Merge K sorted linked lists | Set 1 - GeeksforGeeks

23 Feb, 2022
Given K sorted linked lists of size N each, merge them and print the sorted output.
Examples: 
Input: k = 3, n =  4
list1 = 1->3->5->7->NULL
list2 = 2->4->6->8->NULL
list3 = 0->9->10->11->NULL

Output: 0->1->2->3->4->5->6->7->8->9->10->11
Merged lists in a sorted order 
where every element is greater 
than the previous element.

Input: k = 3, n =  3
list1 = 1->3->7->NULL
list2 = 2->4->8->NULL
list3 = 9->10->11->NULL

Output: 1->2->3->4->7->8->9->10->11
Merged lists in a sorted order 
where every element is greater 
than the previous element.
Method 1 (Simple) 
Approach: A Simple Solution is to initialize the result as the first list. Now traverse all lists starting from the second list. Insert every node of the currently traversed list into result in a sorted way.  
C++
Java
Python3
C#
Javascript
// C++ program to merge k sorted// arrays of size n each#include <bits/stdc++.h>using namespace std; // A Linked List nodestruct Node {    int data;    Node* next;}; /* Function to print nodes in   a given linked list */void printList(Node* node){    while (node != NULL) {        printf("%d ", node->data);        node = node->next;    }} // The main function that// takes an array of lists// arr[0..last] and generates// the sorted outputNode* mergeKLists(Node* arr[], int last){     // Traverse form second list to last    for (int i = 1; i <= last; i++) {        while (true) {            // head of both the lists,            // 0 and ith list.            Node *head_0 = arr[0], *head_i = arr[i];             // Break if list ended            if (head_i == NULL)                break;             // Smaller than first element            if (head_0->data >= head_i->data) {                arr[i] = head_i->next;                head_i->next = head_0;                arr[0] = head_i;            }            else                // Traverse the first list                while (head_0->next != NULL) {                    // Smaller than next element                    if (head_0->next->data                        >= head_i->data) {                        arr[i] = head_i->next;                        head_i->next = head_0->next;                        head_0->next = head_i;                        break;                    }                    // go to next node                    head_0 = head_0->next;                     // if last node                    if (head_0->next == NULL) {                        arr[i] = head_i->next;                        head_i->next = NULL;                        head_0->next = head_i;                        head_0->next->next = NULL;                        break;                    }                }        }    }     return arr[0];} // Utility function to create a new node.Node* newNode(int data){    struct Node* temp = new Node;    temp->data = data;    temp->next = NULL;    return temp;} // Driver program to test// above functionsint main(){    // Number of linked lists    int k = 3;     // Number of elements in each list    int n = 4;     // an array of pointers storing the    // head nodes of the linked lists    Node* arr[k];     arr[0] = newNode(1);    arr[0]->next = newNode(3);    arr[0]->next->next = newNode(5);    arr[0]->next->next->next = newNode(7);     arr[1] = newNode(2);    arr[1]->next = newNode(4);    arr[1]->next->next = newNode(6);    arr[1]->next->next->next = newNode(8);     arr[2] = newNode(0);    arr[2]->next = newNode(9);    arr[2]->next->next = newNode(10);    arr[2]->next->next->next = newNode(11);     // Merge all lists    Node* head = mergeKLists(arr, k - 1);     printList(head);     return 0;}
// Java program to merge k sorted// arrays of size n eachimport java.io.*; // A Linked List nodeclass Node{  int data;  Node next;   // Utility function to create a new node.  Node(int key)  {    data = key;    next = null;  }}class GFG {   static Node head;  static Node temp;   /* Function to print nodes in   a given linked list */  static void printList(Node node)  {    while(node != null)    {      System.out.print(node.data + " ");       node = node.next;    }    System.out.println();  }   // The main function that  // takes an array of lists  // arr[0..last] and generates  // the sorted output  static Node mergeKLists(Node arr[], int last)  {     // Traverse form second list to last    for (int i = 1; i <= last; i++)    {      while(true)      {         // head of both the lists,        // 0 and ith list.         