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C - PLUS-PLUS-HEAP SORT CODE




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

  1. Build a Heap.

  2. Use Heapify.

  3. Remove the Minimum element.

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


Example :



#include <iostream>
using namespace std;

class HeapSort {

	public:
    void heapify(int arr[],int root, int length) {

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

        if (left < length && right <= length && arr[right] < arr[left]) {

            smallest = right;
        }
        else if (left <= length){

            smallest = left;
        }

        if (arr[root] > arr[smallest]) {

            int temp = arr[root];
            arr[root] = arr[smallest];
            arr[smallest] = temp;

            heapify(arr, smallest, length);
        }
    }
    
    void buildHeap(int arr[], int size) {

        int len = size-1;

        for (int parent = (len-1)/ 2; parent >= 0; parent--)
            heapify(arr, parent, len);
    }
    
    void deleteMin(int arr[], int size) {

        int len = size-1;

        // Extract elements one by one from Heap
        for (int last=len; last>0; last--) {
            
            int temp = arr[0];
            arr[0] = arr[last];
            arr[last] = temp;

            
            heapify(arr, 0, last-1);
        }
    }

    void sort(int arr[], int size) {

        // Build heap (rearrange array)   
        buildHeap(arr, size);

        deleteMin(arr, size);
        
    }
};

int main() {
        
    int arr[] = {4,78,32,2,45,11};
        
    int size = sizeof(arr) / sizeof(arr[0]);

    cout <<  "Array before Sorting\n";

    for (int i=0; i < size; i++)
        cout << arr[i] << " ";

    HeapSort heapSort;
        
    heapSort.sort(arr, size);

    cout <<  "\nArray after Sorting\n";

    for (int i=0; i < size; i++)
        cout << arr[i] << " ";
}


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:

  1. void buildHeap(int arr[], int size)

  2. void heapify(int arr[],int root, int length)

  3. void deleteMin(int arr[], int size)

  4. void sort(int arr[], int size)

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


As we know, the process involved in HeapSort are :

  1. Build a Heap.

  2. Use Heapify.

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

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


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


Then deleteMin(...) 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,


int arr[] = {4,78,32,2,45,11};

And its size,


int size = sizeof(arr) / sizeof(arr[0]);

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


heapSort.sort(arr, size);

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


public void sort(int arr[], size) {

	// Build heap (rearrange array)
	buildHeap(arr, size);

	deleteMin(arr, size);

}

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


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


Let us discuss void buildHeap(int arr[], size) method in detail:


public void buildHeap(int arr[], size) {

	int len = size-1;

	for (int parent = (len-1)/ 2; parent >= 0; parent--)
		heapify(arr, parent, len);
}

Let us understand the above method with the existing structure.


We have an array of 6 numbers,

java_Collections

Let us form a logical binary tree with it,

java_Collections

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,


for (int parent = (len-1)/ 2; parent >= 0; parent--)
		heapify(arr, parent, len);

We have taken the last element and calculated its parent.


i.e. The formula to calculate the parent is


parent = (element-1)/2

The position of last element is 5.


So,


parent = (len-1)/2
parent = (5-1)/2 = 2

And the parent of arr[5] (i.e. 11) is arr[2] (i.e. 32).


So, the for loop becomes,


for (int parent = 2; parent >= 0; parent--)
		heapify(arr, parent, len);

Now, is it clear?


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,

java_Collections

Then apply Heapify on,

java_Collections

And loop goes on until it reaches the root.


Now, let us discuss void heapify(int arr[],int root, int length) method in detail:


void heapify(int arr[],int root, int length) {

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

	if (left < length && right <= length && arr[right] < arr[left]) {

		smallest = right;
	}
	else if (left <= length){

		smallest = left;
	}

	if (arr[root] > arr[smallest]) {

		int temp = arr[root];
		arr[root] = arr[smallest];
		arr[smallest] = temp;

		heapify(arr, smallest, length);
	}
}

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

java_Collections

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

java_Collections

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


void heapify(int arr[],int root, int length)

Similarly, left variable would be 1.


int left = (2*root)+1;

As,


left = (2*0)+1 = 1

And, right variable would be 2.


int right = (2*root)+2;

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


int smallest = root;

The smallest variable denotes the position.


Now, let us revisit the logic of Heapify.

java_Collections
  1. 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 && right <= length && arr[right] < arr[left]) {

		smallest = right;
	}
	else if (left <= length){

		smallest = left;
	}

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,

  1. 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]) {

		int temp = arr[root];
		arr[root] = arr[smallest];
		arr[smallest] = temp;

		heapify(arr, smallest, length);
	}

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.


		int temp = arr[root];
		arr[root] = arr[smallest];
		arr[smallest] = temp;

And the new Heap becomes.

java_Collections
java_Collections

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.

java_Collections

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,


void heapify(int arr[],int root, int 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.

java_Collections

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


int left = (2*root)+1;
	left = (2*1)+1
	left = 3

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


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

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


And the below condition comes to rescue at this point.


	if (left < length && right <= length && arr[right] < arr[left]) {

		smallest = right;
	}
	else if (left <= length){

		smallest = left;
	}

The condition left < length && 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 && right <= length fails.


And it comes to the else statement,


else if (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.


int 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]) {

		int temp = arr[root];
		arr[root] = arr[smallest];
		arr[smallest] = temp;

		heapify(arr, smallest, length);
	}

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


Let us see the bigger picture combining buildHeap(...) 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,

java_Collections

As we have seen, the buildHeap(...) 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 && right <= length && arr[right] < arr[left]) {

		smallest = right;
	}
	else if (left <= length){

		smallest = left;
	}

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.

java_Collections

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


And 78 is dealt with.


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

java_Collections

Now, the buildHeap(...) 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.

java_Collections

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 deleteMin(...) method for sorting.


Let us discuss void deleteMin(int arr[]) method in detail:


public void deleteMin(int arr[]) {

	int len = arr.Length-1;

	// Extract elements one by one from Heap
	for (int last=len; last>0; last--) {

		int temp = arr[0];
		arr[0] = arr[last];
		arr[last] = temp;

		heapify(arr, 0, last-1);
	}
}

Let us take the above Heap,

java_Collections
  1. We will start with the last element and go till the top of the Heap.

    for (int last=len; last>0; last--)

  2. Then swap the root element i.e. arr[0] with the last element i.e. arr[last].

    int temp = arr[0];
    arr[0] = arr[last];
    arr[last] = temp;

    ruby

    ruby


    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.

  3. 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.
    ruby

    ruby

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

java_Collections
java_Collections

Running Time of Heap Sort


There is a three step process involved in Heap Sort:

  1. Build a Heap.

  2. Remove the Minimum element.

  3. 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:


Total running time for Heap Sort : O(n log n)