Find pairs with given sum in doubly linked list

Last Updated : 04 Dec, 2024

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 the given value x in sorted order.

Examples:

Input:

Output: (1, 6), (2,5)
Explanation: We can see that there are two pairs (1, 6) and (2, 5) with sum 7.
Input:

Output: (1,5)
Explanation: We can see that there is one pair (1, 5) with a sum of 6.

Table of Content

[Naive Approach] Using Hashing - O(nlogn) Time and O(n) Space

This approach finds pairs of nodes in a doubly linked list that sum up to a given target. It uses a hash map to track the values of nodes as the list is traversed. For each node, it calculates the complement (target - node value) and checks if it exists in the map. If found, a valid pair is stored. After traversal, the pairs are sorted to maintain the order.

C++

// C++ program to find a pair with given sum x
// using map
#include <bits/stdc++.h>
using namespace std;

class Node {
  public:
    int data;
    Node *next;
    Node(int value) {
        data = value;
        next = nullptr;
    }
};

// Function to find pairs in the doubly linked list
// whose sum equals the given value x
vector<pair<int, int>> findPairsWithGivenSum(Node *head, int target) {
    vector<pair<int, int>> ans;
    unordered_map<int, Node *> visited;
    Node *currNode = head;

    // Traverse the doubly linked list
    while (currNode != nullptr) {
      
        int x = target - currNode->data;

        // Check if the target exists in the map
        if (visited.find(x) != visited.end()) {
          
            // Pair found
            ans.push_back({x, currNode->data});
        }

        // Store the current node's value in the map
        visited[currNode->data] = currNode;
        currNode = currNode->next;
    }

    sort(ans.begin(), ans.end());

    return ans;
}

int main() {
  
    // Create a doubly linked list: 1 <-> 2 <-> 4 <-> 5
    Node *head = new Node(1);
    head->next = new Node(2);
    head->next->next = new Node(4);
    head->next->next->next = new Node(5);

    int target = 7;
    vector<pair<int, int>> pairs = findPairsWithGivenSum(head, target);

    if (pairs.empty()) {
        cout << "No pairs found." << endl;
    }
    else {
        for (auto &pair : pairs) {
            cout << pair.first << " " << pair.second << endl;
        }
    }

    return 0;
}

Java

// Java program to find a pair with given sum x
// using map
import java.util.*;

class Node {
    int data;
    Node next;
    Node(int value) {
        data = value;
        next = null;
    }
}

class GfG {

    static ArrayList<ArrayList<Integer> >
    findPairsWithGivenSum(int target, Node head) {
        ArrayList<ArrayList<Integer> > ans
            = new ArrayList<>();
        HashSet<Integer> visited = new HashSet<>();
        Node currNode = head;

        // Traverse the doubly linked list
        while (currNode != null) {
            int x = target - currNode.data;

            // Check if the target exists in the visited set
            if (visited.contains(x)) {
              
                // Pair found, add it to the answer
                ArrayList<Integer> pair = new ArrayList<>();
                pair.add(x);
                pair.add(currNode.data);
                ans.add(pair);
            }

            // Add the current node's value to the visited
            // set
            visited.add(currNode.data);
            currNode = currNode.next;
        }

        // Sort the pairs by the first element of the pair
        Collections.sort(
            ans, (a, b) -> a.get(0).compareTo(b.get(0)));

        return ans;
    }

    public static void main(String[] args) {
      
        // Create a doubly linked list: 1 <-> 2 <-> 4 <-> 5
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(4);
        head.next.next.next = new Node(5);

        int target = 7;
        ArrayList<ArrayList<Integer> > pairs
            = findPairsWithGivenSum(target, head);

        if (pairs.isEmpty()) {
            System.out.println("No pairs found.");
        }
        else {
            for (ArrayList<Integer> pair : pairs) {
                System.out.println(pair.get(0) + " "
                                   + pair.get(1));
            }
        }
    }
}

Python

# Python program to find a pair with given sum x
# using map
class Node:
    def __init__(self, value):
        self.data = value
        self.next = None


def findPairsWithGivenSum(target, head):
    ans = []
    visited = set()
    currNode = head

