Given array representation of min Heap, convert it to max Heap in O(n) time.
Example :
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("
Max 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("
Max Heap array : "); printArray(arr, n); } } // Contributed by Pramod Kumar |
Python3
# A Python3 program to convert min Heap # to max Heap # to heapify a subtree with root # at given index def MaxHeapify(arr, i, n): l = 2 * i + 1 r = 2 * i + 2 largest = i if l < n and arr[l] > arr[i]: largest = l if r < n and arr[r] > arr[largest]: largest = r if largest != i: arr[i], arr[largest] = arr[largest], arr[i] MaxHeapify(arr, largest, n) # This function basically builds max heap def convertMaxHeap(arr, n): # Start from bottommost and rightmost # internal mode and heapify all # internal modes in bottom up way for i in range(int((n - 2) / 2), -1, -1): MaxHeapify(arr, i, n) # A utility function to print a # given array of given size def printArray(arr, size): for i in range(size): print(arr[i], end = " ") print() # Driver Code if __name__ == '__main__': # array representing Min Heap arr = [3, 5, 9, 6, 8, 20, 10, 12, 18, 9] n = len(arr) print("Min Heap array : ") printArray(arr, n) convertMaxHeap(arr, n) print("Max Heap array : ") printArray(arr, n) # This code is contributed by PranchalK |
C#
// C# program to convert // min Heap to max Heap using System; 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) Console.Write(arr[i]+" "); } // Driver Code public static void Main () { // array representing Min Heap int []arr = {3, 5, 9, 6, 8, 20, 10, 12, 18, 9}; int n = arr.Length; Console.Write("Min Heap array : "); printArray(arr, n); convertMaxHeap(arr, n); Console.Write("
Max Heap array : "); printArray(arr, n); } } // This code is contributed by nitin mittal. |
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.
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