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Flattening a Linked List

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  • Difficulty Level : Hard
  • Last Updated : 17 Jun, 2022

Given a linked list where every node represents a linked list and contains two pointers of its type: 

  1. Pointer to next node in the main list (we call it ‘right’ pointer in the code below) 
  2. Pointer to a linked list where this node is headed (we call it the ‘down’ pointer in the code below). 

All linked lists are sorted. See the following example  

       5 -> 10 -> 19 -> 28
       |    |     |     |
       V    V     V     V
       7    20    22    35
       |          |     |
       V          V     V
       8          50    40
       |                |
       V                V
       30               45

Write a function flatten() to flatten the lists into a single linked list. The flattened linked list should also be sorted. For example, for the above input list, output list should be 5->7->8->10->19->20->22->28->30->35->40->45->50. 

The idea is to use the Merge() process of merge sort for linked lists. We use merge() to merge lists one by one. We recursively merge() the current list with the already flattened list. 
The down pointer is used to link nodes of the flattened list.

Below is the implementation of the above approach:

C++




// C++ program for flattening a Linked List
#include <bits/stdc++.h>
using namespace std;
 
// Link list node
class Node
{
    public:
        int data;
        Node *right, *down;
};
 
Node* head = NULL;
 
// An utility function to merge two sorted
// linked lists
Node* merge(Node* a, Node* b)
{
     
    // If first linked list is empty then second
    // is the answer
    if (a == NULL)
        return b;
 
    // If second linked list is empty then first
    // is the result
    if (b == NULL)
        return a;
 
    // Compare the data members of the two linked
    // lists and put the larger one in the result
    Node* result;
 
    if (a->data < b->data)
    {
        result = a;
        result->down = merge(a->down, b);
    }
 
    else
    {
        result = b;
        result->down = merge(a, b->down);
    }
    result->right = NULL;
    return result;
}
 
Node* flatten(Node* root)
{
     
    // Base Cases
    if (root == NULL || root->right == NULL)
        return root;
 
    // Recur for list on right
    root->right = flatten(root->right);
 
    // Now merge
    root = merge(root, root->right);
 
    // Return the root
    // it will be in turn merged with its left
    return root;
}
 
// Utility function to insert a node at
// beginning of the linked list
Node* push(Node* head_ref, int data)
{
     
    // Allocate the Node & Put in the data
    Node* new_node = new Node();
 
    new_node->data = data;
    new_node->right = NULL;
 
    // Make next of new Node as head
    new_node->down = head_ref;
 
    // Move the head to point to new Node
    head_ref = new_node;
 
    return head_ref;
}
 
void printList()
{
    Node* temp = head;
    while (temp != NULL)
    {
        cout << temp->data << " ";
        temp = temp->down;
    }
    cout << endl;
}
 
// Driver code
int main()
{
     
    /* Let us create the following linked list
        5 -> 10 -> 19 -> 28
        |    |     |     |
        V    V     V     V
        7    20    22    35
        |          |     |
        V          V     V
        8          50    40
        |                |
        V                V
        30               45
    */
    head = push(head, 30);
    head = push(head, 8);
    head = push(head, 7);
    head = push(head, 5);
 
    head->right = push(head->right, 20);
    head->right = push(head->right, 10);
 
    head->right->right = push(head->right->right, 50);
    head->right->right = push(head->right->right, 22);
    head->right->right = push(head->right->right, 19);
 
    head->right->right->right = push(head->right->right->right, 45);
    head->right->right->right = push(head->right->right->right, 40);
    head->right->right->right = push(head->right->right->right, 35);
    head->right->right->right = push(head->right->right->right, 20);
 
    // Flatten the list
    head = flatten(head);
 
    printList();
    return 0;
}
 
// This code is contributed by rajsanghavi9.


Java




// Java program for flattening a Linked List
class LinkedList
{
    Node head;  // head of list
 
    /* Linked list Node*/
    class Node
    {
        int data;
        Node right, down;
        Node(int data)
        {
            this.data = data;
            right = null;
            down = null;
        }
    }
 
    // An utility function to merge two sorted linked lists
    Node merge(Node a, Node b)
    {
        // if first linked list is empty then second
        // is the answer
        if (a == null)     return b;
 
        // if second linked list is empty then first
        // is the result
        if (b == null)      return a;
 
        // compare the data members of the two linked lists
        // and put the larger one in the result
        Node result;
 
        if (a.data < b.data)
        {
            result = a;
            result.down =  merge(a.down, b);
        }
 
        else
        {
            result = b;
            result.down = merge(a, b.down);
        }
 
        result.right = null;
        return result;
    }
 
    Node flatten(Node root)
    {
        // Base Cases
        if (root == null || root.right == null)
            return root;
 
