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 * k * log k), where, ‘N’ is the total number of elements among all the linked lists and ‘k’ is the total number of lists.
Insertion and deletion in a min-heap requires log k time. So the overall time complexity is O(N * log k).
- 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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