Given a Linked List of size n, where every node represents a sub-linked-list and contains two pointers:
- next pointer to the next node
- bottom pointer to a linked list where this node is head.
Each of the sub-linked-list is in sorted order. Flatten the Link List such that all the nodes appear in a single level while maintaining the sorted order.
Note: The flattened list will be printed using the bottom pointer instead of next pointer.
Examples:
Input:
Output: 5->7->8->10->19->22->28->30->50
Input:
Output: 5->7->8->10->19->20->22->28->30->35->40->45->50
Flattening a Linked List by Sorting nodes in array - O(n * m * log(n * m)) Time and O(n*m) Space
The idea is to traverse the linked list and push values of all the nodes in an array. Now, we can sort the array in ascending order, and create a new linked list by traversing the sorted array. While creating the new linked list, we will append the nodes using the bottom pointer of flattened linked list.
C++
// C++ program for flattening a Linked List
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node *next, *bottom;
Node(int newdata) {
data = newdata;
next = bottom = nullptr;
}
};
// function to flatten the linked list
Node* flatten(Node* root) {
vector<int> values;
// Push values of all nodes into an array
while (root != nullptr) {
// Push the head node of the sub-linked-list
values.push_back(root->data);
// Push all the nodes of the sub-linked-list
Node* temp = root->bottom;
while (temp != nullptr) {
values.push_back(temp->data);
temp = temp->bottom;
}
// Move to the next head node
root = root->next;
}
// Sort the node values in ascending order
sort(values.begin(), values.end());
// Construct the new flattened linked list
Node* tail = nullptr;
Node* head = nullptr;
for (int i = 0; i < values.size(); i++) {
Node* newNode = new Node(values[i]);
// If this is the first node of the linked list,
// make the node as head
if (head == nullptr) {
head = newNode;
}
else {
tail->bottom = newNode;
}
tail = newNode;
}
return head;
}
void printList(Node* head) {
Node* temp = head;
while (temp != nullptr) {
cout << temp->data << " ";
temp = temp->bottom;
}
cout << endl;
}
int main() {
// Create a hard-coded linked list:
// 5 -> 10 -> 19 -> 28
// | | |
// V V V
// 7 20 22
// | |
// V V
// 8 50
// |
// V
// 30
Node* head = new Node(5);
head->bottom = new Node(7);
head->bottom->bottom = new Node(8);
head->bottom->bottom->bottom = new Node(30);
head->next = new Node(10);
head->next->bottom = new Node(20);
head->next->next = new Node(19);
head->next->next->bottom = new Node(22);
head->next->next->bottom->bottom = new Node(50);
head->next->next->next = new Node(28);
head = flatten(head);
printList(head);
return 0;
}
// Java Program for flattening a linked list
import java.util.*;
class Node {
int data;
Node next, bottom;
Node(int newData) {
data = newData;
next = bottom = null;
}
}
class GfG {
// Function to flatten the linked list
static Node flatten(Node root) {
List<Integer> values = new ArrayList<>();
// Push values of all nodes into a list
while (root != null) {
// Push the head node of the sub-linked-list
values.add(root.data);
// Push all the nodes of the sub-linked-list
Node temp = root.bottom;
while (temp != null) {
values.add(temp.data);
temp = temp.bottom;
}
// Move to the next head node
root = root.next;
}
// Sort the node values in ascending order
Collections.sort(values);
// Construct the new flattened linked list
Node tail = null;
Node head = null;
for (int value : values) {
Node newNode = new Node(value);
// If this is the first node of the linked list,
// make the node as head
if (head == null)
head = newNode;
else
tail.bottom = newNode;
tail = newNode;
}
return head;
}
// Function to print the linked list
static void printList(Node head) {
Node temp = head;
while (temp != null) {
System.out.print(temp.data + " ");
temp = temp.bottom;
}
System.out.println();
}
public static void main(String[] args) {
// Create a hard-coded linked list:
// 5 -> 10 -> 19 -> 28
// | | |
// V V V
// 7 20 22
// | |
// V V
// 8 50
// |
// V
// 30
Node head = new Node(5);
head.bottom = new Node(7);
head.bottom.bottom = new Node(8);
head.bottom.bottom.bottom = new Node(30);
head.next = new Node(10);
head.next.bottom = new Node(20);
head.next.next = new Node(19);
head.next.next.bottom = new Node(22);
head.next.next.bottom.bottom = new Node(50);
head.next.next.next = new Node(28);
head = flatten(head);
