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Find pairs with given sum in doubly linked list

  • Difficulty Level : Easy
  • Last Updated : 13 Aug, 2021

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

Example:  

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

The 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 a second element whose sum is equal to x in the remaining list by traversing in the forward direction. The time complexity for this problem will be O(n^2), n is the total number of nodes in the doubly linked list.

An efficient solution for this problem is the 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 the start of the doubly linked list i.e; first=head and initialize second with the last node of the 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 the second pointer, we traverse the list to initialize the 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 two pointers cross each other (second->next = first), or they become the same (first == second).
  • The case when no pairs are present will be handled by the condition “first==second”

C++




// 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 two pointers
    // cross each other (second->next
    // == first), or they become same (first == second)
    while (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;
}

Java




// 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 
    // they cross each other (second.next
    // == first), or they become same
    // (first == second)
    while ( 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

Python3




# Python3 program to find a pair with
# given sum x.
 
# Structure of node of doubly linked list
class Node:
     
    def __init__(self, x):
         
        self.data = x
        self.next = None
        self.prev = None
 
# Function to find pair whose sum
# equal to given value x.
def pairSum(head, x):
     
    # Set two pointers, first to the
    # beginning of DLL and second to
    # the end of DLL.
    first = head
    second = head
     
    while (second.next != None):
        second = second.next
 
    # To track if we find a pair or not
    found = False
 
    # The loop terminates when they
    # cross each other (second.next ==
    # first), or they become same
    # (first == second)
    while (first != second and second.next != first):
             
        # Pair found
        if ((first.data + second.data) == x):
            found = True
            print("(", 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):
        print("No pair found")
 
# A utility function to insert a new node
# at the beginning of doubly linked list
def insert(head, data):
     
    temp = Node(data)
     
    if not head:
        head = temp
    else:
        temp.next = head
        head.prev = temp
        head = temp
         
    return head
 
# Driver code
if __name__ == '__main__':
     
    head = None
    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)
    x = 7
 
    pairSum(head, x)
 
# This code is contributed by mohit kumar 29

C#




// 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 
        // they cross each other (second.next
        // == first), or they become same
        // (first == second)
        while (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

Javascript




<script>
// Javascript program to find a
// pair with given sum x.
 
// structure of node of
// doubly linked list
class Node
{
    constructor()
    {
        this.data = 0;
        this.next = this.prev = null;
    }
}
 
// Function to find pair whose
// sum equal to given value x.
function pairSum(head, x)
{
 
    // Set two pointers, first
    // to the beginning of DLL
    // and second to the end of DLL.
    let first = head;
    let second = head;
    while (second.next != null)
        second = second.next;
  
    // To track if we find a pair or not
    let found = false;
  
    // The loop terminates when
    // they cross each other (second.next
    // == first), or they become same
    // (first == second)
    while ( first != second && second.next != first)
    {
     
        // pair found
        if ((first.data + second.data) == x)
        {
            found = true;
            document.write( "(" + first.data +
                                ", "+ second.data + ")<br>" );
  
            // 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)
        document.write("No pair found<br>");
}
 
// A utility function to insert
// a new node at the beginning
// of doubly linked list
function insert(head,data)
{
    let 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
let 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);
let x = 7;
 
pairSum(head, x);
 
// This code is contributed by avanitrachhadiya2155
</script>

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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 




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