Convert min Heap to max Heap

2.6

Given array representation of min Heap, convert it to max Heap in O(n) time.

Input: arr[] = [3 5 9 6 8 20 10 12 18 9]
         3
      /     \
     5       9
   /   \    /  \
  6     8  20   10
 /  \   /
12   18 9 


Output: arr[] = [20 18 10 12 9 9 3 5 6 8] OR 
       [any Max Heap formed from input elements]

         20
       /    \
     18      10
   /    \    /  \
  12     9  9    3
 /  \   /
5    6 8 

The problem might look complex at first look. But our final goal is to only build the max heap. The idea is very simple – we simply build Max Heap without caring about the input. We start from bottom-most and rightmost internal mode of min Heap and heapify all internal modes in bottom up way to build the Max heap.

Below is its implementation

C++

// A C++ program to convert min Heap to max Heap
#include<bits/stdc++.h>
using namespace std;

// to heapify a subtree with root at given index
void MaxHeapify(int arr[], int i, int n)
{
    int l = 2*i + 1;
    int r = 2*i + 2;
    int largest = i;
    if (l < n && arr[l] > arr[i])
        largest = l;
    if (r < n && arr[r] > arr[largest])
        largest = r;
    if (largest != i)
    {
        swap(arr[i], arr[largest]);
        MaxHeapify(arr, largest, n);
    }
}

// This function basically builds max heap
void convertMaxHeap(int arr[], int n)
{
    // Start from bottommost and rightmost
    // internal mode and heapify all internal
    // modes in bottom up way
    for (int i = (n-2)/2; i >= 0; --i)
        MaxHeapify(arr, i, n);
}

// A utility function to print a given array
// of given size
void printArray(int* arr, int size)
{
    for (int i = 0; i < size; ++i)
        printf("%d ", arr[i]);
}

// Driver program to test above functions
int main()
{
    // array representing Min Heap
    int arr[] = {3, 5, 9, 6, 8, 20, 10, 12, 18, 9};
    int n = sizeof(arr)/sizeof(arr[0]);

    printf("Min Heap array : ");
    printArray(arr, n);

    convertMaxHeap(arr, n);

    printf("\nMax Heap array : ");
    printArray(arr, n);

    return 0;
}

Java

// Java program to convert min Heap to max Heap

class GFG 
{
    // To heapify a subtree with root at given index
    static void MaxHeapify(int arr[], int i, int n)
    {
        int l = 2*i + 1;
        int r = 2*i + 2;
        int largest = i;
        if (l < n && arr[l] > arr[i])
            largest = l;
        if (r < n && arr[r] > arr[largest])
            largest = r;
        if (largest != i)
        {
            // swap arr[i] and arr[largest]
            int temp = arr[i];
            arr[i] = arr[largest];
            arr[largest] = temp;
            MaxHeapify(arr, largest, n);
        }
    }
 
    // This function basically builds max heap
    static void convertMaxHeap(int arr[], int n)
    {
        // Start from bottommost and rightmost
        // internal mode and heapify all internal
        // modes in bottom up way
        for (int i = (n-2)/2; i >= 0; --i)
            MaxHeapify(arr, i, n);
    }
 
    // A utility function to print a given array
    // of given size
    static void printArray(int arr[], int size)
    {
        for (int i = 0; i < size; ++i)
            System.out.print(arr[i]+" ");
    }
    
    // driver program
	public static void main (String[] args) 
	{
		// array representing Min Heap
        int arr[] = {3, 5, 9, 6, 8, 20, 10, 12, 18, 9};
        int n = arr.length;
 
        System.out.print("Min Heap array : ");
        printArray(arr, n);
 
        convertMaxHeap(arr, n);
 
        System.out.print("\nMax Heap array : ");
        printArray(arr, n);
	}
}

// Contributed by Pramod Kumar


Output :
Min Heap array : 3 5 9 6 8 20 10 12 18 9 
Max Heap array : 20 18 10 12 9 9 3 5 6 8 

The complexity of above solution might looks like O(nLogn) but it is O(n). Refer this G-Fact for more details.

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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