printList(head);
}
}
# Python program for flattening a Linked List
class Node:
def __init__(self, new_data):
self.data = new_data
self.next = None
self.bottom = None
# Function to flatten the linked list
def flatten(root):
values = []
# Push values of all nodes into an array
while root is not None:
# Push the head node of the sub-linked-list
values.append(root.data)
# Push all the nodes of the sub-linked-list
temp = root.bottom
while temp is not None:
values.append(temp.data)
temp = temp.bottom
# Move to the next head node
root = root.next
# Sort the node values in ascending order
values.sort()
# Construct the new flattened linked list
tail = None
head = None
for value in values:
newNode = Node(value)
# If this is the first node of the linked list,
# make the node as head
if head is None:
head = newNode
else:
tail.bottom = newNode
tail = newNode
return head
def printList(head):
temp = head
while temp is not None:
print(temp.data, end=" ")
temp = temp.bottom
print()
if __name__ == "__main__":
# Create a hard-coded linked list:
# 5 -> 10 -> 19 -> 28
# | | |
# V V V
# 7 20 22
# | |
# V V
# 8 50
# |
# V
# 30
head = Node(5)
head.bottom = Node(7)
head.bottom.bottom = Node(8)
head.bottom.bottom.bottom = Node(30)
head.next = Node(10)
head.next.bottom = Node(20)
head.next.next = Node(19)
head.next.next.bottom = Node(22)
head.next.next.bottom.bottom = Node(50)
head.next.next.next = Node(28)
head = flatten(head)
printList(head)
// C# Program for flattening a linked list
using System;
using System.Collections.Generic;
using System.Linq;
class Node {
public int Data;
public Node Next, Bottom;
public Node(int newData) {
Data = newData;
Next = Bottom = null;
}
}
class GfG {
// Function to flatten the linked list
static Node Flatten(Node root) {
List<int> values = new List<int>();
// Push values of all nodes into a list
while (root != null) {
// Push the head node of the sub-linked-list
values.Add(root.Data);
// Push all the nodes of the sub-linked-list
Node temp = root.Bottom;
while (temp != null) {
values.Add(temp.Data);
temp = temp.Bottom;
}
// Move to the next head node
root = root.Next;
}
// Sort the node values in ascending order
values.Sort();
// Construct the new flattened linked list
Node tail = null;
Node head = null;
foreach(int value in values) {
Node newNode = new Node(value);
// If this is the first node of the linked list,
// make the node as head
if (head == null) {
head = newNode;
}
else {
tail.Bottom = newNode;
}
tail = newNode;
}
return head;
}
// Function to print the linked list
static void PrintList(Node head) {
Node temp = head;
while (temp != null) {
Console.Write(temp.Data + " ");
temp = temp.Bottom;
}
Console.WriteLine();
}
static void Main() {
/* Create a hard-coded linked list:
5 -> 10 -> 19 -> 28
| | |
V V V
7 20 22
| |
V V
8 50
|
V
30
*/
Node head = new Node(5);
head.Bottom = new Node(7);
head.Bottom.Bottom = new Node(8);
head.Bottom.Bottom.Bottom = new Node(30);
head.Next = new Node(10);
head.Next.Bottom = new Node(20);
head.Next.Next = new Node(19);
head.Next.Next.Bottom = new Node(22);
head.Next.Next.Bottom.Bottom = new Node(50);
head.Next.Next.Next = new Node(28);
head = Flatten(head);
PrintList(head);
}
}
// Javascript Program for flattening a linked list
class Node {
constructor(newData) {
this.data = newData;
this.next = null;
this.bottom = null;
}
}
// Function to flatten the linked list
function flatten(root) {
let values = [];
// Push values of all nodes into an array
while (root !== null) {
// Push the head node of the sub-linked-list
values.push(root.data);
// Push all the nodes of the sub-linked-list
let temp = root.bottom;
while (temp !== null) {
values.push(temp.data);
temp = temp.bottom;
}
// Move to the next head node
root = root.next;
}
// Sort the node values in ascending order
values.sort((a, b) => a - b);
// Construct the new flattened linked list
let head = null;
let tail = null;
for (let value of values) {
let newNode = new Node(value);
// If this is the first node of the linked list,
// make the node as head
if (head === null)
head = newNode;
else
tail.bottom = newNode;
tail = newNode;
}
return head;
}
// Function to print the linked list
function printList(head) {
let temp = head;
while (temp !== null) {
console.log(temp.data + " ");
temp = temp.bottom;
}
console.log();
}
/* Create a hard-coded linked list:
5 -> 10 -> 19 -> 28
| | |
V V V
7 20 22
| |
V V
8 50
|
V
30
*/
let head = new Node(5);
