# Merge K sorted linked lists | Set 1

• Difficulty Level : Medium
• Last Updated : 30 Sep, 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.

Implementation:

## C++

 `// C++ program to merge k sorted``// arrays of size n each``#include ``using` `namespace` `std;` `// A Linked List node``struct` `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 output``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, *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 = 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;``}` `// 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 functions``int` `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 = newNode(1);``    ``arr->next = newNode(3);``    ``arr->next->next = newNode(5);``    ``arr->next->next->next = newNode(7);` `    ``arr = newNode(2);``    ``arr->next = newNode(4);``    ``arr->next->next = newNode(6);``    ``arr->next->next->next = newNode(8);` `    ``arr = newNode(0);``    ``arr->next = newNode(9);``    ``arr->next->next = newNode(10);``    ``arr->next->next->next = newNode(11);` `    ``// Merge all lists``    ``Node* head = mergeKLists(arr, k - 1);` `    ``printList(head);` `    ``return` `0;``}`

## Java

 `// Java program to merge k sorted``// arrays of size n each``import` `java.io.*;` `// A Linked List node``class` `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

 `# Python3 program to merge k``# sorted arrays of size n each` `# A Linked List node``class` `Node:``  ` `    ``def` `__init__(``self``, x):``      ` `        ``self``.data ``=` `x``        ``self``.``next` `=` `None` `# Function to print nodes in``# a given linked list``def` `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 output``def` `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 code``if` `__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#

 `// C# program to merge k sorted``// arrays of size n each``using` `System;` `// A Linked List node``public` `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;``        ``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 = 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;``  ``}``  ``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 = ``new` `Node(1);``    ``arr.next = ``new` `Node(3);``    ``arr.next.next = ``new` `Node(5);``    ``arr.next.next.next = ``new` `Node(7);` `    ``arr = ``new` `Node(2);``    ``arr.next = ``new` `Node(4);``    ``arr.next.next = ``new` `Node(6);``    ``arr.next.next.next = ``new` `Node(8);` `    ``arr = ``new` `Node(0);``    ``arr.next = ``new` `Node(9);``    ``arr.next.next = ``new` `Node(10);``    ``arr.next.next.next = ``new` `Node(11);` `    ``// Merge all lists``    ``head = mergeKLists(arr, k - 1);``    ``printList(head);``  ``}``}` `// This code is contributed by rag2127`

## Javascript

 ``

Output

`0 1 2 3 4 5 6 7 8 9 10 11 `

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.

1. The idea is to pair up K lists and merge each pair in linear time using O(n) space.
2. 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.
3. Repeat the procedure until we have only one list left.

Below is the implementation of the above idea.

## C++

 `// C++ program to merge k sorted``// arrays of size n each``#include ``using` `namespace` `std;` `// A Linked List node``struct` `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 output``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;``}` `// 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 functions``int` `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 = newNode(1);``    ``arr->next = newNode(3);``    ``arr->next->next = newNode(5);``    ``arr->next->next->next = newNode(7);` `    ``arr = newNode(2);``    ``arr->next = newNode(4);``    ``arr->next->next = newNode(6);``    ``arr->next->next->next = newNode(8);` `    ``arr = newNode(0);``    ``arr->next = newNode(9);``    ``arr->next->next = newNode(10);``    ``arr->next->next->next = newNode(11);` `    ``// Merge all lists``    ``Node* head = mergeKLists(arr, k - 1);` `    ``printList(head);` `    ``return` `0;``}`

## Java

 `// Java program to merge k sorted arrays of size n each``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``) {``            ``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

 `# Python3 program to merge k sorted``# arrays of size n each`` ` `# A Linked List node``class` `Node:``    ` `    ``def` `__init__(``self``):``        ` `        ``self``.data ``=` `0``        ``self``.``next` `=` `None` `# Function to print nodes in a``# given linked list``def` `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 space``def` `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 output``def` `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 code``if` `__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#

 `// C# program to merge k sorted arrays of size n each``using` `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;``    ``}` `    ``/* 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 = ``new` `Node(1);``        ``arr.next = ``new` `Node(3);``        ``arr.next.next = ``new` `Node(5);``        ``arr.next.next.next = ``new` `Node(7);` `        ``arr = ``new` `Node(2);``        ``arr.next = ``new` `Node(4);``        ``arr.next.next = ``new` `Node(6);``        ``arr.next.next.next = ``new` `Node(8);` `        ``arr = ``new` `Node(0);``        ``arr.next = ``new` `Node(9);``        ``arr.next.next = ``new` `Node(10);``        ``arr.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 */`

## Javascript

 ``

Output

`0 1 2 3 4 5 6 7 8 9 10 11 `

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

Approach : Select the minimum of top elements iteratively store that in a new node and increment the pointer of the minimum element.

Implementation:

## C++

 `// C++ program to merge k sorted arrays of size n each``#include ``using` `namespace` `std;` `// A Linked List node``struct` `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 functions``int` `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`

Output

` `

Time complexity: O(n*k2
Auxiliary Space: O(n)

Let’s take an example to understand the Time Complexity

arr = 0->1->2->NULL

arr = 3->4->5->NULL

arr = 6->7->8->NULL

N = 3 K = 3

I have taken this example to make you understand the complexity in an easy way.

According to the approach, we find 0 as the minimum element in the first k nodes, to find this we will take K time.

Similarly, we found 1 as a minimum in K time, this process takes runs for N nodes in the first list, So the time taken will be O(NK) for arr, after this, our minimum element come in the remaining two lists so our loop works for K-1 and for 2nd list time taken will be N*K-1. This process will continue till we merge all nodes of all lists.

So the equation can be

NK + N*K-1 + N*K-2 + …… + N * 1

N(K+ K-1 + K-2 + … + 1)

Which is equal to O (N*K^2)

My Personal Notes arrow_drop_up