Node head_0 = arr[0];        Node head_i = arr[i];         // Break if list ended        if (head_i == null)          break;         // Smaller than first element        if(head_0.data >= head_i.data)        {          arr[i] = head_i.next;          head_i.next = head_0;          arr[0] = head_i;        }        else        {           // Traverse the first list          while (head_0.next != null)          {             // Smaller than next element            if (head_0.next.data >= head_i.data)            {              arr[i] = head_i.next;              head_i.next = head_0.next;              head_0.next = head_i;              break;            }             // go to next node            head_0 = head_0.next;             // if last node            if (head_0.next == null)            {              arr[i] = head_i.next;              head_i.next = null;              head_0.next = head_i;              head_0.next.next = null;              break;            }          }        }      }    }    return arr[0];  }   // Driver program to test  // above functions   public static void main (String[] args)  {     // Number of linked lists    int k = 3;     // Number of elements in each list    int n = 4;     // an array of pointers storing the    // head nodes of the linked lists     Node[] arr = new Node[k];     arr[0] = new Node(1);    arr[0].next = new Node(3);    arr[0].next.next = new Node(5);    arr[0].next.next.next = new Node(7);     arr[1] = new Node(2);    arr[1].next = new Node(4);    arr[1].next.next = new Node(6);    arr[1].next.next.next = new Node(8);     arr[2] = new Node(0);    arr[2].next = new Node(9);    arr[2].next.next = new Node(10);    arr[2].next.next.next = new Node(11);     // Merge all lists    head = mergeKLists(arr, k - 1);    printList(head);   }} // This code is contributed by avanitrachhadiya2155
# Python3 program to merge k# sorted arrays of size n each # A Linked List nodeclass Node:       def __init__(self, x):               self.data = x        self.next = None # Function to print nodes in# a given linked listdef printList(node):       while (node != None):        print(node.data,              end = " ")        node = node.next # The main function that# takes an array of lists# arr[0..last] and generates# the sorted outputdef mergeKLists(arr, last):     # Traverse form second    # list to last    for i in range(1, last + 1):        while (True):            # head of both the lists,            # 0 and ith list.            head_0 = arr[0]            head_i = arr[i]             # Break if list ended            if (head_i == None):                break             # Smaller than first            # element            if (head_0.data >=                head_i.data):                arr[i] = head_i.next                head_i.next = head_0                arr[0] = head_i            else:                # Traverse the first list                while (head_0.next != None):                    # Smaller than next                    # element                    if (head_0.next.data >=                        head_i.data):                        arr[i] = head_i.next                        head_i.next = head_0.next                        head_0.next = head_i                        break                    # go to next node                    head_0 = head_0.next                    # if last node                    if (head_0.next == None):                        arr[i] = head_i.next                        head_i.next = None                        head_0.next = head_i                        head_0.next.next = None                        break    return arr[0] # Driver codeif __name__ == '__main__':       # Number of linked    # lists    k = 3         # Number of elements    # in each list    n = 4     # an array of pointers    # storing the head nodes    # of the linked lists    arr = [None for i in range(k)]     arr[0] = Node(1)    arr[0].next = Node(3)    arr[0].next.next = Node(5)    arr[0].next.next.next = Node(7)     arr[1] = Node(2)    arr[1].next = Node(4)    arr[1].next.next = Node(6)    arr[1].next.next.next = Node(8)     arr[2] = Node(0)    arr[2].next = Node(9)    arr[2].next.next = Node(10)    arr[2].next.next.next = Node(11)     # Merge all lists    head = mergeKLists(arr, k - 1)     printList(head) # This code is contributed by Mohit Kumar 29
// C# program to merge k sorted// arrays of size n eachusing System; // A Linked List nodepublic class Node{  public int data;  public Node next;   // Utility function to create a new node.  public Node(int key)  {    data = key;    next = null;  }} public class GFG{  static Node head;   /* Function to print nodes in   a given linked list */  static void printList(Node node)  {    while(node != null)    {      Console.Write(node.data + " ");      node = node.next;    }    Console.WriteLine();  }   // The main function that  // takes an array of lists  // arr[0..last] and generates  // the sorted output  static Node mergeKLists(Node[] arr, int last)  {     // Traverse form second list to last    for (int i = 1; i <= last; i++)    {      while(true)      {         // head of both the lists,        // 0 and ith list.         