        // recur for list on right
        root.right = flatten(root.right);
 
        // now merge
        root = merge(root, root.right);
 
        // return the root
        // it will be in turn merged with its left
        return root;
    }
 
    /* Utility function to insert a node at beginning of the
       linked list */
    Node push(Node head_ref, int data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(data);
 
        /* 3. Make next of new Node as head */
        new_node.down = head_ref;
 
        /* 4. Move the head to point to new Node */
        head_ref = new_node;
 
        /*5. return to link it back */
        return head_ref;
    }
 
    void printList()
    {
        Node temp = head;
        while (temp != null)
        {
            System.out.print(temp.data + " ");
            temp = temp.down;
        }
        System.out.println();
    }
 
    /* Driver program to test above functions */
    public static void main(String args[])
    {
        LinkedList L = new LinkedList();
 
        /* Let us create the following linked list
            5 -> 10 -> 19 -> 28
            |    |     |     |
            V    V     V     V
            7    20    22    35
            |          |     |
            V          V     V
            8          50    40
            |                |
            V                V
            30               45
        */
 
        L.head = L.push(L.head, 30);
        L.head = L.push(L.head, 8);
        L.head = L.push(L.head, 7);
        L.head = L.push(L.head, 5);
 
        L.head.right = L.push(L.head.right, 20);
        L.head.right = L.push(L.head.right, 10);
 
        L.head.right.right = L.push(L.head.right.right, 50);
        L.head.right.right = L.push(L.head.right.right, 22);
        L.head.right.right = L.push(L.head.right.right, 19);
 
        L.head.right.right.right = L.push(L.head.right.right.right, 45);
        L.head.right.right.right = L.push(L.head.right.right.right, 40);
        L.head.right.right.right = L.push(L.head.right.right.right, 35);
        L.head.right.right.right = L.push(L.head.right.right.right, 20);
 
        // flatten the list
        L.head = L.flatten(L.head);
 
        L.printList();
    }
} /* This code is contributed by Rajat Mishra */


Python3




# Python program for flattening a Linked List
 
class Node():
    def __init__(self,data):
        self.data = data
        self.right = None
        self.down = None
 
class LinkedList():
    def __init__(self):
 
        # head of list
        self.head = None
 
    # Utility function to insert a node at beginning of the
    #   linked list
    def push(self,head_ref,data):
 
        # 1 & 2: Allocate the Node &
        # Put in the data
        new_node = Node(data)
 
        # Make next of new Node as head
        new_node.down = head_ref
 
        # 4. Move the head to point to new Node
        head_ref = new_node
 
        # 5. return to link it back
        return head_ref
 
    def printList(self):
 
        temp = self.head
        while(temp != None):
            print(temp.data,end=" ")
            temp = temp.down
 
        print()
 
    # An utility function to merge two sorted linked lists
    def merge(self, a, b):
        # if first linked list is empty then second
        # is the answer
        if(a == None):
            return b
         
        # if second linked list is empty then first
        # is the result
        if(b == None):
            return a
 
        # compare the data members of the two linked lists
        # and put the larger one in the result
        result = None
 
        if (a.data < b.data):
            result = a
            result.down = self.merge(a.down,b)
        else:
            result = b
            result.down = self.merge(a,b.down)
 
        result.right = None
        return result
 
    def flatten(self, root):
 
        # Base Case
        if(root == None or root.right == None):
            return root
        # recur for list on right
 
        root.right = self.flatten(root.right)
 
        # now merge
        root = self.merge(root, root.right)
 
        # return the root
        # it will be in turn merged with its left
        return root
 
# Driver program to test above functions
L = LinkedList()
 
'''
Let us create the following linked list
            5 -> 10 -> 19 -> 28
            |    |     |     |
            V    V     V     V
            7    20    22    35
            |          |     |
            V          V     V
            8          50    40
            |                |
            V                V
            30               45
'''
L.head = L.push(L.head, 30);
L.head = L.push(L.head, 8);
L.head = L.push(L.head, 7);
L.head = L.push(L.head, 5);
 
L.head.right = L.push(L.head.right, 20);
L.head.right = L.push(L.head.right, 10);
 
L.head.right.right = L.push(L.head.right.right, 50);
L.head.right.right = L.push(L.head.right.right, 22);
L.head.right.right = L.push(L.head.right.right, 19);
 
L.head.right.right.right = L.push(L.head.right.right.right, 45);
L.head.right.right.right = L.push(L.head.right.right.right, 40);
L.head.right.right.right = L.push(L.head.right.right.right, 35);
L.head.right.right.right = L.push(L.head.right.right.right, 20);
 