    # Traverse the doubly linked list
    while currNode is not None:
        x = target - currNode.data

        # Check if the target exists in the visited set
        if x in visited:
          
            # Pair found, add it to the answer
            ans.append([x, currNode.data])

        # Add the current node's value to the visited set
        visited.add(currNode.data)
        currNode = currNode.next

    # Sort the pairs by the first element of the pair
    ans.sort(key=lambda pair: pair[0])

    return ans

# Helper function to create a linked list
def create_linked_list(values):
    head = Node(values[0])
    current = head
    for value in values[1:]:
        current.next = Node(value)
        current = current.next
    return head


if __name__ == "__main__":
  
    # Create a doubly linked list: 1 -> 2 -> 4 -> 5
    values = [1, 2, 4, 5]
    head = create_linked_list(values)

    target = 7
    pairs = findPairsWithGivenSum(target, head)

    if not pairs:
        print("No pairs found.")
    else:
        for pair in pairs:
            print(pair[0], pair[1])

// C# program to find a pair with given sum x
// using map
using System;
using System.Collections.Generic;

class Node {
    public int data;
    public Node next;
    
    public Node(int value) {
        data = value;
        next = null;
    }
}

class GfG {
     static List<Tuple<int, int>> FindPairsWithGivenSum(Node head, int target) {
        List<Tuple<int, int>> ans = new List<Tuple<int, int>>();
        HashSet<int> visited = new HashSet<int>();
        Node currNode = head;

        // Traverse the doubly linked list
        while (currNode != null) {
            int x = target - currNode.data;

            // Check if the target exists in the visited set
            if (visited.Contains(x)) {
              
                // Pair found
                ans.Add(new Tuple<int, int>(x, currNode.data));
            }

            // Add the current node's value to the visited set
            visited.Add(currNode.data);
            currNode = currNode.next;
        }

        // Sort the pairs
        ans.Sort((a, b) => a.Item1.CompareTo(b.Item1));

        return ans;
    }

    static void Main() {
      
        // Create a doubly linked list: 1 <-> 2 <-> 4 <-> 5
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(4);
        head.next.next.next = new Node(5);

        int target = 7;
        List<Tuple<int, int>> pairs = FindPairsWithGivenSum(head, target);

        if (pairs.Count == 0) {
            Console.WriteLine("No pairs found.");
        } else {
            foreach (var pair in pairs) {
                Console.WriteLine(pair.Item1 + " " + pair.Item2);
            }
        }
    }
}

JavaScript

// JavaScript program to find a pair with given sum x
// using map
class Node {
    constructor(value) {
        this.data = value;
        this.next = null;
    }
}

function findPairsWithGivenSum(head, target) {
    let ans = [];
    let visited = new Set();
    let currNode = head;

    // Traverse the doubly linked list
    while (currNode !== null) {
        let x = target - currNode.data;

        // Check if the target exists in the visited set
        if (visited.has(x)) {
            // Pair found
            ans.push([ x, currNode.data ]);
        }

        // Add the current node's value to the visited set
        visited.add(currNode.data);
        currNode = currNode.next;
    }

    // Sort the pairs
    ans.sort((a, b) => a[0] - b[0]);

    return ans;
}

let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(4);
head.next.next.next = new Node(5);

let target = 7;
let pairs = findPairsWithGivenSum(head, target);

if (pairs.length === 0) {
    console.log("No pairs found.");
}
else {
    for (let pair of pairs) {
        console.log(pair[0] + " " + pair[1]);
    }
}

Output

2 5

[Expected Approach] Using Two Pointer Technique - O(n) Time and O(1) Space

The idea is to use two pointers, one starting at the head and the other at the end of the sorted doubly linked list. Move the first pointer forward if the sum of the two pointer's values is less than the target, or move the second pointer backward if the sum is greater. Continue until the pointers meet or cross each other.

An efficient solution for this problem is the same as Pair with given Sum (Two Sum) article.

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 and second with the last node.
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.
The loop terminates when two pointers cross each other (second->next = first), or they become the same (first == second).

