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 nodeclass 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 outputdef mergeKLists(arr, last): # Traverse from 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 codeif __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.
- 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
# Python3 program to merge k sorted# arrays of size n each # A Linked List nodeclass Node: def __init__(self): self.data = 0 self.next = None# Function to print nodes in a# given linked listdef 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 outputdef 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 codeif __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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