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Comments on Regarding the heap sort algorithm.

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Regarding the heap sort algorithm.

+1
−3

I get the concept of the heap sort algorithm and its like first you have a heap(ordered binary tree) then we have the Max heap which has the highest element value in the array at the top of the tree. The parent nodes will be basically > than the child nodes. But I don't get the heapify sample code here. Can someone explain? Thank you.

#include <stdio.h>

void swap(int *a, int *b) {
    int tmp = *a;
    *a = *b;
    *b = tmp;
}

//can someone explain the heapify function of the coding?
void heapify(int arr[], int n, int i) {
    int max = I;  
    int leftChild = 2 * i + 1; coding
    int rightChild = 2 * i + 2;

    //If left child is greater than root
    if (leftChild < n && arr[leftChild] > arr[max])
        max = leftChild;

    //If right child is greater than max
    if (rightChild < n && arr[rightChild] > arr[max])
        max = rightChild;

        //If max is not root
    if (max != i) {
        swap(&arr[i], &arr[max]);
        //heapify the affected sub-tree recursively
        heapify(arr, n, max);
    }
}

//Main function to perform heap sort
void heapSort(int arr[], int n) {
    //Rearrange array (building heap)
    for (int i = n / 2 - 1; i >= 0; i--)
        heapify(arr, n, i);

    //Extract elements from heap one by one
    for (int i = n - 1; i >= 0; i--) {
        swap(&arr[0], &arr[i]); //Current root moved to the end
        heapify(arr, i, 0); //calling max heapify on the heap reduced
    }
}

//print size of array n using utility function
//print size of array n using utility function
void display(int arr[], int n) {
    for (int i = 0; i < n; ++i)
    printf("%d ", arr[i]);
    printf("\n");
}

//main function coding not included```
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2 comment threads

"Heapify" is just the process of building the "max heap"... (3 comments)
Please fix the formatting (1 comment)
Please fix the formatting
Lundin‭ wrote over 2 years ago

Please edit the post and fix the code formatting - you have non-existent indention.