Node head_0 = arr[0];        Node head_i = arr[i];         // Break if list ended        if (head_i == null)          break;         // Smaller than first element        if(head_0.data >= head_i.data)        {          arr[i] = head_i.next;          head_i.next = head_0;          arr[0] = head_i;        }        else        {           // Traverse the first list          while (head_0.next != null)          {             // Smaller than next element            if (head_0.next.data >= head_i.data)            {              arr[i] = head_i.next;              head_i.next = head_0.next;              head_0.next = head_i;              break;            }             // go to next node            head_0 = head_0.next;             // if last node            if (head_0.next == null)            {              arr[i] = head_i.next;              head_i.next = null;              head_0.next = head_i;              head_0.next.next = null;              break;            }          }        }      }    }    return arr[0];  }  static public void Main ()  {     // Number of linked lists    int k = 3;     // an array of pointers storing the    // head nodes of the linked lists    Node[] arr = new Node[k];     arr[0] = new Node(1);    arr[0].next = new Node(3);    arr[0].next.next = new Node(5);    arr[0].next.next.next = new Node(7);     arr[1] = new Node(2);    arr[1].next = new Node(4);    arr[1].next.next = new Node(6);    arr[1].next.next.next = new Node(8);     arr[2] = new Node(0);    arr[2].next = new Node(9);    arr[2].next.next = new Node(10);    arr[2].next.next.next = new Node(11);     // Merge all lists    head = mergeKLists(arr, k - 1);    printList(head);  }} // This code is contributed by rag2127
<script>// Javascript program to merge k sorted// arrays of size n each         // A Linked List node    class Node    {        // Utility function to create a new node.        constructor(key)        {            this.data=key;            this.next=null;        }    }         let head;    let temp; /* Function to print nodes ina given linked list */function printList(node){    while(node != null)    {      document.write(node.data + " ");        node = node.next;    }    document.write("<br>");         } // The main function that// takes an array of lists// arr[0..last] and generates// the sorted outputfunction mergeKLists(arr,last){    // Traverse form second list to last    for (let i = 1; i <= last; i++)    {      while(true)      {          // head of both the lists,        // 0 and ith list.        let head_0 = arr[0];        let head_i = arr[i];          // Break if list ended        if (head_i == null)          break;          // Smaller than first element        if(head_0.data >= head_i.data)        {          arr[i] = head_i.next;          head_i.next = head_0;          arr[0] = head_i;        }        else        {            // Traverse the first list          while (head_0.next != null)          {              // Smaller than next element            if (head_0.next.data >= head_i.data)            {              arr[i] = head_i.next;              head_i.next = head_0.next;              head_0.next = head_i;              break;            }              // go to next node            head_0 = head_0.next;              // if last node            if (head_0.next == null)            {              arr[i] = head_i.next;              head_i.next = null;              head_0.next = head_i;              head_0.next.next = null;              break;            }          }        }      }    }    return arr[0];} // Driver program to test// above functions// Number of linked listslet k = 3;// Number of elements in each listlet n = 4;// an array of pointers storing the// head nodes of the linked listslet  arr = new Array(k);arr[0] = new Node(1);arr[0].next = new Node(3);arr[0].next.next = new Node(5);arr[0].next.next.next = new Node(7); arr[1] = new Node(2);arr[1].next = new Node(4);arr[1].next.next = new Node(6);arr[1].next.next.next = new Node(8); arr[2] = new Node(0);arr[2].next = new Node(9);arr[2].next.next = new Node(10);arr[2].next.next.next = new Node(11); // Merge all listshead = mergeKLists(arr, k - 1);printList(head);       // This code is contributed by unknown2108 </script>
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Complexity Analysis: 
Time complexity: O(nk2)
Auxiliary Space: O(1). As no extra space is required.
Method 2: Min Heap. A Better solution is to use Min Heap-based solution which is discussed here for arrays. The time complexity of this solution would be O(nk Log k)  Method 3: Divide and Conquer. In this post, Divide and Conquer approach is discussed. This approach doesn’t require extra space for heap and works in O(nk Log k)It is known that merging of two linked lists can be done in O(n) time and O(n) space. 
The idea is to pair up K lists and merge each pair in linear time using O(n) space.After the first cycle, K/2 lists are left each of size 2*N. After the second cycle, K/4 lists are left each of size 4*N and so on.Repeat the procedure until we have only one list left.
The idea is to pair up K lists and merge each pair in linear time using O(n) space.
After the first cycle, K/2 lists are left each of size 2*N. After the second cycle, K/4 lists are left each of size 4*N and so on.
Repeat the procedure until we have only one list left.
Below is the implementation of the above idea. 
