Merge K Sorted Arrays

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.

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Complete Code:


Output:

[1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 8, 9, 9, 9, 10, 11, 14, 15, 16, 19]

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
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  • Rohit Dumbre

    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.

    • tutorialhorizon

      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.

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