to next node    
        else
            head1 = head1->next;
    }        
        
    // required count of pairs     
    return count;
}

// Driver program to test above
int main()
{
    struct Node* head1 = NULL;
    struct Node* head2 = NULL;
    
    // create linked list1 1->3->5->7
    // assumed to be in ascending order
    push(&head1, 7);
    push(&head1, 5);
    push(&head1, 3);
    push(&head1, 1);    
    
    // create linked list2 8->5->3->2
    // assumed to be in descending order
    push(&head2, 2);
    push(&head2, 3);
    push(&head2, 5);
    push(&head2, 8);
    
    int x = 10;
    
    cout << "Count = "
         << countPairs(head1, head2, x);
    return 0;
}

Java

// Java implementation to count pairs from both linked 
// lists  whose sum is equal to a given value

// Note : here we use java.util.LinkedList for 
// linked list implementation

import java.util.Arrays;
import java.util.Collections;
import java.util.Iterator;
import java.util.LinkedList;

class GFG 
{
	// method to count all pairs from both the linked lists
	// whose sum is equal to a given value
	static int countPairs(LinkedList<Integer> head1, LinkedList<Integer> head2, int x)
	{
	    int count = 0;
	     
	    // sort head1 in ascending order and
		// head2 in descending order
	    Collections.sort(head1);
	    Collections.sort(head2,Collections.reverseOrder());
	    
	    // traverse both the lists from left to right
	    Iterator<Integer> itr1 = head1.iterator();
	    Iterator<Integer> itr2 = head2.iterator();
	    
	    Integer num1 = itr1.hasNext() ? itr1.next() : null;
        Integer num2 = itr2.hasNext() ? itr2.next() : null;
	    
	    while(num1 != null && num2 != null)
	    {	 
	        
	    	// if this sum is equal to 'x', then move both 
			// the lists to next nodes and increment 'count'
	    	
			if ((num1 + num2) == x)
			{
				num1 = itr1.hasNext() ? itr1.next() : null;
				num2 = itr2.hasNext() ? itr2.next() : null;
				
				count++; 
			} 
			
			// if this sum is greater than x, then
			// move itr2 to next node
			else if ((num1 + num2) > x)
				num2 = itr2.hasNext() ? itr2.next() : null;
			
			// else move itr1 to next node 
			else
				num1 = itr1.hasNext() ? itr1.next() : null;
			
	    }
	                       
	    // required count of pairs     
	    return count;
	}
	
	// Driver method
	public static void main(String[] args) 
	{
		Integer arr1[] = {3, 1, 5, 7};
		Integer arr2[] = {8, 2, 5, 3};
		
		// create linked list1 3->1->5->7
		LinkedList<Integer> head1 = new LinkedList<>(Arrays.asList(arr1));
		
		// create linked list2 8->2->5->3
	    LinkedList<Integer> head2 = new LinkedList<>(Arrays.asList(arr2));
	   
	    int x = 10;
	     
	    System.out.println("Count = " + countPairs(head1, head2, x));
	}	
}


Output:

Count = 2

Time Complexity: O(n1*logn1) + O(n2*logn2)
Auxiliary Space: O(1)
Sorting will change the order of nodes. If order is important, then copy of the linked lists can be created and used.

Method 3 (Hashing): Hash table is implemented using unordered_set in C++. We store all first linked list elements in hash table. For elements of second linked list, we subtract every element from x and check the result in hash table. If result is present, we increment the count.

C++

// C++ implementation to count pairs from both linked  
// lists whose sum is equal to a given value
#include <bits/stdc++.h>
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 sum is equal to a given value
int countPairs(struct Node* head1, struct Node* head2, 
                                               int x)
{
    int count = 0;
    
    unordered_set<int> us;
    
    // insert all the elements of 1st list
    // in the hash table(unordered_set 'us')
    while (head1 != NULL)
    {
        us.insert(head1->data);    
        
        // move to next node    
        head1 = head1->next;
    }
    
    // for each element of 2nd list
    while (head2 != NULL)    
    {
        // find (x - head2->data) in 'us'
        if (us.find(x - head2->data) != us.end())
            count++;
        
        // move to next node
        head2 = head2->next;    
    }
    // required count of pairs     
    return count;
}

// Driver program to test above
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

// Java implementation to count pairs from both linked 
// lists  whose sum is equal to a given value

// Note : here we use java.util.LinkedList for 
// linked list implementation

import java.util.Arrays;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;

class GFG 
{
	// method to count all pairs from both the linked lists
	// whose sum is equal to a given value
	static int countPairs(LinkedList<Integer> head1, LinkedList<Integer> head2, int x)
	{
		int count = 0;
	     
	    HashSet<Integer> us = new HashSet<Integer>();
	     
	    // insert all the elements of 1st list
	    // in the hash table(unordered_set 'us')
	    Iterator<Integer> itr1 = head1.iterator();
	    while (itr1.hasNext())
	    {
	        us.add(itr1.next());    
	       
	    }
	    
	    Iterator<Integer> itr2 = head2.iterator();
	    // for each element of 2nd list
	    while (itr2.hasNext())    
	    {
	        // find (x - head2->data) in 'us'
	        if (us.add(x - itr2.next()))
	            count++;
	           
	    }
	    
	    // required count of pairs     
	    return count;
	}
	
	// Driver method
	public static void main(String[] args) 
	{
		Integer arr1[] = {3, 1, 5, 7};
		Integer arr2[] = {8, 2, 5, 3};
		
		// create linked list1 3->1->5->7
		LinkedList<Integer> head1 = new LinkedList<>(Arrays.asList(arr1));
		
		// create linked list2 8->2->5->3
	    LinkedList<Integer> head2 = new LinkedList<>(Arrays.asList(arr2));
	   
	    int x = 10;
	     
	    System.out.println("Count = " + countPairs(head1, head2, x));
	}	
}


Output:

Count = 2

Time Complexity: O(n1 + n2)
Auxiliary Space: O(n1), hash table should be created of the array having smaller size so as to reduce the space complexity.

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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