# Merge k sorted linked lists | Set 2 (Using Min Heap)

Given k sorted linked lists each of size n, 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.
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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

Approach: This solution is based on the MIN HEAP approach used to solve the problem ‘merge k sorted arrays’ which is discussed here.
MinHeap: A Min-Heap is a complete binary tree in which the value in each internal node is smaller than or equal to the values in the children of that node. 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.

The process must start with creating a MinHeap and inserting the first element of all the k linked list. Remove the root element of Minheap and insert it in the output 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 in the MinHeap left.

## C++

 `// C++ implementation to merge k ` `// sorted linked lists ` `// | Using MIN HEAP method ` `#include ` `using` `namespace` `std; ` ` `  `struct` `Node { ` `    ``int` `data; ` `    ``struct` `Node* next; ` `}; ` ` `  `// '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) ` `{ ` `    ``struct` `Node *head = NULL, *last; ` ` `  `    ``// 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]); ` ` `  `    ``// loop till 'pq' is not empty ` `    ``while` `(!pq.empty()) { ` ` `  `        ``// get the top element of 'pq' ` `        ``struct` `Node* top = pq.top(); ` `        ``pq.pop(); ` ` `  `        ``// check if there is a node ` `        ``// next to the 'top' node ` `        ``// in the list of which 'top' ` `        ``// node is a member ` `        ``if` `(top->next != NULL) ` `            ``// push the next node in 'pq' ` `            ``pq.push(top->next); ` ` `  `        ``// if final merged list is empty ` `        ``if` `(head == NULL) { ` `            ``head = top; ` ` `  `            ``// points to the last node so far of ` `            ``// the final merged list ` `            ``last = top; ` `        ``} ` ` `  `        ``else` `{ ` `            ``// insert 'top' at the end ` `            ``// of the merged list so far ` `            ``last->next = top; ` ` `  `            ``// update the 'last' pointer ` `            ``last = top; ` `        ``} ` `    ``} ` ` `  `    ``// head node of the required merged list ` `    ``return` `head; ` `} ` ` `  `// function to print the singly linked list ` `void` `printList(``struct` `Node* head) ` `{ ` `    ``while` `(head != NULL) { ` `        ``cout << head->data << ``" "``; ` `        ``head = head->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; ` `} ` ` `  `// Driver program to test above ` `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]; ` ` `  `    ``// creating k = 3 sorted lists ` `    ``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 the k sorted lists ` `    ``struct` `Node* head = mergeKSortedLists(arr, k); ` ` `  `    ``// print the merged list ` `    ``printList(head); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation to merge ` `// k sorted linked lists ` `// Using MIN HEAP method ` `import` `java.util.PriorityQueue; ` `import` `java.util.Comparator; ` `public` `class` `MergeKLists { ` ` `  `    ``// function to merge k ` `    ``// sorted linked lists ` `    ``public` `static` `Node mergeKSortedLists( ` `        ``Node arr[], ``int` `k) ` `    ``{ ` `        ``Node head = ``null``, last = ``null``; ` ` `  `        ``// priority_queue 'pq' implemeted ` `        ``// as min heap with the ` `        ``// help of 'compare' function ` `        ``PriorityQueue pq ` `            ``= ``new` `PriorityQueue<>( ` `                ``new` `Comparator() { ` `                    ``public` `int` `compare(Node a, Node b) ` `                    ``{ ` `                        ``return` `a.data - b.data; ` `                    ``} ` `                ``}); ` ` `  `        ``// push the head nodes of all ` `        ``// the k lists in 'pq' ` `        ``for` `(``int` `i = ``0``; i < k; i++) ` `            ``if` `(arr[i] != ``null``) ` `                ``pq.add(arr[i]); ` ` `  `        ``// loop till 'pq' is not empty ` `        ``while` `(!pq.isEmpty()) { ` `            ``// get the top element of 'pq' ` `            ``Node top = pq.peek(); ` `            ``pq.remove(); ` ` `  `            ``// check if there is a node ` `            ``// next to the 'top' node ` `            ``// in the list of which 'top' ` `            ``// node is a member ` `            ``if` `(top.next != ``null``) ` `                ``// push the next node in 'pq' ` `                ``pq.add(top.next); ` ` `  `            ``// if final merged list is empty ` `            ``if` `(head == ``null``) { ` `                ``head = top; ` `                ``// points to the last node so far of ` `                ``// the final merged list ` `                ``last = top; ` `            ``} ` `            ``else` `{ ` `                ``// insert 'top' at the end ` `                ``// of the merged list so far ` `                ``last.next = top; ` ` `  `                ``// update the 'last' pointer ` `                ``last = top; ` `            ``} ` `        ``} ` `        ``// head node of the required merged list ` `        ``return` `head; ` `    ``} ` ` `  `    ``// function to print the singly linked list ` `    ``public` `static` `void` `printList(Node head) ` `    ``{ ` `        ``while` `(head != ``null``) { ` `            ``System.out.print(head.data + ``" "``); ` `            ``head = head.next; ` `        ``} ` `    ``} ` ` `  `    ``// Utility function to create a new node ` `    ``public` `Node push(``int` `data) ` `    ``{ ` `        ``Node newNode = ``new` `Node(data); ` `        ``newNode.next = ``null``; ` `        ``return` `newNode; ` `    ``} ` ` `  `    ``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 = mergeKSortedLists(arr, k); ` `        ``printList(head); ` `    ``} ` `} ` ` `  `class` `Node { ` `    ``int` `data; ` `    ``Node next; ` `    ``Node(``int` `data) ` `    ``{ ` `        ``this``.data = data; ` `        ``next = ``null``; ` `    ``} ` `} ` `// This code is contributed by Gaurav Tiwari `

Output:

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

Complexity Analysis:

• Time Complexity: O( n * k * log k).
Insertion and deletion in a Min Heap requires log k time. So the Overall time complexity is O( n * k * log k)
• Auxiliary Space: O(k).
k is the space required to store the priority queue.

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.

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