Javascript Program To Merge K Sorted Linked Lists – Set 1

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.  

Javascript

<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 in a
   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 output
function mergeKLists(arr, last)
{
    // Traverse from 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 code
// Number of linked lists
let k = 3;

// Number of elements in each list
let n = 4;

// An array of pointers storing the
// head nodes of the linked lists
let  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 lists
head = mergeKLists(arr, k - 1);
printList(head);
// This code is contributed by unknown2108
</script>

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. 

Javascript

<script>
// JavaScript program to merge k sorted
// arrays of size n each
class 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;
    }
}

// Number of linked lists
var k = 3;

// Number of elements in each list
var n = 4;

// An array of pointers storing the
// head nodes of the linked lists
var 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 lists
var head = mergeKLists(arr, k - 1);
printList(head);
// This code is contributed by gauravrajput1
</script>

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.

Please refer complete article on Merge K sorted linked lists | Set 1 for more details!