Find pairs with given sum in doubly linked list

06 Jul, 2022
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:  
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”
Implementation:
C++
Java
Python3
C#
Javascript
// C++ program to find a pair with given sum x.#include<bits/stdc++.h>using namespace std; // structure of node of doubly linked liststruct 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 listvoid 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 programint 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 program to find a// pair with given sum x.class GFG{ // structure of node of// doubly linked liststatic 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 liststatic 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 Codepublic 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 program to find a pair with# given sum x. # Structure of node of doubly linked listclass 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 listdef insert(head, data):         temp = Node(data)         if not head:        head = temp    else:        temp.next = head        head.prev = temp        head = temp             return head # Driver codeif __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# 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
<script>// Javascript program to find a// pair with given sum x. // structure of node of// doubly linked listclass 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 listfunction 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 Codelet 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>
(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. 
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29AjayKumar
abhishek gupta 47
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mohit kumar 29
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