Given k linked lists each of size n and each list is sorted in non-decreasing order, merge them into a single sorted (non-decreasing order) linked list and print the sorted linked list as output.
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.
An efficient solution for the problem has been discussed 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.
- Create a min-heap and insert the first element of all the ‘k’ linked lists.
- As long as the min-heap is not empty, perform the following steps:
- Remove the top element of the min-heap (which is the current minimum among all the elements in the min-heap) and add it to the result list.
- If there exists an element (in the same linked list) next to the element popped out in previous step, insert it into the min-heap.
- Return the head node address of the merged list.
Below is the implementation of the above approach:
0 1 2 3 4 5 6 7 8 9 10 11
- Time Complexity: O(N * log k) or O(n * k * log k), where, ‘N’ is the total number of elements among all the linked lists, ‘k’ is the total number of lists, and ‘n’ is the size of each linked list.
Insertion and deletion operation will be performed in min-heap for all N nodes.
Insertion and deletion in a min-heap require log k time.
- Auxiliary Space: O(k).
The priority queue will have atmost ‘k’ number of elements at any point of time, hence the additional space required for our algorithm is O(k).
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