C++
Java
Python3
C#
Javascript
// C++ program to merge k sorted// arrays of size n each#include <bits/stdc++.h>using namespace std; // A Linked List nodestruct Node {    int data;    Node* next;}; /* Function to print nodes in   a given linked list */void printList(Node* node){    while (node != NULL) {        printf("%d ", node->data);        node = node->next;    }} /* Takes two lists sorted in increasing order, and merge   their nodes together to make one big sorted list. Below   function takes O(n) extra space for recursive calls,    */Node* SortedMerge(Node* a, Node* b){    Node* result = NULL;     /* Base cases */    if (a == NULL)        return (b);    else if (b == NULL)        return (a);     /* Pick either a or b, and recur */    if (a->data <= b->data) {        result = a;        result->next = SortedMerge(a->next, b);    }    else {        result = b;        result->next = SortedMerge(a, b->next);    }     return result;} // The main function that takes an array of lists// arr[0..last] and generates the sorted outputNode* mergeKLists(Node* arr[], int last){    // repeat until only one list is left    while (last != 0) {        int i = 0, j = last;         // (i, j) forms a pair        while (i < j) {            // merge List i with List j and store            // merged list in List i            arr[i] = SortedMerge(arr[i], arr[j]);             // consider next pair            i++, j--;             // If all pairs are merged, update last            if (i >= j)                last = j;        }    }     return arr[0];} // Utility function to create a new node.Node* newNode(int data){    struct Node* temp = new Node;    temp->data = data;    temp->next = NULL;    return temp;} // Driver program to test above functionsint main(){    int k = 3; // Number of linked lists    int n = 4; // Number of elements in each list     // an array of pointers storing the head nodes    // of the linked lists    Node* arr[k];     arr[0] = newNode(1);    arr[0]->next = newNode(3);    arr[0]->next->next = newNode(5);    arr[0]->next->next->next = newNode(7);     arr[1] = newNode(2);    arr[1]->next = newNode(4);    arr[1]->next->next = newNode(6);    arr[1]->next->next->next = newNode(8);     arr[2] = newNode(0);    arr[2]->next = newNode(9);    arr[2]->next->next = newNode(10);    arr[2]->next->next->next = newNode(11);     // Merge all lists    Node* head = mergeKLists(arr, k - 1);     printList(head);     return 0;}
// Java program to merge k sorted arrays of size n eachpublic class MergeKSortedLists {     /* Takes two lists sorted in increasing order, and merge    their nodes together to make one big sorted list. Below    function takes O(Log n) extra space for recursive calls,    but it can be easily modified to work with same time and    O(1) extra space  */    public static Node SortedMerge(Node a, Node b)    {        Node result = null;        /* Base cases */        if (a == null)            return b;        else if (b == null)            return a;         /* Pick either a or b, and recur */        if (a.data <= b.data) {            result = a;            result.next = SortedMerge(a.next, b);        }        else {            result = b;            result.next = SortedMerge(a, b.next);        }         return result;    }     // The main function that takes an array of lists    // arr[0..last] and generates the sorted output    public static Node mergeKLists(Node arr[], int last)    {        // repeat until only one list is left        while (last != 0) {            int i = 0, j = last;             // (i, j) forms a pair            while (i < j) {                // merge List i with List j and store                // merged list in List i                arr[i] = SortedMerge(arr[i], arr[j]);                 // consider next pair                i++;                j--;                 // If all pairs are merged, update last                if (i >= j)                    last = j;            }        }         return arr[0];    }     /* Function to print nodes in a given linked list */    public static void printList(Node node)    {        while (node != null) {            System.out.print(node.data + " ");            node = node.next;        }    }     public static void main(String args[])    {        int k = 3; // Number of linked lists        int n = 4; // Number of elements in each list         // an array of pointers storing the head nodes        // of the linked lists        Node arr[] = new Node[k];         arr[0] = new Node(1);        arr[0].next = new Node(3);        arr[0].next.next = new Node(5);        arr[0].next.next.next = new Node(7);         arr[1] = new Node(2);        arr[1].next = new Node(4);        arr[1].next.next = new Node(6);        arr[1].next.next.next = new Node(8);         arr[2] = new Node(0);        arr[2].next = new Node(9);        arr[2].next.next = new Node(10);        arr[2].next.next.next = new Node(11);         // Merge all lists        Node head = mergeKLists(arr, k - 1);        printList(head);    }} class Node {    int data;    Node next;    Node(int data)    {        this.data = data;    }}// This code is contributed by Gaurav Tiwari