head.bottom = new Node(7);
head.bottom.bottom = new Node(8);
head.bottom.bottom.bottom = new Node(30);
head.next = new Node(10);
head.next.bottom = new Node(20);
head.next.next = new Node(19);
head.next.next.bottom = new Node(22);
head.next.next.bottom.bottom = new Node(50);
head.next.next.next = new Node(28);
head = flatten(head);
printList(head);
Output5 7 8 10 19 20 22 28 30 50
Flattening a Linked List by Merging lists recursively - O(n * n * m) Time and O(n) Space
The idea is similar to Merge two sorted linked lists. Lets assume that we have only two linked list, head1 and head2 which we need to flatten. We can maintain a dummy node as the start of the result list and a pointer, say tail which always points to the last node in the result list, so appending new nodes is easy. Initially tail node points to the dummy node. Now, traverse till both either head1 or head2 does not reach NULL.
- If value of head1 is smaller than or equal to value of head2, then, append head1 after the tail node and update tail to head1.
- Similarly, if value of head2 is smaller than value of head1, then append head2 after the tail node and update tail to head2.
When either head1 or head2 reaches NULL, simply append the remaining nodes after the tail node.
We can use the above idea to recursively merge the linked list, starting from the last two sub-linked-lists. At each node, we can have two cases:
- If the node is the last node of linked list, simply return the head.
- Otherwise, call the recursive function with the next of head to get the flattened linked list and merge the flattened linked list with the current sub-linked-list.
C++
// C++ program for flattening a Linked List
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node *next, *bottom;
Node(int newdata) {
data = newdata;
next = bottom = nullptr;
}
};
// Utility function to merge two sorted linked lists
// using their bottom pointers
Node* merge(Node* head1, Node* head2) {
// A dummy first node to store the result list
Node dummy(-1);
// Tail points to the last result node to add new nodes
// to the result
Node* tail = &dummy;
// Iterate till either head1 or head2 does not reach NULL
while (head1 && head2) {
if (head1->data <= head2->data) {
// append head1 to the result
tail->bottom = head1;
head1 = head1->bottom;
}
else {
// append head2 to the result
tail->bottom = head2;
head2 = head2->bottom;
}
// Move tail pointer to the next node
tail = tail->bottom;
}
// Append the remaining nodes of non-null linked list
if (!head1)
tail->bottom = head2;
else
tail->bottom = head1;
return (dummy.bottom);
}
// function to flatten the linked list
Node* flatten(Node* root) {
// Base Cases
if (root == nullptr || root->next == nullptr)
return root;
// Recur for next list
root->next = flatten(root->next);
// Now merge the current and next list
root = merge(root, root->next);
// Return the root
return root;
}
void printList(Node* head) {
Node* temp = head;
while (temp != nullptr) {
cout << temp->data << " ";
temp = temp->bottom;
}
cout << endl;
}
int main() {
/* Create a hard-coded linked list:
5 -> 10 -> 19 -> 28
| | |
V V V
7 20 22
| |
V V
8 50
|
V
30
*/
Node* head = new Node(5);
head->bottom = new Node(7);
head->bottom->bottom = new Node(8);
head->bottom->bottom->bottom = new Node(30);
head->next = new Node(10);
head->next->bottom = new Node(20);
head->next->next = new Node(19);
head->next->next->bottom = new Node(22);
head->next->next->bottom->bottom = new Node(50);
head->next->next->next = new Node(28);
head = flatten(head);
printList(head);
return 0;
}
// C program for flattening a Linked List
#include <stdio.h>
struct Node {
int data;
struct Node *next, *bottom;
};
// Utility function to merge two sorted linked lists using
// their bottom pointers
struct Node* merge(struct Node* head1, struct Node* head2) {
// A dummy first node to store the result list
struct Node dummy;
dummy.data = -1;
dummy.bottom = NULL;
// Tail points to the last result node to add new nodes
// to the result
struct Node* tail = &dummy;
// Iterate till either head1 or head2 does not reach
// NULL
while (head1 && head2) {
if (head1->data <= head2->data) {
// append head1 to the result
tail->bottom = head1;
head1 = head1->bottom;
}
else {
// append head2 to the result
tail->bottom = head2;
head2 = head2->bottom;
}
// Move tail pointer to the next node
tail = tail->bottom;
}
// Append the remaining nodes of non-null linked list
