Python Program To Merge K Sorted Linked Lists – Set 1
Last Updated :
15 Feb, 2023
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
class Node:
def __init__( self , x):
self .data = x
self . next = None
def printList(node):
while (node ! = None ):
print (node.data,
end = " " )
node = node. next
def mergeKLists(arr, last):
for i in range ( 1 , last + 1 ):
while ( True ):
head_0 = arr[ 0 ]
head_i = arr[i]
if (head_i = = None ):
break
if (head_0.data > =
head_i.data):
arr[i] = head_i. next
head_i. next = head_0
arr[ 0 ] = head_i
else :
while (head_0. next ! = None ):
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
head_0 = head_0. next
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 ]
if __name__ = = '__main__' :
k = 3
n = 4
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 )
head = mergeKLists(arr, k - 1 )
printList(head)
|
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.
- The idea is to pair up K lists and merge each pair in linear time using O(n) space.
- 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.
- Repeat the procedure until we have only one list left.
Below is the implementation of the above idea.
Python3
class Node:
def __init__( self ):
self .data = 0
self . next = None
def printList(node):
while (node ! = None ):
print (node.data, end = ' ' )
node = node. next
def SortedMerge(a, b):
result = None
if (a = = None ):
return (b)
elif (b = = None ):
return (a)
if (a.data < = b.data):
result = a
result. next =
SortedMerge(a. next , b)
else :
result = b
result. next =
SortedMerge(a, b. next )
return result
def mergeKLists(arr, last):
while (last ! = 0 ):
i = 0
j = last
while (i < j):
arr[i] = SortedMerge(arr[i], arr[j])
i + = 1
j - = 1
if (i > = j):
last = j
return arr[ 0 ]
def newNode(data):
temp = Node()
temp.data = data
temp. next = None
return temp
if __name__ = = '__main__' :
k = 3
n = 4
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 )
head = mergeKLists(arr, k - 1 )
printList(head)
|
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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