# Python3 program to merge k sorted# arrays of size n each  # A Linked List nodeclass Node:         def __init__(self):                 self.data = 0        self.next = None # Function to print nodes in a# given linked listdef printList(node):     while (node != None):        print(node.data, end = ' ')        node = node.next     # Takes two lists sorted in increasing order,# and merge their nodes together to make one# big sorted list. Below function takes# O(Log n) extra space for recursive calls,# but it can be easily modified to work with# same time and O(1) extra spacedef SortedMerge(a, b):     result = None      # Base cases    if (a == None):        return(b)    elif (b == None):        return(a)      # Pick either a or b, and recur    if (a.data <= b.data):        result = a        result.next = SortedMerge(a.next, b)    else:        result = b        result.next = SortedMerge(a, b.next)         return result # The main function that takes an array of lists# arr[0..last] and generates the sorted outputdef mergeKLists(arr, last):     # Repeat until only one list is left    while (last != 0):        i = 0        j = last          # (i, j) forms a pair        while (i < j):                         # Merge List i with List j and store            # merged list in List i            arr[i] = SortedMerge(arr[i], arr[j])              # Consider next pair            i += 1            j -= 1                         # If all pairs are merged, update last            if (i >= j):                last = j      return arr[0] # Utility function to create a new node.def newNode(data):     temp = Node()    temp.data = data    temp.next = None    return temp # Driver codeif __name__=='__main__':         # Number of linked lists    k = 3         # Number of elements in each list    n = 4      # An array of pointers storing the    # head nodes of the linked lists    arr = [0 for i in range(k)]      arr[0] = newNode(1)    arr[0].next = newNode(3)    arr[0].next.next = newNode(5)    arr[0].next.next.next = newNode(7)      arr[1] = newNode(2)    arr[1].next = newNode(4)    arr[1].next.next = newNode(6)    arr[1].next.next.next = newNode(8)      arr[2] = newNode(0)    arr[2].next = newNode(9)    arr[2].next.next = newNode(10)    arr[2].next.next.next = newNode(11)      # Merge all lists    head = mergeKLists(arr, k - 1)      printList(head) # This code is contributed by rutvik_56
// C# program to merge k sorted arrays of size n eachusing System; public class MergeKSortedLists {     /* Takes two lists sorted in    increasing order, and merge    their nodes together to make    one big sorted list. Below    function takes O(Log n) extra    space for recursive calls,    but it can be easily modified    to work with same time and    O(1) extra space */    public static Node SortedMerge(Node a, Node b)    {        Node result = null;         /* Base cases */        if (a == null)            return b;        else if (b == null)            return a;         /* Pick either a or b, and recur */        if (a.data <= b.data) {            result = a;            result.next = SortedMerge(a.next, b);        }        else {            result = b;            result.next = SortedMerge(a, b.next);        }         return result;    }     // The main function that takes    // an array of lists arr[0..last]    // and generates the sorted output    public static Node mergeKLists(Node[] arr, int last)    {        // repeat until only one list is left        while (last != 0) {            int i = 0, j = last;             // (i, j) forms a pair            while (i < j) {                // merge List i with List j and store                // merged list in List i                arr[i] = SortedMerge(arr[i], arr[j]);                 // consider next pair                i++;                j--;                 // If all pairs are merged, update last                if (i >= j)                    last = j;            }        }         return arr[0];    }     /* Function to print nodes in a given linked list */    public static void printList(Node node)    {        while (node != null) {            Console.Write(node.data + " ");            node = node.next;        }    }     public static void Main()    {        int k = 3; // Number of linked lists        //int n = 4; // Number of elements in each list         // An array of pointers storing the head nodes        // of the linked lists        Node[] arr = new Node[k];         arr[0] = new Node(1);        arr[0].next = new Node(3);        arr[0].next.next = new Node(5);        arr[0].next.next.next = new Node(7);         arr[1] = new Node(2);        arr[1].next = new Node(4);        arr[1].next.next = new Node(6);        arr[1].next.next.next = new Node(8);         arr[2] = new Node(0);        arr[2].next = new Node(9);        arr[2].next.next = new Node(10);        arr[2].next.next.next = new Node(11);         // Merge all lists        Node head = mergeKLists(arr, k - 1);        printList(head);    }} public class Node {    public int data;    public Node next;    public Node(int data)    {        this.data = data;    }} /* This code contributed by PrinciRaj1992 */