# flatten the list
L.head = L.flatten(L.head);
 
L.printList()
# This code is contributed by maheshwaripiyush9


C#




// C# program for flattening a Linked List
using System;
public class List {
    Node head; // head of list
 
    /* Linked list Node */
    public
 
 
 class Node {
        public
 
 
 int data;
        public
 
 
 Node right, down;
 
        public
 
 
 Node(int data) {
            this.data = data;
            right = null;
            down = null;
        }
    }
 
    // An utility function to merge two sorted linked lists
    Node merge(Node a, Node b) {
        // if first linked list is empty then second
        // is the answer
        if (a == null)
            return b;
 
        // if second linked list is empty then first
        // is the result
        if (b == null)
            return a;
 
        // compare the data members of the two linked lists
        // and put the larger one in the result
        Node result;
 
        if (a.data < b.data) {
            result = a;
            result.down = merge(a.down, b);
        }
 
        else {
            result = b;
            result.down = merge(a, b.down);
        }
 
        result.right = null;
        return result;
    }
 
    Node flatten(Node root) {
        // Base Cases
        if (root == null || root.right == null)
            return root;
 
        // recur for list on right
        root.right = flatten(root.right);
 
        // now merge
        root = merge(root, root.right);
 
        // return the root
        // it will be in turn merged with its left
        return root;
    }
 
    /*
     * Utility function to insert a node at beginning of the linked list
     */
    Node Push(Node head_ref, int data) {
        /*
         * 1 & 2: Allocate the Node & Put in the data
         */
        Node new_node = new Node(data);
 
        /* 3. Make next of new Node as head */
        new_node.down = head_ref;
 
        /* 4. Move the head to point to new Node */
        head_ref = new_node;
 
        /* 5. return to link it back */
        return head_ref;
    }
 
    void printList() {
        Node temp = head;
        while (temp != null) {
            Console.Write(temp.data + " ");
            temp = temp.down;
        }
        Console.WriteLine();
    }
 
    /* Driver program to test above functions */
    public static void Main(String []args) {
        List L = new List();
 
        /*
         * Let us create the following linked list 5 -> 10 -> 19 -> 28 | | | | V V V V 7
         * 20 22 35 | | | V V V 8 50 40 | | V V 30 45
         */
 
        L.head = L.Push(L.head, 30);
        L.head = L.Push(L.head, 8);
        L.head = L.Push(L.head, 7);
        L.head = L.Push(L.head, 5);
 
        L.head.right = L.Push(L.head.right, 20);
        L.head.right = L.Push(L.head.right, 10);
 
        L.head.right.right = L.Push(L.head.right.right, 50);
        L.head.right.right = L.Push(L.head.right.right, 22);
        L.head.right.right = L.Push(L.head.right.right, 19);
 
        L.head.right.right.right = L.Push(L.head.right.right.right, 45);
        L.head.right.right.right = L.Push(L.head.right.right.right, 40);
        L.head.right.right.right = L.Push(L.head.right.right.right, 35);
        L.head.right.right.right = L.Push(L.head.right.right.right, 20);
 
        // flatten the list
        L.head = L.flatten(L.head);
 
        L.printList();
    }
}
 
// This code is contributed by umadevi9616


Javascript




<script>
// javascript program for flattening a Linked List
var head; // head of list
 
    /* Linked list Node */
      
     class Node {
            constructor(val) {
                this.data = val;
                this.down = null;
                this.next = null;
            }
        }
 
    // An utility function to merge two sorted linked lists
    function merge(a,  b) {
        // if first linked list is empty then second
        // is the answer
        if (a == null)
            return b;
 
        // if second linked list is empty then first
        // is the result
        if (b == null)
            return a;
 
        // compare the data members of the two linked lists
        // and put the larger one in the result
        var result;
 
        if (a.data < b.data) {
            result = a;
            result.down = merge(a.down, b);
        }
 
        else {
            result = b;
            result.down = merge(a, b.down);
        }
 
        result.right = null;
        return result;
    }
 
    function flatten(root) {
        // Base Cases
        if (root == null || root.right == null)
            return root;
 
        // recur for list on right
        root.right = flatten(root.right);
 
        // now merge
        root = merge(root, root.right);
 
        // return the root
        // it will be in turn merged with its left
        return root;
    }
 
    /*
     * Utility function to insert a node at beginning of the linked list
     */
    function push(head_ref , data) {
        /*
         * 1 & 2: Allocate the Node & Put in the data
         */
         var new_node = new Node(data);
 
        /* 3. Make next of new Node as head */
        new_node.down = head_ref;
 
