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1. Overview

In this tutorial, we’ll talk about the problem of merging two max binary heaps.

First, we’ll define the problem and provide an example to explain it. Then, we’ll present a solution for this problem and work through its implementation and time complexity.

2. Defining the Problem

Suppose we have two max binary heaps H_1 and H_2, and we want to merge them into a single max heap H. Recall the max binary heap is a binary tree where each node in this tree has a value greater than all the nodes’ values in its subtree.

Let’s take a look at the following example for a better understanding, Given two max binary heaps H_1 and H_2 in array representation. Recall we can represent a binary tree in an array format where the node with index \mathbf{pos} has its left and right children located on \mathbf{2 \times pos + 1} and \mathbf{2 \times pos + 2} respectively:

  • H_1: [7, 5, 3, 2, 1]
  • H_2: [9, 4, 8]

Heap1

After merging these two max binary heaps, we’ll get the following one:

  • H: [9, 8, 7, 5, 4, 3, 2, 1]
Heap2

3. Solution

3.1. Main Idea

The main idea is to merge the array representation of the given max binary heaps; then we build the new max heap from the merged array.

First, we fix one of the given max heaps as a solution. Then, we’ll append the elements of the other max heap to it. Second, we’ll build a max heap on the merged array.

Finally, the max heap will be the merged version of the two given max heaps.

3.2. Max Heapify

Let’s take a look at the implementation of the algorithm:

Rendered by QuickLaTeX.com

Initially, we have the recursive function MaxHeapify that will rearrange the elements of a given array to make the heap valid. The function will have two parameters: the array of the heap we want to build H and the current position pos.

First, we declare left and right children of the current node, which will be equal to 2 \times pos + 1 and 2 \times pos + 2 respectively. Then, we’ll declare max, which represents the index of the node that has the maximum value.

Second, we check if there’s a left node. If there’s one and its value exceeds the current max node, we update the max to equal left. Also, we do the same for the righ child.

Finally, we check if max isn’t equal to the current pos, which means there’s a child of pos node with a value greater than its value. If so, we swap them to make the heap valid and call our function to the new max to check if it’s still valid.

3.3. Algorithm

Let’s take a look at the implementation of the algorithm:

Rendered by QuickLaTeX.com

Initially, we have the MergeHeaps function that will return the merged max heap from the given two max heaps. The function will have two parameters, the first and the second max heaps.

First, we declare merged \_ heap that will store the array representation of the merged version of the given max heap. Initially, it’s equal to the first max heap H_1.

Second, we iterate over the elements of the second max heap’s array H_2 and append each element to the merged \_ heap.

Third, we iterate over the nodes of the current heap in bottom-up order and call MaxHeapify to fix the invalid nodes that don’t meet the definition of max heap.

Finally, we return the merged \_ heap.

3.4. Complexity

The time complexity of this approach is \mathbf{O(N + M)}, where N is the number of vertices of the first heap H_1, and M is the number of vertices of the second heap H_2. The reason behind this complexity is the same as the complexity of building a max heap.

The space complexity here is \mathbf{O(N + M)}. since we’re storing the vertices of the new heap.

4. Conclusion

In this article, we discussed the problem of merging two max binary heaps. We defined the problem and provided an example to explain it.

Finally, we provided a simple approach to solve it and walked through its implementation and time complexity.

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