<script> // JavaScript program to merge k sorted// arrays of size n eachclass Node {    constructor(val) {        this.data = val;        this.next = null;    }}     /*      Takes two lists sorted in increasing order,     and merge their nodes together to     make one big sorted list.     Below function takes O(Log n) extra space for     * recursive calls, but it can be easily modified     to work with same time and     * O(1) extra space     */    function SortedMerge(a,  b) {var result = null;        /* Base cases */        if (a == null)            return b;        else if (b == null)            return a;         /* Pick either a or b, and recur */        if (a.data <= b.data) {            result = a;            result.next = SortedMerge(a.next, b);        } else {            result = b;            result.next = SortedMerge(a, b.next);        }         return result;    }     // The main function that takes an array of lists    // arr[0..last] and generates the sorted output    function mergeKLists(arr , last) {        // repeat until only one list is left        while (last != 0) {            var i = 0, j = last;             // (i, j) forms a pair            while (i < j) {                // merge List i with List j and store                // merged list in List i                arr[i] = SortedMerge(arr[i], arr[j]);                 // consider next pair                i++;                j--;                 // If all pairs are merged, update last                if (i >= j)                    last = j;            }        }         return arr[0];    }     /* Function to print nodes in a given linked list */    function printList(node) {        while (node != null) {            document.write(node.data + " ");            node = node.next;        }    }              var k = 3; // Number of linked lists        var n = 4; // Number of elements in each list         // an array of pointers storing the head nodes        // of the linked listsvar arr = Array(k);         arr[0] = new Node(1);        arr[0].next = new Node(3);        arr[0].next.next = new Node(5);        arr[0].next.next.next = new Node(7);         arr[1] = new Node(2);        arr[1].next = new Node(4);        arr[1].next.next = new Node(6);        arr[1].next.next.next = new Node(8);         arr[2] = new Node(0);        arr[2].next = new Node(9);        arr[2].next.next = new Node(10);        arr[2].next.next.next = new Node(11);         // Merge all listsvar head = mergeKLists(arr, k - 1);        printList(head); // This code contributed by gauravrajput1 </script>
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Complexity Analysis: 
     Assuming N(n*k) is the total number of nodes, n is the size of each linked list, and k is the total number of linked lists.
Time Complexity: O(N*log k) or O(n*k*log k)As outer while loop in function mergeKLists() runs log k times and every time it processes n*k elements.
Auxiliary Space: O(N) or O(n*k)Because recursion is used in SortedMerge() and to merge the final 2 linked lists of size N/2, N recursive calls will be made.
Merge k sorted linked lists | Set 2 (Using Min Heap)
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 Method 4:   Selecting min of top element iteratively
The time complexity of this solution would be O(n*k) and the space complexity of this solution would be O(n) 
Approach :
Select the minimum of top elements iteratively store that in a new node and increment the pointer of the minimum element .
Code :
C++
// C++ program to merge k sorted arrays of size n each#include <bits/stdc++.h>using namespace std; // A Linked List nodestruct Node{    int data;    Node* next;    Node(int x)    {      data = x;        next = NULL;    }}; /* Function to print nodes in a given linked list */void printList(Node* node){    while (node != NULL)    {        printf("%d ", node->data);        node = node->next;    }    cout << endl;} /*Linked list Node structure struct Node{int data;Node* next;Node(int x){    data = x;    next = NULL;}};*/ class Solution{public:       // Function to merge K sorted linked list.    Node* mergeKLists(Node* arr[], int K)    {        Node* head = NULL;        while (1)        {            int a = 0;            int z;            Node* curr;            int min = INT_MAX;            for (int i = 0; i < K; i++)            {                if (arr[i] != NULL)                {                    a++;                    if (arr[i]->data < min)                    {                        min = arr[i]->data;                        z = i;                    }                }            }            if (a != 0)            {                arr[z] = arr[z]->next;                Node* temp = new Node(min);                if (head == NULL)                {                    head = temp;                    curr = temp;                }                else                {                    curr->next = temp;                    curr = temp;                }            }            else            {                return head;            }        }    }}; // { Driver Code Starts. // Driver program to test above functionsint main(){    int t;    cin >> t;    while (t--)    {        int N;        cin >> N;        struct Node* arr[N];        for (int j = 0; j < N; j++)        {            int n;            cin >> n;            int x;            cin >> x;            arr[j] = new Node(x);            Node* curr = arr[j];            n--;            for (int i = 0; i < n; i++)            {                cin >> x;                Node* temp = new Node(x);                curr->next = temp;                curr = temp;            }        }        Solution obj;        Node* res = obj.mergeKLists(arr, N);        printList(res);    }    return 0;} // } Driver Code Ends
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Amazon
Merge Sort
VMWare
Heap
Linked List
VMWare
Amazon
Linked List
Heap
Merge Sort
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