if (!head1)
tail->bottom = head2;
else
tail->bottom = head1;
return dummy.bottom;
}
// Function to flatten the linked list
struct Node* flatten(struct Node* root) {
// Base Cases
if (root == NULL || root->next == NULL)
return root;
// Recur for next list
root->next = flatten(root->next);
// Now merge the current and next list
root = merge(root, root->next);
// Return the root
return root;
}
void printList(struct Node* head) {
struct Node* temp = head;
while (temp != NULL) {
printf("%d ", temp->data);
temp = temp->bottom;
}
printf("\n");
}
// Function to create a new node
struct Node* createNode(int newdata) {
struct Node* newnode
= (struct Node*)malloc(sizeof(struct Node));
newnode->data = newdata;
newnode->next = NULL;
newnode->bottom = NULL;
return newnode;
}
int main() {
/* Create a hard-coded linked list:
5 -> 10 -> 19 -> 28
| | |
V V V
7 20 22
| |
V V
8 50
|
V
30
*/
struct Node* head = createNode(5);
head->bottom = createNode(7);
head->bottom->bottom = createNode(8);
head->bottom->bottom->bottom = createNode(30);
head->next = createNode(10);
head->next->bottom = createNode(20);
head->next->next = createNode(19);
head->next->next->bottom = createNode(22);
head->next->next->bottom->bottom = createNode(50);
head->next->next->next = createNode(28);
head = flatten(head);
printList(head);
return 0;
}
// Java program for flattening a Linked List
class Node {
int data;
Node next, bottom;
Node(int newData) {
data = newData;
next = bottom = null;
}
}
class GfG {
// Utility function to merge two sorted linked lists
// using their bottom pointers
static Node merge(Node head1, Node head2) {
// A dummy first node to store the result list
Node dummy = new Node(-1);
// Tail points to the last result node to add new
// nodes to the result
Node tail = dummy;
// Iterate till either head1 or head2 does not reach
// null
while (head1 != null && head2 != null) {
if (head1.data <= head2.data) {
// Append head1 to the result
tail.bottom = head1;
head1 = head1.bottom;
}
else {
// Append head2 to the result
tail.bottom = head2;
head2 = head2.bottom;
}
// Move tail pointer to the next node
tail = tail.bottom;
}
// Append the remaining nodes of the non-null list
if (head1 != null)
tail.bottom = head1;
else
tail.bottom = head2;
return dummy.bottom;
}
// Function to flatten the linked list
static Node flatten(Node root) {
// Base Cases
if (root == null || root.next == null)
return root;
// Recur for next list
root.next = flatten(root.next);
// Now merge the current and next list
root = merge(root, root.next);
// Return the root
return root;
}
static void printList(Node head) {
Node temp = head;
while (temp != null) {
System.out.print(temp.data + " ");
temp = temp.bottom;
}
System.out.println();
}
public static void main(String[] args) {
/* Create a hard-coded linked list:
5 -> 10 -> 19 -> 28
| | |
V V V
7 20 22
| |
V V
8 50
|
V
30
*/
Node head = new Node(5);
head.bottom = new Node(7);
head.bottom.bottom = new Node(8);
head.bottom.bottom.bottom = new Node(30);
head.next = new Node(10);
head.next.bottom = new Node(20);
head.next.next = new Node(19);
head.next.next.bottom = new Node(22);
head.next.next.bottom.bottom = new Node(50);
head.next.next.next = new Node(28);
head = flatten(head);
printList(head);
}
}
# Python3 program for flattening a Linked List
class Node:
def __init__(self, new_data):
self.data = new_data
self.next = None
self.bottom = None
# Utility function to merge two sorted linked lists
# using their bottom pointers
def merge(head1, head2):
# A dummy first node to store the result list
dummy = Node(-1)
tail = dummy
# Iterate till either head1 or head2 does not reach None
while head1 and head2:
if head1.data <= head2.data:
# Append head1 to the result
tail.bottom = head1
head1 = head1.bottom
else:
# Append head2 to the result
tail.bottom = head2
head2 = head2.bottom
# Move tail pointer to the next node
tail = tail.bottom
# Append the remaining nodes of the non-null linked list
if head1:
tail.bottom = head1
else:
tail.bottom = head2
return dummy.bottom
# Function to flatten the linked list
def flatten(root):
# Base Cases
if root is None or root.next is None:
return root
# Recur for next list
root.next = flatten(root.next)
# Now merge the current and next list
root = merge(root, root.next)
# Return the root
return root
def printList(head):
temp = head
while temp is not None:
print(temp.data, end=" ")
temp = temp.bottom
print()