C++

// C++ program to find a pair with given sum target.
#include <bits/stdc++.h>
using namespace std;

class Node {
  public:
    int data;
    Node *next, *prev;

    Node(int value) {
        data = value;
        next = nullptr;
        prev = nullptr;
    }
};

// Function to find pairs in the doubly
// linked list whose sum equals the given value x
vector<pair<int, int>> findPairsWithGivenSum(Node *head, int target) {

    vector<pair<int, int>> res;

    // Set two pointers, first to the beginning of DLL
    // and second to the end of DLL.
    Node *first = head;
    Node *second = head;

    // Move second to the end of the DLL
    while (second->next != nullptr)
        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 the same (first == second)
    while (first != second && second->next != first) {

        // If the sum of the two nodes is equal to x, print the pair
        if ((first->data + second->data) == target) {
            found = true;
            res.push_back({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) < target)
                first = first->next;
            else
                second = second->prev;
        }
    }

    // If no pair is found
    return res;
}

int main() {

    // Create a doubly linked list: 1 <-> 2 <-> 4 <-> 5
    Node *head = new Node(1);
    head->next = new Node(2);
    head->next->prev = head;
    head->next->next = new Node(4);
    head->next->next->prev = head->next;
    head->next->next->next = new Node(5);
    head->next->next->next->prev = head->next->next;

    int target = 7;
    vector<pair<int, int>> pairs = findPairsWithGivenSum(head, target);

    if (pairs.empty()) {
        cout << "No pairs found." << endl;
    }
    else {
        for (auto &pair : pairs) {
            cout << pair.first << " " << pair.second << endl;
        }
    }
    return 0;
}

Java

// Java program to find a pair with given sum target.
import java.util.ArrayList;

class Node {
    int data;
    Node next, prev;

    Node(int value) {
        data = value;
        next = prev = null;
    }
}

class GfG {

    // Function to find pairs in the doubly linked list
    // whose sum equals the given value target
    static ArrayList<ArrayList<Integer> >
    findPairsWithGivenSum(int target, Node head) {
        ArrayList<ArrayList<Integer> > res
            = new ArrayList<>();

        // Set two pointers, first to the beginning of DLL
        // and second to the end of DLL.
        Node first = head;
        Node second = head;

        // Move second to the end of the DLL
        while (second.next != null)
            second = second.next;

        // Iterate through the list using two pointers to
        // find pairs
        while (first != second && second.next != first) {

            // If the sum of the two nodes is equal to
            // target, add the pair
            if ((first.data + second.data) == target) {
                ArrayList<Integer> pair = new ArrayList<>();
                pair.add(first.data);
                pair.add(second.data);
                res.add(pair);

                // Move first in forward direction
                first = first.next;

                // Move second in backward direction
                second = second.prev;
            }
            else {
                if ((first.data + second.data) < target)
                    first = first.next;
                else
                    second = second.prev;
            }
        }

        return res;
    }

    public static void main(String[] args) {

        // Create a doubly linked list: 1 <-> 2 <-> 4 <-> 5
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.prev = head;
        head.next.next = new Node(4);
        head.next.next.prev = head.next;
        head.next.next.next = new Node(5);
        head.next.next.next.prev = head.next.next;

        int target = 7;
        ArrayList<ArrayList<Integer> > pairs
            = findPairsWithGivenSum(target, head);

        if (pairs.isEmpty()) {
            System.out.println("No pairs found.");
        }
        else {
            for (ArrayList<Integer> pair : pairs) {
                System.out.println(pair.get(0)
                                   + " " + pair.get(1));
            }
        }
    }
}

Python

# Python program to find a pair with given sum target.
class Node:
    def __init__(self, value):
        self.data = value
        self.next = None
        self.prev = None


# Function to find pairs in the doubly linked list
# whose sum equals the given value target
def find_pairs_with_given_sum(head, target):
    res = []
    
    # Set two pointers, first to the beginning of DLL
    # and second to the end of DLL.
    first = head
    second = head
    
    # Move second to the end of the DLL
    while second.next is not None:
        second = second.next
    
    # The loop terminates when two pointers
    # cross each other (second.next == first),
    # or they become the same (first == second)
    while first != second and second.next != first:
        if (first.data + second.data) == target:
          
