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Python Program To Merge K Sorted Linked Lists – Set 1

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  • Last Updated : 28 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.

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


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. 

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


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!


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