if __name__ == "__main__":
# Create a hard-coded linked list:
# 5 -> 10 -> 19 -> 28
# | | |
# V V V
# 7 20 22
# | |
# V V
# 8 50
# |
# V
# 30
head = Node(5)
head.bottom = Node(7)
head.bottom.bottom = Node(8)
head.bottom.bottom.bottom = Node(30)
head.next = Node(10)
head.next.bottom = Node(20)
head.next.next = Node(19)
head.next.next.bottom = Node(22)
head.next.next.bottom.bottom = Node(50)
head.next.next.next = Node(28)
head = flatten(head)
printList(head)
// C# program for flattening a Linked List
using System;
class Node {
public int Data;
public Node Next;
public Node Bottom;
public Node(int newData) {
Data = newData;
Next = null;
Bottom = null;
}
}
class GfG {
// Utility function to merge two sorted linked lists
// using their bottom pointers
static Node Merge(Node head1, Node head2) {
// A dummy first node to store the result list
Node dummy = new Node(-1);
// Tail points to the last result node to add new
// nodes to the result
Node tail = dummy;
// Iterate till either head1 or head2 does not reach
// null
while (head1 != null && head2 != null) {
if (head1.Data <= head2.Data) {
// Append head1 to the result
tail.Bottom = head1;
head1 = head1.Bottom;
}
else {
// Append head2 to the result
tail.Bottom = head2;
head2 = head2.Bottom;
}
// Move tail pointer to the next node
tail = tail.Bottom;
}
// Append the remaining nodes of the
// non-null linked list
if (head1 == null)
tail.Bottom = head2;
else
tail.Bottom = head1;
return dummy.Bottom;
}
// Function to flatten the linked list
static Node Flatten(Node root) {
// Base Cases
if (root == null || root.Next == null)
return root;
// Recur for next list
root.Next = Flatten(root.Next);
// Now merge the current and next list
root = Merge(root, root.Next);
// Return the root
return root;
}
static void PrintList(Node head) {
Node temp = head;
while (temp != null) {
Console.Write(temp.Data + " ");
temp = temp.Bottom;
}
Console.WriteLine();
}
static void Main() {
/* Create a hard-coded linked list:
5 -> 10 -> 19 -> 28
| | |
V V V
7 20 22
| |
V V
8 50
|
V
30
*/
Node head = new Node(5);
head.Bottom = new Node(7);
head.Bottom.Bottom = new Node(8);
head.Bottom.Bottom.Bottom = new Node(30);
head.Next = new Node(10);
head.Next.Bottom = new Node(20);
head.Next.Next = new Node(19);
head.Next.Next.Bottom = new Node(22);
head.Next.Next.Bottom.Bottom = new Node(50);
head.Next.Next.Next = new Node(28);
head = Flatten(head);
PrintList(head);
}
}
// Javascript program for flattening a Linked List
class Node {
constructor(newData) {
this.data = newData;
this.next = null;
this.bottom = null;
}
}
// Utility function to merge two sorted linked lists
// using their bottom pointers
function merge(head1, head2) {
// A dummy first node to store the result list
let dummy = new Node(-1);
// Tail points to the last result node to add new nodes to the result
let tail = dummy;
// Iterate till either head1 or head2 does not reach null
while (head1 !== null && head2 !== null) {
if (head1.data <= head2.data) {
// Append head1 to the result
tail.bottom = head1;
head1 = head1.bottom;
} else {
// Append head2 to the result
tail.bottom = head2;
head2 = head2.bottom;
}
// Move tail pointer to the next node
tail = tail.bottom;
}
// Append the remaining nodes of non-null linked list
if (head1 === null)
tail.bottom = head2;
else
tail.bottom = head1;
return dummy.bottom;
}
// Function to flatten the linked list
function flatten(root) {
// Base Cases
if (root === null || root.next === null) {
return root;
}
// Recur for next list
root.next = flatten(root.next);
// Now merge the current and next list
root = merge(root, root.next);
// Return the root
return root;
}
function printList(head) {
let temp = head;
while (temp !== null) {
console.log(temp.data, " ");
temp = temp.bottom;
}
console.log();
}
/* Create a hard-coded linked list:
5 -> 10 -> 19 -> 28
| | |
V V V
7 20 22
| |
V V
8 50
|
V
30
*/
let head = new Node(5);
head.bottom = new Node(7);
head.bottom.bottom = new Node(8);
head.bottom.bottom.bottom = new Node(30);
head.next = new Node(10);
head.next.bottom = new Node(20);
head.next.next = new Node(19);
head.next.next.bottom = new Node(22);
head.next.next.bottom.bottom = new Node(50);
head.next.next.next = new Node(28);
head = flatten(head);
printList(head);
Output5 7 8 10 19 20 22 28 30 50
Time Complexity: O(n * n * m) - where n is the no of nodes in the main linked list and m is the no of nodes in a single sub-linked list.