        /* 4. Move the head to point to new Node */
        head_ref = new_node;
 
        /* 5. return to link it back */
        return head_ref;
    }
 
    function printList() {
    var temp = head;
        while (temp != null) {
            document.write(temp.data + " ");
            temp = temp.down;
        }
        document.write();
    }
 
    /* Driver program to test above functions */
     
         
 
        /*
         * Let us create the following linked list 5 -> 10 -> 19 -> 28 | | | | V V V V 7
         * 20 22 35 | | | V V V 8 50 40 | | V V 30 45
         */
 
        head = push(head, 30);
        head = push(head, 8);
        head = push(head, 7);
        head = push(head, 5);
 
        head.right = push(head.right, 20);
        head.right = push(head.right, 10);
 
        head.right.right = push(head.right.right, 50);
        head.right.right = push(head.right.right, 22);
        head.right.right = push(head.right.right, 19);
 
        head.right.right.right = push(head.right.right.right, 45);
        head.right.right.right = push(head.right.right.right, 40);
        head.right.right.right = push(head.right.right.right, 35);
        head.right.right.right = push(head.right.right.right, 20);
 
        // flatten the list
        head = flatten(head);
 
        printList();
 
// This code contributed by aashish1995
</script>


Output

5 7 8 10 19 20 20 22 30 35 40 45 50 

Time Complexity: O(N*N*M) – where N is the no of nodes in main linked list (reachable using right pointer) and M is the no of node in a single sub linked list (reachable using down pointer). 

Explanation: As we are merging 2 lists at a time,

  • After adding first 2 lists, 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).
  • Then we will merge another list -> time = O(3M + M).
  • 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
  • Using arithmetic sum formula: time = O((N*N + N – 2)*M/2)
  • Above expression is roughly equal to O(N*N*M) for large value of N

Space Complexity: O(N*M) – because of the recursion. The recursive functions will use recursive stack of size equivalent to total number of elements in the lists, which is N*M.

Method 2: Using Priority Queues

The idea is, to build min-heap and  use Extract-min property  to get minimum element from priority queue.

C++




#include <bits/stdc++.h>
 
struct Node {
    int data;
    struct Node* next;
    struct Node* bottom;
 
    Node(int x)
    {
        data = x;
        next = NULL;
        bottom = NULL;
    }
};
 
using namespace std;
 
void flatten(Node* root);
 
int main(void)
{
  //this code builds the flattened linked list
  //of first picture in this article ;
  Node* head=new Node(5);
  auto temp=head;
  auto bt=head;
  bt->bottom=new Node(7);
  bt->bottom->bottom=new Node(8);
  bt->bottom->bottom->bottom=new Node(30);
  temp->next=new Node(10);
   
  temp=temp->next;
  bt=temp;
  bt->bottom=new Node(20);
  temp->next=new Node(19);
  temp=temp->next;
  bt=temp;
  bt->bottom=new Node(22);
  bt->bottom->bottom=new Node(50);
  temp->next=new Node(28);
  temp=temp->next;
  bt=temp;
  bt->bottom=new Node(35);
  bt->bottom->bottom=new Node(40);
  bt->bottom->bottom->bottom=new Node(45);
   
  flatten(head);
  cout << endl;
  return 0;
}
 
 
 
struct mycomp {
    bool operator()(Node* a, Node* b)
    {
        return a->data > b->data;
    }
};
void flatten(Node* root)
{
    priority_queue<Node*, vector<Node*>, mycomp> p;
  //pushing main link nodes into priority_queue.
    while (root!=NULL) {
        p.push(root);
        root = root->next;
    }
   
    while (!p.empty()) {
      //extracting min
        auto k = p.top();
        p.pop();
      //printing  least element
        cout << k->data << " ";
        if (k->bottom)
            p.push(k->bottom);
    }
    
}
//this code is contributed by user_990i


Output

5 7 8 10 19 20 22 28 30 35 40 45 50 

Time Complexity: O(N*M*log(N)) – where N is the no of nodes in main linked list (reachable using next pointer) and M is the no of node in a single sub linked list (reachable using bottom pointer).

Explanation:

1. Push all the nodes that are reachable using next pointer into priority queue till we encounter NULL

2.while(priority_queue is not empty)

        (i) Extract Node which contains minimum-value(k) from priority_queue and print k->data.

        (ii) If  k->bottom!=NULL ,insert k->bottom into priority_queue. Else our priority_queue size will decrease.

Space Complexity: O(N)-where N is the no of nodes in main linked list (reachable using next pointer).

NOTE: In above explanation, k means the Node which contains minimum-element.


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