Find pairs with given sum in doubly linked list

Given a sorted doubly linked list of positive distinct elements, the task is to find pairs in doubly linked list whose sum is equal to given value x, without using any extra space ?
Example:

Input : head : 1 <-> 2 <-> 4 <-> 5 <-> 6 <-> 8 <-> 9
        x = 7
Output: (6, 1), (5,2)

Expected time complexity is O(n) and auxiliary space is O(1).



A simple approach for this problem is to one by one pick each node and find second element whose sum is equal to x in the remaining list by traversing in forward direction.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 sum of first and second is less than x, then we move first in forward direction. If current sum 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)

C++

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// C++ program to find a pair with given sum x.
#include<bits/stdc++.h>
using namespace std;
  
// structure of node of doubly linked list
struct Node
{
    int data;
    struct Node *next, *prev;
};
  
// Function to find pair whose sum equal to given value x.
void pairSum(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 program
int main()
{
    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 = 7;
  
    pairSum(head, x);
  
    return 0;
}

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Java

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// Java program to find a
// pair with given sum x.
class GFG
{
  
// structure of node of
// doubly linked list
static class Node
{
    int data;
    Node next, prev;
};
  
// Function to find pair whose 
// sum equal to given value x.
static void pairSum( 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 temp;
}
  
// Driver Code
public static void main(String args[])
{
    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 = 7;
  
    pairSum(head, x);
}
}
  
// This code is contributed
// by Arnab Kundu

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C#

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// C# program to find a 
// pair with given sum x.
using System; 
  
class GFG 
  
    // structure of node of 
    // doubly linked list 
    class Node 
    
        public int data; 
        public Node next, prev; 
    }; 
  
    // Function to find pair whose 
    // sum equal to given value x. 
    static void pairSum( 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 temp; 
    
  
    // Driver Code 
    public static void Main(String []args) 
    
        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 = 7; 
  
        pairSum(head, x); 
    
  
// This code is contributed by 29AjayKumar

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Output:

(1,6) 
(2,5)

Time complexity : O(n)
Auxiliary space : O(1)

If linked list is not sorted, then we can sort the list as a first step. But in that case overall time complexity would become O(n Log n). We can use Hashing in such cases if extra space is not a constraint. The hashing based solution is same as method 2 here.

This article is contributed by Shashank Mishra ( Gullu ). 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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Improved By : andrew1234, 29AjayKumar



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