- After adding the first 2 lists, the time taken will be O(m+m) = O(2m).
- Then we will merge another list to above merged list -> time = O(2m + m) = O(3m).
- We will keep merging lists to previously merged lists until all lists are merged.
- Total time taken will be O(2m + 3m + 4m + .... n*m) = (2 + 3 + 4 + ... + n) * m = O(n * n * m)
Auxiliary Space: O(n), the recursive functions will use a recursive stack of a size equivalent to a total number of nodes in the main linked list.
Flattening a Linked List using Priority Queues - O(n * m * log(n)) Time and O(n) Space
The idea is to use a Min Heap and push all the nodes that are on the first level into it. While the priority queue is not empty, pop the smallest element from the priority queue and append it to the flattened linked list. Also, check if there is a node at the bottom of the popped node then push that bottom node to the priority queue. Since all the sub-linked -lists are sorted, all the nodes will be popped in sorted order.
C++
// C++ program for flattening a Linked List
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node *next, *bottom;
Node(int newdata) {
data = newdata;
next = bottom = nullptr;
}
};
// comparator function for priority queue
struct mycomp {
bool operator()(Node* a, Node* b) {
return a->data > b->data;
}
};
// function to flatten the given linked list
Node* flatten(Node* root) {
priority_queue<Node*, vector<Node*>, mycomp> pq;
Node* head = nullptr;
Node* tail = nullptr;
// pushing main link nodes into priority_queue
while (root != nullptr) {
pq.push(root);
root = root->next;
}
// Extracting the minimum node
// while priority queue is not empty
while (!pq.empty()) {
// extracting min
auto minNode = pq.top();
pq.pop();
if (head == nullptr) {
head = minNode;
tail = minNode;
}
else {
tail->bottom = minNode;
tail = tail->bottom;
}
// If we have another node at the bottom of the
// popped node, push that node into the priority
// queue
if (minNode->bottom) {
pq.push(minNode->bottom);
minNode->bottom = nullptr;
}
}
return head;
}
void printList(Node* head) {
Node* temp = head;
while (temp != nullptr) {
cout << temp->data << " ";
temp = temp->bottom;
}
cout << endl;
}
int main() {
/* Create a hard-coded linked list:
5 -> 10 -> 19 -> 28
| | |
V V V
7 20 22
| |
V V
8 50
|
V
30
*/
Node* head = new Node(5);
head->bottom = new Node(7);
head->bottom->bottom = new Node(8);
head->bottom->bottom->bottom = new Node(30);
head->next = new Node(10);
head->next->bottom = new Node(20);
head->next->next = new Node(19);
head->next->next->bottom = new Node(22);
head->next->next->bottom->bottom = new Node(50);
head->next->next->next = new Node(28);
head = flatten(head);
printList(head);
return 0;
}
// Java program for flattening a Linked List
import java.util.PriorityQueue;
class Node {
int data;
Node next, bottom;
Node(int newData) {
data = newData;
next = bottom = null;
}
}
// Comparator class for the priority queue
class NodeComparator implements java.util.Comparator<Node> {
@Override
public int compare(Node a, Node b) {
return Integer.compare(a.data, b.data);
}
}
// Function to flatten the linked list
class GfG {
static Node flatten(Node root) {
// Create a priority queue with custom comparator
PriorityQueue<Node> pq = new PriorityQueue<>(new NodeComparator());
Node head = null;
Node tail = null;
// Pushing main link nodes into the priority_queue.
