# Merge K Sorted Linked Lists using Min Heap

• Difficulty Level : Medium
• Last Updated : 19 Jan, 2023

Given K linked lists each of size N and each list is sorted in non-decreasing order, merge them into a single sorted (non-decreasing order) linked list and print the sorted linked list as 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.

An efficient solution for the problem has been discussed in Method 3 of this post.

Approach (Min-Heap): To solve the problem using this approach follow the below idea:

This solution is based on the MIN HEAP approach. The process must start with creating a MinHeap and inserting the first element of all the k Linked Lists. Remove the root element of Minheap and put it in the output Linked List and insert the next element from the Linked List of removed element. To get the result the step must continue until there is no element left in the MinHeap.
Note: Mapping the elements of a heap into an array is trivial, If a node is stored at index k, then its left child is stored at index 2k + 1 and its right child at index 2k + 2.

Follow the given steps to solve the problem:

• Create a min-heap and insert the first element of all the ‘k’ linked lists.
• As long as the min-heap is not empty, perform the following steps:
• Remove the Root of the min-heap (which is the current minimum among all the elements in the min-heap) and add it to the result list.
• If there exists an element (in the same linked list) next to the element that popped out in the previous step, then insert it into the min-heap.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to merge k``// sorted linked lists``// Using MIN HEAP method` `#include ``using` `namespace` `std;` `struct` `Node``{``    ``int` `data;``    ``struct` `Node* next;``};` `// Utility function to create``// a new node``struct` `Node* newNode(``int` `data)``{``    ``// Allocate node``    ``struct` `Node* new_node = ``new` `Node();` `    ``// Put in the data``    ``new_node->data = data;``    ``new_node->next = NULL;` `    ``return` `new_node;``}` `// 'compare' function used to build``// up the priority queue``struct` `compare``{``    ``bool` `operator()(``         ``struct` `Node* a, ``struct` `Node* b)``    ``{``        ``return` `a->data > b->data;``    ``}``};` `// Function to merge k sorted linked lists``struct` `Node* mergeKSortedLists(``             ``struct` `Node* arr[], ``int` `K)``{``    ``// Priority_queue 'pq' implemented``    ``// as min heap with the help of``    ``// 'compare' function``    ``priority_queue, compare> pq;` `    ``// Push the head nodes of all``    ``// the k lists in 'pq'``    ``for` `(``int` `i = 0; i < K; i++)``        ``if` `(arr[i] != NULL)``            ``pq.push(arr[i]);``    ` `      ``// Handles the case when k = 0``      ``// or lists have no elements in them   ``      ``if` `(pq.empty())``        ``return` `NULL;``  ` `      ``struct` `Node *dummy = newNode(0);``      ``struct` `Node *last = dummy;``  ` `    ``// Loop till 'pq' is not empty``    ``while` `(!pq.empty()) ``    ``{``        ``// Get the top element of 'pq'``        ``struct` `Node* curr = pq.top();``        ``pq.pop();` `        ``// Add the top element of 'pq'``        ``// to the resultant merged list``        ``last->next = curr;``        ``last = last->next; ``      ` `        ``// Check if there is a node``         ``// next to the 'top' node``        ``// in the list of which 'top'``        ``// node is a member``        ``if` `(curr->next != NULL)``            ` `        ``// Push the next node of top node``        ``// in 'pq'``        ``pq.push(curr->next);``    ``}` `    ``// Address of head node of the required``    ``// merged list``    ``return` `dummy->next;``}` `// Function to print the singly``// linked list``void` `printList(``struct` `Node* head)``{``    ``while` `(head != NULL)``    ``{``        ``cout << head->data << ``" "``;``        ``head = head->next;``    ``}``}` `// Driver code``int` `main()``{``    ``// Number of linked lists``    ``int` `N = 3;` `    ``// Number of elements in each list``    ``int` `K = 4;` `    ``// An array of pointers storing the``    ``// head nodes of the linked lists``    ``Node* arr[N];` `    ``// Creating k = 3 sorted lists``    ``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 the k sorted lists``    ``struct` `Node* head = mergeKSortedLists(arr, N);` `    ``// Print the merged list``    ``printList(head);` `    ``return` `0;``}`

