RUBY - HEAP SORT CODE

JAVA PYTHON GO KOTLIN C# RUBY C++ Java Script

As we know, there is a three step process involved in Heap Sort:

Build a Heap.

Use Heapify.

Remove the Minimum element.

We will be using arrays for implementing Heap and sort the values in descending order.

Example :

def heapify(arr, root, length) 

	left = (2*root)+1
	right = (2*root)+2
	smallest = root

	if (left < length and right <= length and arr[right] < arr[left]) 
		smallest = right
	elsif (left <= length) 
		smallest = left
	end	

	if (arr[root] > arr[smallest]) 
		temp = arr[root]
		arr[root] = arr[smallest]
		arr[smallest] = temp
		heapify(arr, smallest, length)
	end
end		

def build_heap(arr) 
	myLen = ((arr.length-1-1)/ 2).to_int
	(0..myLen).reverse_each do |parent|
		heapify(arr, parent, arr.length-1)
	end
end
		
    
def delete_min(arr) 
	#Extract elements one by one from Heap
	(0..arr.length-1).reverse_each do |last|
		temp = arr[0]
		arr[0] = arr[last]
		arr[last] = temp
		heapify(arr, 0, last-1)
	end
end

def sort(arr) 
	# Build heap (rearrange array) 	  
	build_heap(arr)
	delete_min(arr)
end	


arr = [4,78,32,2,45,11]

puts "Array before Sorting"

for i in 0..arr.length-1
	puts arr[i]
end	

sort(arr)

puts "Array after Sorting"

for i in 0..arr.length-1
	puts arr[i]
end

Output :

Array before Sorting
4
78
32
2
45
11
Array after Sorting
78
45
32
11
4
2

Code Explanation

So, in the above code we have four methods:

def build_heap(arr)

def heapify(arr, root, length)

def delete_min(arr)

def sort(arr)

The best thing about the methods are, they are quite self explanatory.

As we know, the process involved in HeapSort are :

Build a Heap.

Use Heapify.

Remove the Minimum element and place it at the end of the array.

And that's exactly what the above methods are doing.

The build_heap(...) method is building the Heap. Also, build_heap(...) method takes the help of heapify(...) method to construct the Heap.

Then delete_min(...) method comes into picture, which takes the minimum element from the root and places at the end of the array. Then uses heapify(...) method to correct the Heap.

And, finally we have the sort(...) method, that clubs all the methods and gives us a sorted result.

Let us understand the entire flow :

Initially, we take the array,

arr = [4,78,32,2,45,11]

And pass it to the sort(...) method,

sort(arr)

The sort(...) method is quite simple.

def sort(arr)
	# Build heap (rearrange array)
	build_heap(arr)
	delete_min(arr)
end

It calls the build_heap(...) method for building the array.

Then calls the delete_min(...) method for deleting the minimum element from the the root and place it at the end of the array.

Let us discuss def build_heap(arr) method in detail:

def build_heap(arr)
	myLen = ((arr.length-1-1)/ 2).to_int
	(0..myLen).reverse_each do |parent|
		heapify(arr, parent, arr.length-1)
	end
end

Let us understand the above method with the existing structure.

We have an array of 6 numbers,

Let us form a logical binary tree with it,

As we know, the elements in Level 3 satisfies the Heap property, as they don't have any child Nodes.

So, we need to target the Nodes of Level 2 and apply Heapify on them.

And that's exactly what we are doing,

(0..myLen).reverse_each do |parent|
	heapify(arr, parent, arr.length-1)
end

We have taken the last element and calculated its parent.

We are running the for loop, starting with the parents and applying Heapify on each of them.

Say for example, we will apply Heapify on 32 at first,

Then apply Heapify on,

And loop goes on until it reaches the root.

Now, let us discuss def heapify(arr, root, length) method in detail:

def heapify(arr, root, length)

	left = (2*root)+1
	right = (2*root)+2
	smallest = root

	if (left < length and right <= length and arr[right] < arr[left])
		smallest = right
	elsif (left <= length)
		smallest = left
	end

	if (arr[root] > arr[smallest])
		temp = arr[root]
		arr[root] = arr[smallest]
		arr[smallest] = temp
		heapify(arr, smallest, length)
	end
end

Let us take a small chunk of the Heap and understand the above Heapify code:

So, the array representation of the above Heap would be,

So the root variable would contain 0 and length variable would contain 2.

def heapify(arr, root, length)

Similarly, left variable would be 1.

left = (2*root)+1

As,

left = (2*0)+1 = 1

And, right variable would be 2.

right = (2*root)+2

Next, we have initialised the smallest variable with the root variable(We will explain this soon).

smallest = root

The smallest variable denotes the position.

Now, let us revisit the logic of Heapify.

If the right child is less than the left child, right child is smallest. Else left child is the smallest.

That's exactly we have done in the code.

if (left < length and right <= length and arr[right] < arr[left])
	smallest = right
elsif (left <= length)
	smallest = left
end

We are checking, if arr[right] is less than arr[left]. If so, we can consider right element is the smallest. And can initialise,

smallest = right

Else, we can suppose left element is the smallest element and can initialise,

smallest = left

To simplify,

arr[smallest] = 2 (Since, the left value is smaller than the right)

But what about the condition left < length && right <= length in the if statement and left <= length. Don't worry! We will explain it soon.