while (root != null) {
pq.add(root);
root = root.next;
}
// Extracting the minimum node while the priority
// queue is not empty
while (!pq.isEmpty()) {
// Extracting min
Node minNode = pq.poll();
if (head == null) {
head = minNode;
tail = minNode;
} else {
tail.bottom = minNode;
tail = tail.bottom;
}
// If we have another node at the bottom of the
// popped node, push that node into the priority queue
if (minNode.bottom != null) {
pq.add(minNode.bottom);
minNode.bottom = null;
}
}
return head;
}
// Function to print the linked list
static void printList(Node head) {
Node temp = head;
while (temp != null) {
System.out.print(temp.data + " ");
temp = temp.bottom;
}
System.out.println();
}
public static void main(String[] args) {
/* Create a hard-coded linked list:
5 -> 10 -> 19 -> 28
| | |
V V V
7 20 22
| |
V V
8 50
|
V
30
*/
Node head = new Node(5);
head.bottom = new Node(7);
head.bottom.bottom = new Node(8);
head.bottom.bottom.bottom = new Node(30);
head.next = new Node(10);
head.next.bottom = new Node(20);
head.next.next = new Node(19);
head.next.next.bottom = new Node(22);
head.next.next.bottom.bottom = new Node(50);
head.next.next.next = new Node(28);
head = flatten(head);
printList(head);
}
}
# Python program for flattening a Linked List
from heapq import heappush, heappop
class Node:
def __init__(self, data):
self.data = data
self.next = self.bottom = None
# Utility function to insert a node at beginning
# of the linked list
def push(head, data):
# 1 & 2: Allocate the Node & Put in the data
newNode = Node(data)
# Make next of newNode as head
newNode.bottom = head
# Move the head to point to newNode
head = newNode
# Return to link it back
return head
def printList(head):
temp = head
while(temp is not None):
print(temp.data, end=" ")
temp = temp.bottom
print()
# Class to compare two node objects
class Cmp:
def __init__(self, node):
self.node = node
def __lt__(self, other):
return self.node.data < other.node.data
def flatten(root):
pq = []
head = None
tail = None
# Pushing main link nodes into priority_queue
while root:
heappush(pq, Cmp(root))
root = root.next
# Extracting the minimum node while the priority
# queue is not empty
while pq:
minNode = heappop(pq).node
if head is None:
head = minNode
tail = minNode
else:
tail.bottom = minNode
tail = tail.bottom
# If we have another node at the bottom of the popped
# node, push that node into the priority queue
if minNode.bottom:
heappush(pq, Cmp(minNode.bottom))
minNode.bottom = None
return head
if __name__ == '__main__':
head = Node(5)
head.bottom = Node(7)
head.bottom.bottom = Node(8)
head.bottom.bottom.bottom = Node(30)
head.next = Node(10)
head.next.bottom = Node(20)
head.next.next = Node(19)
head.next.next.bottom = Node(22)
head.next.next.bottom.bottom = Node(50)
head.next.next.next = Node(28)
head = flatten(head)
printList(head)
// C# implementation for above approach
using System;
using System.Collections.Generic;
public class Node {
public int Data;
public Node Next;
public Node Bottom;
public Node(int newData) {
Data = newData;
Next = null;
Bottom = null;
}
}
// Comparator class for the priority queue
public class NodeComparer : IComparer<Node> {
public int Compare(Node a, Node b) {
return a.Data.CompareTo(b.Data);
}
}
class GfG {
static Node Flatten(Node root) {
// Priority queue (min-heap) using a sorted list for
// simplicity
SortedList<int, Node> pq
= new SortedList<int, Node>();
Node head = null;
Node tail = null;
// Push main link nodes into the priority queue
while (root != null) {
pq.Add(root.Data, root);
root = root.Next;
}
// Extracting the minimum node while priority queue
// is not empty
while (pq.Count > 0) {
// Extracting min
Node minNode = pq.Values[0];
pq.RemoveAt(0);
if (head == null) {
head = minNode;
tail = minNode;
}
else {
tail.Bottom = minNode;