## Java

 `// Java code for the above approach` `import` `java.io.*;``import` `java.util.*;` `class` `Node {``    ``int` `data;``    ``Node next;` `    ``Node(``int` `key)``    ``{``        ``data = key;``        ``next = ``null``;``    ``}``}` `// Class implements Comparator to compare Node data``class` `NodeComparator ``implements` `Comparator {` `    ``public` `int` `compare(Node k1, Node k2)``    ``{``        ``if` `(k1.data > k2.data)``            ``return` `1``;``        ``else` `if` `(k1.data < k2.data)``            ``return` `-``1``;``        ``return` `0``;``    ``}``}``class` `GFG {``    ``// Function to merge k sorted linked lists``    ``static` `Node mergeKList(Node[] arr, ``int` `K)``    ``{``        ``// Priority_queue 'queue' implemented``        ``// as min heap with the help of``        ``// 'compare' function``        ``PriorityQueue queue``            ``= ``new` `PriorityQueue<>(``new` `NodeComparator());``        ``Node at[] = ``new` `Node[K];``        ``Node head = ``new` `Node(``0``);``        ``Node last = head;``        ``// Push the head nodes of all``        ``// the k lists in 'queue'``        ``for` `(``int` `i = ``0``; i < K; i++) {``            ``if` `(arr[i] != ``null``) {``                ``queue.add(arr[i]);``            ``}``        ``}``        ``// Handles the case when k = 0``        ``// or lists have no elements in them``        ``if` `(queue.isEmpty())``            ``return` `null``;``        ``// Loop till 'queue' is not empty``        ``while` `(!queue.isEmpty()) {``            ``// Get the top element of 'queue'``            ``Node curr = queue.poll();` `            ``// Add the top element of 'queue'``            ``// to the resultant merged list``            ``last.next = curr;``            ``last = last.next;``            ``// Check if there is a node``            ``// next to the 'top' node``            ``// in the list of which 'top'``            ``// node is a member``            ``if` `(curr.next != ``null``) {``                ``// Push the next node of top node``                ``// in 'queue'``                ``queue.add(curr.next);``            ``}``        ``}``        ``// Address of head node of the required``        ``// merged list``        ``return` `head.next;``    ``}``    ``// Print 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` `N = ``3``;``      ` `        ``// array to store head of linkedlist``        ``Node[] a = ``new` `Node[N];``      ` `        ``// Linkedlist1``        ``Node head1 = ``new` `Node(``1``);``        ``a[``0``] = head1;``        ``head1.next = ``new` `Node(``3``);``        ``head1.next.next = ``new` `Node(``5``);``          ``head1.next.next.next = ``new` `Node(``7``);``      ` `        ``// Limkedlist2``        ``Node head2 = ``new` `Node(``2``);``        ``a[``1``] = head2;``        ``head2.next = ``new` `Node(``4``);``          ``head2.next.next = ``new` `Node(``6``);``          ``head2.next.next.next = ``new` `Node(``8``);``      ` `        ``// Linkedlist3``        ``Node head3 = ``new` `Node(``0``);``        ``a[``2``] = head3;``        ``head3.next = ``new` `Node(``9``);``          ``head3.next.next = ``new` `Node(``10``);``          ``head3.next.next.next = ``new` `Node(``11``);` `        ``Node res = mergeKList(a, N);` `        ``if` `(res != ``null``)``            ``printList(res);``        ``System.out.println();``    ``}``}`

## Python3

 `# Python implementation to merge k sorted linked lists``# Using MIN HEAP method` `import` `heapq` `class` `Node:``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.``next` `=` `None` `def` `mergeKList(arr, K):``    ``# Priority_queue 'queue' implemented``    ``# as min heap with the help of``    ``# 'compare' function``    ``queue ``=` `[]``    ``for` `i ``in` `range``(K):``        ``if` `arr[i] !``=` `None``:``            ``heapq.heappush(queue, (arr[i].data, arr[i]))``    ``dummy ``=` `Node(``0``)``    ``last ``=` `dummy``    ``while` `queue:``        ``curr ``=` `heapq.heappop(queue)[``1``]``        ``last.``next` `=` `curr``        ``last ``=` `last.``next``        ``if` `curr.``next` `!``=` `None``:``            ``heapq.heappush(queue, (curr.``next``.data, curr.``next``))``    ``return` `dummy.``next` `def` `printList(node):``    ``while` `node !``=` `None``:``        ``print``(node.data, end``=``" "``)``        ``node ``=` `node.``next``    ``print``()` `if` `__name__ ``=``=` `"__main__"``:``    ``N ``=` `3``    ``a ``=` `[``None``] ``*` `N``    ``# Linkedlist1``    ``head1 ``=` `Node(``1``)``    ``a[``0``] ``=` `head1``    ``head1.``next` `=` `Node(``3``)``    ``head1.``next``.``next` `=` `Node(``5``)``    ``head1.``next``.``next``.``next` `=` `Node(``7``)``    ``# Limkedlist2``    ``head2 ``=` `Node(``2``)``    ``a[``1``] ``=` `head2``    ``head2.``next` `=` `Node(``4``)``    ``head2.``next``.``next` `=` `Node(``6``)``    ``head2.``next``.``next``.``next` `=` `Node(``8``)``    ``# Linkedlist3``    ``head3 ``=` `Node(``0``)``    ``a[``2``] ``=` `head3``    ``head3.``next` `=` `Node(``9``)``    ``head3.``next``.``next` `=` `Node(``10``)``    ``head3.``next``.``next``.``next` `=` `Node(``11``)``    ``res ``=` `mergeKList(a, N)``    ``if` `res !``=` `None``:``        ``printList(res)`  `# This code is contributed by lokesh.`

Output

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

Time Complexity: O(N * K * log K)
Auxiliary Space: O(K)

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.