Find quadruplets with given sum in a Doubly Linked List
Last Updated :
23 Jul, 2025
Given a sorted doubly linked list and an integer X, the task is to print all the quadruplets in the doubly linked list whose sum is X.
Examples:
Input: LL: -3 ? 1 ? 2 ? 3 ? 5 ? 6, X = 7
Output:
-3 2 3 5
-3 3 1 6
Explanation: The quadruplets having sum 7( = X) are: {-3, 2, 3, 5}, {-3, 3, 1, 6}.
Input: LL: -2 ? -1 ? 0 ? 0 ? 1 ? 2, X = 0
Output:
-2 -1 1 2
-2 0 0 2
-1 0 0 1
Approach: The given problem can be solved by using the idea discussed in this article using the 4 pointer technique. Follow the steps below to solve the problem:
- Initialize four variables, say first as the start of the doubly linked list i.e.; first = head, second to the next of the first pointer, third to the next of second pointer, and fourth to the last node of a doubly linked list that stores all the 4 elements in the sorted doubly linked list.
- Iterate a loop until the first node and the fourth node is not NULL, and they are not equal and these 2 nodes do not cross each other and perform the following steps:
- Initialize the second pointer to the next of the first.
- Iterate a loop until the second and fourth nodes are not NULL, they are not equal and do not cross each other.
- Initialize a variable, say sum as (X - (first?data + second?data)), point the third pointer to the next of second pointer, and take another temp pointer initialized to the last node i.e., pointer fourth.
- Iterate a loop while temp and third are not NULL, they are not equal and do not cross each other
- If the value of the sum is third?data + temp?data, then print the quadruple and increment the third pointer to the next of current third pointer and temp to the previous of current temp.
- If the value of sum is less than the third?data + temp?data, then increment the third pointer, i.e., third = third?next.
- Otherwise, decrement the temp pointer i.e temp = temp?prev.
- Move the second pointer to its next pointer.
- Move the first pointer to its next pointer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Structure of node of a doubly
// linked list
struct Node {
int data;
struct Node *next, *prev;
};
// Function to insert a new node at
// the beginning of the doubly linked
// list
void insert(struct Node** head, int data)
{
// Allocate the node
struct Node* temp = new Node();
// Fill in the data value
temp->data = data;
temp->next = temp->prev = NULL;
if ((*head) == NULL)
(*head) = temp;
else {
temp->next = *head;
(*head)->prev = temp;
(*head) = temp;
}
}
// Function to print the quadruples
// having sum equal to x
void PrintFourSum(struct Node* head, int x)
{
// First pointer to the head node
struct Node* first = head;
// Pointer to point to the second
// node for the required sum
struct Node* second;
// Pointer to point to the third
// node for the required sum
struct Node* third;
// Fourth points to the last node
struct Node* fourth = head;
// Update the fourth pointer to
// the end of the DLL
while (fourth->next != NULL) {
fourth = fourth->next;
}
// Node to point to the fourth node
// of the required sum
struct Node* temp;
while (first != NULL
&& fourth != NULL
&& first != fourth
&& fourth->next != first) {
// Point the second node to the
// second element of quadruple
second = first->next;
while (second != NULL
&& fourth != NULL
&& second != fourth
&& fourth->next != second) {
int reqsum = x - (first->data
+ second->data);
// Points to the 3rd element
// of quadruple
third = second->next;
// Points to the tail of the DLL
temp = fourth;
while (third != NULL && temp != NULL
&& third != temp
&& temp->next != third) {
// Store the current sum
int twosum = third->data
+ temp->data;
// If the sum is equal,
// then print quadruple
if (twosum == reqsum) {
cout << "(" << first->data
<< ", "
<< second->data
<< ", "
<< third->data
<< ", "
<< temp->data
<< ")\n";
third = third->next;
temp = temp->prev;
}
// If twosum is less than
// the reqsum then move the
// third pointer to the next
else if (twosum < reqsum) {
third = third->next;
}
// Otherwise move the fourth
// pointer to the previous
// of the fourth pointer
else {
temp = temp->prev;
}
}
// Move to the next of
// the second pointer
second = second->next;
}
// Move to the next of
// the first pointer
first = first->next;
}
}
// Driver Code
int main()
{
struct Node* head = NULL;
insert(&head, 2);
insert(&head, 1);
insert(&head, 0);
insert(&head, 0);
insert(&head, -1);
insert(&head, -2);
int X = 0;
PrintFourSum(head, X);
return 0;
}
// Java program for the above approach
class GFG {
// structure of node of
// doubly linked list
static class Node {
int data;
Node next, prev;
};
// A utility function to insert
// a new node at the beginning
// of the doubly linked list
static Node insert(Node head, int data)
{
// Allocate the node
Node temp = new Node();
// Fill in the data value
