Objective: Given k sorted array, write an algorithm to merge Them into One sorted array.
Example :
int[][] A = new int[5][];
A[0] = new int[] { 1, 5, 8, 9 };
A[1] = new int[] { 2, 3, 7, 10 };
A[2] = new int[] { 4, 6, 11, 15 };
A[3] = new int[] { 9, 14, 16, 19 };
A[4] = new int[] { 2, 4, 6, 9 };
Output:
[1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 8, 9, 9, 9, 10, 11, 14, 15, 16, 19]
Naive Solution: O(nkLognk)
- Create an result[] of size n*k.
- Copy all the elements from k arrays into result array. This will take O(nk).
- Sort the result[] using Merge Sort. This will take O(nkLognk) time.
Better Approach: O(nkLogk)
- Create an result[] of size n*k.
- Create Min-Heap of type HeapNode.( HeapNode- Every Node will store the data and the list no from which it belongs).
- Now take one element from each of the K list and create HeapNode object and insert into min-Heap.
- Extract the minimum Node from the min-Heap, insert the data into result array.
- The extracted node will also contain the list to which it belongs, insert the next element from that list into min-Heap.
- If any point of time any list gets over, insert +∞ into min-Heap.
- Keep repeating until all the K list gets over.
See the animation below for better understanding.
Complete Code:
Output:
[1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 8, 9, 9, 9, 10, 11, 14, 15, 16, 19]
Great Code ! However, I didn’t understand few things here.
1. Why are you starting at heap location 1 instead of 0.
2. During the initialization you are basically inserting the first element of every array into the heap. You will never come across ptrs[i] > n. Because you initialized ptrs[i] to be zero.
1. We start with heap location 1 because ,
For any given node at position i:
– Its Left Child is at [2*i] if available.
– Its Right Child is at [2*i+1] if available.
So “i” cannot be 0. Read this link for more information.
http://algorithms.tutorialhorizon.com/binary-min-max-heap/
2. ptrs[] start with 0 and goes till n-1. So when it becomes n, we will set the value as
Integer.max so it will never become greater than n.
You can start with 0-based index too. In that case, Left child would be 2i+1 and Right Child would be 2i+2