tail = tail.Bottom;
}
// If we have another node at the bottom of the
// popped node, push that node into the priority
// queue
if (minNode.Bottom != null) {
pq.Add(minNode.Bottom.Data, minNode.Bottom);
minNode.Bottom = null;
}
}
return head;
}
static void PrintList(Node head) {
Node temp = head;
while (temp != null) {
Console.Write(temp.Data + " ");
temp = temp.Bottom;
}
Console.WriteLine();
}
static void Main() {
/*
Create a hard-coded linked list:
5 -> 10 -> 19 -> 28
| | |
V V V
7 20 22
| |
V V
8 50
|
V
30
*/
Node head = new Node(5);
head.Bottom = new Node(7);
head.Bottom.Bottom = new Node(8);
head.Bottom.Bottom.Bottom = new Node(30);
head.Next = new Node(10);
head.Next.Bottom = new Node(20);
head.Next.Next = new Node(19);
head.Next.Next.Bottom = new Node(22);
head.Next.Next.Bottom.Bottom = new Node(50);
head.Next.Next.Next = new Node(28);
head = Flatten(head);
PrintList(head);
}
}
// Javascript implementation for above approach
class Node {
constructor(data) {
this.data = data;
this.next = null;
this.bottom = null;
}
}
// Comparator function for the priority queue
class MinHeap {
constructor() { this.heap = []; }
push(node) {
this.heap.push(node);
this._heapifyUp(this.heap.length - 1);
}
pop() {
if (this.size() === 0)
return null;
const root = this.heap[0];
const last = this.heap.pop();
if (this.size() > 0) {
this.heap[0] = last;
this._heapifyDown(0);
}
return root;
}
size() { return this.heap.length; }
_heapifyUp(index) {
const parentIndex = Math.floor((index - 1) / 2);
if (index > 0
&& this.heap[index].data
< this.heap[parentIndex].data) {
[this.heap[index], this.heap[parentIndex]] = [
this.heap[parentIndex], this.heap[index]
];
this._heapifyUp(parentIndex);
}
}
_heapifyDown(index) {
const leftChildIndex = 2 * index + 1;
const rightChildIndex = 2 * index + 2;
let smallest = index;
if (leftChildIndex < this.size()
&& this.heap[leftChildIndex].data
< this.heap[smallest].data) {
smallest = leftChildIndex;
}
if (rightChildIndex < this.size()
&& this.heap[rightChildIndex].data
< this.heap[smallest].data) {
smallest = rightChildIndex;
}
if (smallest !== index) {
[this.heap[index], this.heap[smallest]] =
[ this.heap[smallest], this.heap[index] ];
this._heapifyDown(smallest);
}
}
}
function flatten(root) {
const pq = new MinHeap();
let head = null;
let tail = null;
// Push main linked list nodes into the priority queue
while (root !== null) {
pq.push(root);
root = root.next;
}
// Extracting the minimum node while priority queue is
// not empty
while (pq.size() > 0) {
const minNode = pq.pop();
if (head === null) {
head = minNode;
tail = minNode;
}
else {
tail.bottom = minNode;
tail = tail.bottom;
}
// If we have another node at the bottom of the
// popped node, push that node into the priority
// queue
if (minNode.bottom !== null) {
pq.push(minNode.bottom);
minNode.bottom = null;
}
}
return head;
}
function printList(head) {
let temp = head;
while (temp !== null) {
console.log(temp.data + " ");
temp = temp.bottom;
}
console.log();
}
/*
Create a hard-coded linked list:
5 -> 10 -> 19 -> 28
| | |
V V V
7 20 22
| |
V V
8 50
|
V
30
*/
let head = new Node(5);
head.bottom = new Node(7);
head.bottom.bottom = new Node(8);
head.bottom.bottom.bottom = new Node(30);
head.next = new Node(10);
head.next.bottom = new Node(20);
head.next.next = new Node(19);
head.next.next.bottom = new Node(22);
head.next.next.bottom.bottom = new Node(50);
head.next.next.next = new Node(28);
head = flatten(head);
printList(head);
Output5 7 8 10 19 20 22 28 30 50
Time Complexity: O(n * m * log(n)) – where n is the no of nodes in the main linked list (reachable using the next pointer) and m is the no of nodes in a single sub-linked list (reachable using a bottom pointer).
Auxiliary Space: O(n)
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