temp.data = data;
temp.next = temp.prev = null;
if (head == null)
(head) = temp;
else {
temp.next = head;
(head).prev = temp;
(head) = temp;
}
return temp;
}
// Function to print the quadruples
// having sum equal to x
static void PrintFourSum(Node head, int x)
{
// First pointer to the head node
Node first = head;
// Pointer to point to the second
// node for the required sum
Node second = head;
// Pointer to point to the third
// node for the required sum
Node third = head;
// Fourth points to the last node
Node fourth = head;
// Update the fourth pointer to
// the end of the DLL
while (fourth.next != null) {
fourth = fourth.next;
}
// Node to point to the fourth node
// of the required sum
Node temp;
while (first != null && fourth != null
&& first != fourth && fourth.next != first) {
// Point the second node to the
// second element of quadruple
second = first.next;
while (second != null && fourth != null
&& second != fourth
&& fourth.next != second) {
int reqsum = x - (first.data + second.data);
// Points to the 3rd element
// of quadruple
third = second.next;
// Points to the tail of the DLL
temp = fourth;
while (third != null && temp != null
&& third != temp
&& temp.next != third) {
// Store the current sum
int twosum = third.data + temp.data;
// If the sum is equal,
// then print quadruple
if (twosum == reqsum) {
System.out.println(
"(" + first.data + ", "
+ second.data + ", "
+ third.data + ", " + temp.data
+ ")");
third = third.next;
temp = temp.prev;
}
// If twosum is less than
// the reqsum then move the
// third pointer to the next
else if (twosum < reqsum) {
third = third.next;
}
// Otherwise move the fourth
// pointer to the previous
// of the fourth pointer
else {
temp = temp.prev;
}
}
// Move to the next of
// the second pointer
second = second.next;
}
// Move to the next of
// the first pointer
first = first.next;
}
}
// Driver Code
public static void main(String args[])
{
Node head = null;
head = insert(head, 2);
head = insert(head, 1);
head = insert(head, 0);
head = insert(head, 0);
head = insert(head, -1);
head = insert(head, -2);
int x = 0;
PrintFourSum(head, x);
}
}
// This code is contributed
// by kirtishsurangalikar
# Python3 program for the above approach
# Structure of node of a doubly
# linked list
class Node:
def __init__(self, d):
self.data = d
self.left = None
self.right = None
# Function to insert a new node at
# the beginning of the doubly linked
# list
def insert(head, data):
# Allocate the node
temp = Node(data)
# Fill in the data value
temp.data = data
temp.next = temp.prev = None
if (head == None):
head = temp
else:
temp.next = head
head.prev = temp
head = temp
return head
# Function to print the quadruples
# having sum equal to x
def PrintFourSum(head, x):
# First pointer to the head node
first = head
# Pointer to point to the second
# node for the required sum
second = None
# Pointer to point to the third
# node for the required sum
third = None
# Fourth points to the last node
fourth = head
# Update the fourth pointer to
# the end of the DLL
while (fourth.next != None):
fourth = fourth.next
# Node to point to the fourth node
# of the required sum
temp = None
while (first != None and
fourth != None and
first != fourth and
fourth.next != first):
# Point the second node to the
# second element of quadruple
second = first.next
while (second != None and
fourth != None and
second != fourth and
fourth.next != second):
reqsum = x - (first.data +
second.data)
# Points to the 3rd element
# of quadruple
third = second.next
# Points to the tail of the DLL
temp = fourth
while (third != None and temp != None and
third != temp and temp.next != third):
# Store the current sum
twosum = third.data + temp.data
# If the sum is equal,
# then print quadruple
if (twosum == reqsum):
print("(" + str(first.data) +
", " + str(second.data) +
", " + str(third.data) +
", " + str(temp.data) + ")")
third = third.next
temp = temp.prev
# If twosum is less than
# the reqsum then move the
# third pointer to the next
elif (twosum < reqsum):
third = third.next
# Otherwise move the fourth
# pointer to the previous
# of the fourth pointer
else:
temp = temp.prev
# Move to the next of
# the second pointer
second = second.next
# Move to the next of
# the first pointer
first = first.next
# Driver Code
if __name__ == '__main__':
head = None
head = insert(head, 2)
head = insert(head, 1)
head = insert(head, 0)
head = insert(head, 0)
head = insert(head, -1)
head = insert(head, -2)
X = 0
PrintFourSum(head, X)
# This code is contributed by mohit kumar 29
// C# program for the above approach
using System;
public class GFG {
// structure of node of
// doubly linked list
public class Node {
public int data;
public Node next, prev;
};
// A utility function to insert
// a new node at the beginning
// of the doubly linked list
static Node insert(Node head, int data) {