Now, since we have found out the smallest of left and right. According to the logic of Heapify,

If the root element is greater than the smallest of its right and left element. We need to swap them.

And, so we have done,

if (arr[root] > arr[smallest])
	temp = arr[root]
	arr[root] = arr[smallest]
	arr[smallest] = temp
	heapify(arr, smallest, length)
end

We have checked if the root element arr[root] (i.e. arr[0] or 78) is greater than arr[smallest] (i.e. arr[1] or 2).

If so, we swap it.

temp = arr[root]
arr[root] = arr[smallest]
arr[smallest] = temp

And the new Heap becomes.

Then we make a recursive call to heapify(...) method, to check if there are any more children or not(In this case we have 3 elements but there can be more elements).

So, in the recursive call to heapify(...) method. We substitute the root variable with the smallest variable.

Note : We will understand this recursive call, once we start explaining the entire flow.

So, the heapify(...) method is called with the new set of values,

def heapify(arr, root, length)

Now, root = 1. (Because we have initialised smallest to root in the recursive call).i.e. root is pointing arr[1] where 78 lies.

But the strange thing is, left is referring to a location 3. That does not exist.

	left = (2*1)+1
	left = 3

At the same time right is referring to location 4. Which doesn't exist as well.

	right = (2*root)+2
	right = (2*1)+2
	right = 4

Now, if we refer to arr[3] or arr[4], it will throw an exception.

And the below condition comes to rescue at this point.

	smallest = right
elsif (left <= length)
	smallest = left
end

The condition left < length and right <= length checks, if the left and right element exists or not.

As in this case left and right element doesn't exist.

And the value of left is 3 and right is 4. But the value of length is 2.

So, the condition left < length and right <= length fails.

And it comes to the else statement,

elsif (left <= length)
	smallest = left

Even this condition fails, as left is 3 and length is 2.

Now, the question would be, what would be the value of smallest?(As none of the if conditions met).

And, this is the time, the default initialisation of smallest variable comes into picture.

smallest = root

And the next if statement containing the recursive call to heapify(...) fails (Since, arr[root] is never greater than arr[root]. As root = smallest).

if (arr[root] > arr[smallest])
	temp = arr[root]
	arr[root] = arr[smallest]
	arr[smallest] = temp
	heapify(arr, smallest, length)
end

And the method exits as the recursive call to heapify(...) fails.

Let us see the bigger picture combining build_heap(...) and heapify(...)

So far we have dealt with the numbers 78, 2 and 45.

Let's take the bigger picture of the Binary Tree from the previous tutorial,

As we have seen, the build_heap(...) method calls void heapify(int arr[],int root, int length) method.

Where the value of root variable is 2.

So, we will be applying Heapify on 32.

Just like the above explanation. We will compare left and right element.

if (left < length and right <= length and arr[right] < arr[left])
	smallest = right
elsif (left <= length)
	smallest = left
end

But the thing to be noted is, there is no right element.

And this is when the condition left <= length of else statement comes into picture.

Where smallestvariable is directly initialised with the left variable(Since, there is no right variable).

And the swapping happens and the heapify(...) method ends.

Again, control goes to build_heap(...) method and heapify(...) is called again.

And 78 is dealt with.

Once again the swapping is done and control goes to build_heap(...) method.

Now, the build_heap(...) method calls heapify(...), dealing with the root element 4 finally.

As usual, we check if the left element i.e. 2 or the right element i.e. 11 is smaller and the swapping is done.

And the recursive call to heapify(...) is made making 4 the root element. And no change is made.

And finally, we get a Heap.

Now, comes the delete_min(...) method for sorting.

Let us discuss def delete_min(arr) method in detail:

def delete_min(arr)
	#Extract elements one by one from Heap
	(0..arr.length-1).reverse_each do |last|
		temp = arr[0]
		arr[0] = arr[last]
		arr[last] = temp
		heapify(arr, 0, last-1)
	end
end

Let us take the above Heap,

We will start with the last element and go till the top of the Heap.
```
(0..arr.length-1).reverse_each do |last|
```

Then swap the root element i.e. arr[0] with the last element i.e. arr[last].
```
temp = arr[0]
arr[0] = arr[last]
arr[last] = temp
```
Now, if you look at the above structure, it is no longer a Heap(Since, the root element i.e 32 in Level 1 is greater than its children i.e. 4 and 11).

So we need to call heapify(...) method.

Finally, we call the heapify(...) method to convert the above structure to a Heap. But ignoring the last element as the last element (i.e. 2) will not be a part of heap anymore.
```
heapify(arr, 0, last-1)
```
last-1 tells to ignore the last element.

And, after completing the loop we get the final array in descending order.

Running Time of Heap Sort

There is a three step process involved in Heap Sort:

Build a Heap.

Remove the Minimum element.

Use Heapify.

Now, if we analyse the running time.

Building a Heap can be done in O(n) time.

Removing the element takes O(log n) time.

Heapify takes O(log n) time.

Combining all the above running time, we get:

❮❮ PREV NEXT ❯❯