            # Add pair to the result list
            res.append((first.data, second.data))
            
            # Move first in forward direction
            first = first.next
            
            # Move second in backward direction
            second = second.prev
        elif (first.data + second.data) < target:
            first = first.next
        else:
            second = second.prev
    
    return res


if __name__ == "__main__":
  
    # Create a doubly linked list: 1 <-> 2 <-> 4 <-> 5
    head = Node(1)
    head.next = Node(2)
    head.next.prev = head
    head.next.next = Node(4)
    head.next.next.prev = head.next
    head.next.next.next = Node(5)
    head.next.next.next.prev = head.next.next

    target = 7
    
    pairs = find_pairs_with_given_sum(head, target)

    if not pairs:
        print("No pairs found.")
    else:
        for pair in pairs:
            print(pair[0], pair[1])

// C# program to find a pair with given sum target.
using System;
using System.Collections.Generic;

class Node {
    public int Data;
    public Node Next, Prev;

    public Node(int value) {
        Data = value;
        Next = null;
        Prev = null;
    }
}

class GfG {
  
    // Function to find pairs in the doubly linked list
    // whose sum equals the given value target
    static List<Tuple<int, int> >
    FindPairsWithGivenSum(Node head, int target) {
        List<Tuple<int, int> > result
            = new List<Tuple<int, int> >();

        Node first = head;
        Node second = head;

        // Move second pointer to the last node
        while (second.Next != null)
            second = second.Next;

        // Iterate using two pointers to find pairs
        while (first != second && second.Next != first) {
            if (first.Data + second.Data == target) {
                result.Add(
                    Tuple.Create(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 < target) {
                first = first.Next;
            }
            else {
                second = second.Prev;
            }
        }

        return result;
    }

    static void Main(string[] args) {
      
        // Create a doubly linked list: 1 <-> 2 <-> 4 <-> 5
        Node head = new Node(1);
        head.Next = new Node(2);
        head.Next.Prev = head;
        head.Next.Next = new Node(4);
        head.Next.Next.Prev = head.Next;
        head.Next.Next.Next = new Node(5);
        head.Next.Next.Next.Prev = head.Next.Next;

        int target = 7;
        List<Tuple<int, int> > pairs
            = FindPairsWithGivenSum(head, target);

        if (pairs.Count == 0) {
            Console.WriteLine("No pairs found.");
        }
        else {
            foreach(var pair in pairs) {
               Console.WriteLine($"{pair.Item1} {pair.Item2}");
            }
        }
    }
}

JavaScript

// JavaScript program to find a pair with given sum target.
class Node {
    constructor(value) {
        this.data = value;
        this.next = null;
        this.prev = null;
    }
}

// Function to find pairs in the doubly linked list
// whose sum equals the given value target
function findPairsWithGivenSum(head, target) {
    let res = [];

    let first = head;
    let second = head;

    // Move second pointer to the end of the DLL
    while (second.next !== null) {
        second = second.next;
    }

    // Iterate through the list using two pointers
    while (first !== second && second.next !== first) {
        if (first.data + second.data === target) {
            res.push([ first.data, second.data ]);

            // Move first pointer forward and second pointer
            // backward
            first = first.next;
            second = second.prev;
        }
        else if (first.data + second.data < target) {
            first = first.next;
        }
        else {
            second = second.prev;
        }
    }

    return res;
}

// Function to create a doubly linked list node
function createNode(value) { return new Node(value); }

// Create a doubly linked list: 1 <-> 2 <-> 4 <-> 5
let head = createNode(1);
head.next = createNode(2);
head.next.prev = head;
head.next.next = createNode(4);
head.next.next.prev = head.next;
head.next.next.next = createNode(5);
head.next.next.next.prev = head.next.next;

let target = 7;
let pairs = findPairsWithGivenSum(head, target);

if (pairs.length === 0) {
    console.log("No pairs found.");
}
else {
    pairs.forEach(
        pair => { console.log(`${pair[0]} ${pair[1]}`); });
}