// Allocate the node
Node temp = new Node();
// Fill in the data value
temp.data = data;
temp.next = temp.prev = null;
if (head == null)
(head) = temp;
else {
temp.next = head;
(head).prev = temp;
(head) = temp;
}
return temp;
}
// Function to print the quadruples
// having sum equal to x
static void PrintFourSum(Node head, int x) {
// First pointer to the head node
Node first = head;
// Pointer to point to the second
// node for the required sum
Node second = head;
// Pointer to point to the third
// node for the required sum
Node third = head;
// Fourth points to the last node
Node fourth = head;
// Update the fourth pointer to
// the end of the DLL
while (fourth.next != null) {
fourth = fourth.next;
}
// Node to point to the fourth node
// of the required sum
Node temp;
while (first != null && fourth != null && first != fourth && fourth.next != first) {
// Point the second node to the
// second element of quadruple
second = first.next;
while (second != null && fourth != null && second != fourth && fourth.next != second) {
int reqsum = x - (first.data + second.data);
// Points to the 3rd element
// of quadruple
third = second.next;
// Points to the tail of the DLL
temp = fourth;
while (third != null && temp != null && third != temp && temp.next != third) {
// Store the current sum
int twosum = third.data + temp.data;
// If the sum is equal,
// then print quadruple
if (twosum == reqsum) {
Console.WriteLine(
"(" + first.data + ", " + second.data + ", " + third.data + ", " + temp.data + ")");
third = third.next;
temp = temp.prev;
}
// If twosum is less than
// the reqsum then move the
// third pointer to the next
else if (twosum < reqsum) {
third = third.next;
}
// Otherwise move the fourth
// pointer to the previous
// of the fourth pointer
else {
temp = temp.prev;
}
}
// Move to the next of
// the second pointer
second = second.next;
}
// Move to the next of
// the first pointer
first = first.next;
}
}
// Driver Code
public static void Main(String []args) {
Node head = null;
head = insert(head, 2);
head = insert(head, 1);
head = insert(head, 0);
head = insert(head, 0);
head = insert(head, -1);
head = insert(head, -2);
int x = 0;
PrintFourSum(head, x);
}
}
// This code is contributed by gauravrajput1
<script>
// Javascript program for the above approach
// Structure of node of a doubly
// linked list
class Node {
constructor()
{
this.data = 0;
this.next = null;
this.prev = null;
}
};
// Function to insert a new node at
// the beginning of the doubly linked
// list
function insert(head, data)
{
// Allocate the node
var temp = new Node();
// Fill in the data value
temp.data = data;
temp.next = temp.prev = null;
if ((head) == null)
(head) = temp;
else {
temp.next = head;
(head).prev = temp;
(head) = temp;
}
return head;
}
// Function to print the quadruples
// having sum equal to x
function PrintFourSum(head, x)
{
// First pointer to the head node
var first = head;
// Pointer to point to the second
// node for the required sum
var second;
// Pointer to point to the third
// node for the required sum
var third;
// Fourth points to the last node
var fourth = head;
// Update the fourth pointer to
// the end of the DLL
while (fourth.next != null) {
fourth = fourth.next;
}
// Node to point to the fourth node
// of the required sum
var temp;
while (first != null
&& fourth != null
&& first != fourth
&& fourth.next != first) {
// Point the second node to the
// second element of quadruple
second = first.next;
while (second != null
&& fourth != null
&& second != fourth
&& fourth.next != second) {
var reqsum = x - (first.data
+ second.data);
// Points to the 3rd element
// of quadruple
third = second.next;
// Points to the tail of the DLL
temp = fourth;
while (third != null && temp != null
&& third != temp
&& temp.next != third) {
// Store the current sum
var twosum = third.data
+ temp.data;
// If the sum is equal,
// then print quadruple
if (twosum == reqsum) {
document.write( "(" + first.data
+ ", "
+ second.data
+ ", "
+ third.data
+ ", "
+ temp.data
+ ")<br>");
third = third.next;
temp = temp.prev;
}
// If twosum is less than
// the reqsum then move the
// third pointer to the next
else if (twosum < reqsum) {
third = third.next;
}
// Otherwise move the fourth
// pointer to the previous
// of the fourth pointer
else {
temp = temp.prev;
}
}
// Move to the next of
// the second pointer
second = second.next;
}
// Move to the next of
// the first pointer
first = first.next;
}
}
// Driver Code
var head = null;
head = insert(head, 2);
head = insert(head, 1);
head = insert(head, 0);
head = insert(head, 0);
head = insert(head, -1);
head = insert(head, -2);
var X = 0;
PrintFourSum(head, X);
// This code is contributed by rrrtnx.
</script>
Output(-2, -1, 1, 2)
(-2, 0, 0, 2)
(-1, 0, 0, 1)
Time Complexity: O(N3)
Auxiliary Space: O(1)
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