Merge two binary Max Heaps
Given two binary max heaps as arrays, merge the given heaps.
Examples :
Input : a = {10, 5, 6, 2}, b = {12, 7, 9} Output : {12, 10, 9, 2, 5, 7, 6}
The idea is simple. We create an array to store result. We copy both given arrays one by one to result. Once we have copied all elements, we call standard build heap to construct full merged max heap.
C++
// C++ program to merge two max heaps. #include <bits/stdc++.h> using namespace std; // Standard heapify function to heapify a // subtree rooted under idx. It assumes // that subtrees of node are already heapified. void maxHeapify( int arr[], int n, int idx) { // Find largest of node and its children if (idx >= n) return ; int l = 2 * idx + 1; int r = 2 * idx + 2; int max; if (l < n && arr[l] > arr[idx]) max = l; else max = idx; if (r < n && arr[r] > arr[max]) max = r; // Put maximum value at root and // recur for the child with the // maximum value if (max != idx) { swap(arr[max], arr[idx]); maxHeapify(arr, n, max); } } // Builds a max heap of given arr[0..n-1] void buildMaxHeap( int arr[], int n) { // building the heap from first non-leaf // node by calling max heapify function for ( int i = n / 2 - 1; i >= 0; i--) maxHeapify(arr, n, i); } // Merges max heaps a[] and b[] into merged[] void mergeHeaps( int merged[], int a[], int b[], int n, int m) { // Copy elements of a[] and b[] one by one // to merged[] for ( int i = 0; i < n; i++) merged[i] = a[i]; for ( int i = 0; i < m; i++) merged[n + i] = b[i]; // build heap for the modified array of // size n+m buildMaxHeap(merged, n + m); } // Driver code int main() { int a[] = { 10, 5, 6, 2 }; int b[] = { 12, 7, 9 }; int n = sizeof (a) / sizeof (a[0]); int m = sizeof (b) / sizeof (b[0]); int merged[m + n]; mergeHeaps(merged, a, b, n, m); for ( int i = 0; i < n + m; i++) cout << merged[i] << " " ; return 0; } |
Java
// Java program to merge two max heaps. class GfG { // Standard heapify function to heapify a // subtree rooted under idx. It assumes // that subtrees of node are already heapified. public static void maxHeapify( int [] arr, int n, int i) { // Find largest of node and its children if (i >= n) { return ; } int l = i * 2 + 1 ; int r = i * 2 + 2 ; int max; if (l < n && arr[l] > arr[i]) { max = l; } else max = i; if (r < n && arr[r] > arr[max]) { max = r; } // Put maximum value at root and // recur for the child with the // maximum value if (max != i) { int temp = arr[max]; arr[max] = arr[i]; arr[i] = temp; maxHeapify(arr, n, max); } } // Merges max heaps a[] and b[] into merged[] public static void mergeHeaps( int [] arr, int [] a, int [] b, int n, int m) { for ( int i = 0 ; i < n; i++) { arr[i] = a[i]; } for ( int i = 0 ; i < m; i++) { arr[n + i] = b[i]; } n = n + m; // Builds a max heap of given arr[0..n-1] for ( int i = n / 2 - 1 ; i >= 0 ; i--) { maxHeapify(arr, n, i); } } // Driver Code public static void main(String[] args) { int [] a = { 10 , 5 , 6 , 2 }; int [] b = { 12 , 7 , 9 }; int n = a.length; int m = b.length; int [] merged = new int [m + n]; mergeHeaps(merged, a, b, n, m); for ( int i = 0 ; i < m + n; i++) System.out.print(merged[i] + " " ); System.out.println(); } } |
Python3
# Python3 program to merge two Max heaps. # Standard heapify function to heapify a # subtree rooted under idx. It assumes that # subtrees of node are already heapified. def MaxHeapify(arr, n, idx): # Find largest of node and # its children if idx > = n: return l = 2 * idx + 1 r = 2 * idx + 2 Max = 0 if l < n and arr[l] > arr[idx]: Max = l else : Max = idx if r < n and arr[r] > arr[ Max ]: Max = r # Put Maximum value at root and # recur for the child with the # Maximum value if Max ! = idx: arr[ Max ], arr[idx] = arr[idx], arr[ Max ] MaxHeapify(arr, n, Max ) # Builds a Max heap of given arr[0..n-1] def buildMaxHeap(arr, n): # building the heap from first non-leaf # node by calling Max heapify function for i in range ( int (n / 2 ) - 1 , - 1 , - 1 ): MaxHeapify(arr, n, i) # Merges Max heaps a[] and b[] into merged[] def mergeHeaps(merged, a, b, n, m): # Copy elements of a[] and b[] one # by one to merged[] for i in range (n): merged[i] = a[i] for i in range (m): merged[n + i] = b[i] # build heap for the modified # array of size n+m buildMaxHeap(merged, n + m) # Driver code if __name__ = = '__main__' : a = [ 10 , 5 , 6 , 2 ] b = [ 12 , 7 , 9 ] n = len (a) m = len (b) merged = [ 0 ] * (m + n) mergeHeaps(merged, a, b, n, m) for i in range (n + m): print (merged[i], end = " " ) # This code is contributed by PranchalK |
C#
// C# program to merge two max heaps. using System; class GfG { // Standard heapify function to heapify a // subtree rooted under idx. It assumes // that subtrees of node are already heapified. public static void maxHeapify( int [] arr, int n, int i) { // Find largest of node // and its children if (i >= n) { return ; } int l = i * 2 + 1; int r = i * 2 + 2; int max; if (l < n && arr[l] > arr[i]) { max = l; } else max = i; if (r < n && arr[r] > arr[max]) { max = r; } // Put maximum value at root and // recur for the child with the // maximum value if (max != i) { int temp = arr[max]; arr[max] = arr[i]; arr[i] = temp; maxHeapify(arr, n, max); } } // Merges max heaps a[] and b[] into merged[] public static void mergeHeaps( int [] arr, int [] a, int [] b, int n, int m) { for ( int i = 0; i < n; i++) { arr[i] = a[i]; } for ( int i = 0; i < m; i++) { arr[n + i] = b[i]; } n = n + m; // Builds a max heap of given arr[0..n-1] for ( int i = n / 2 - 1; i >= 0; i--) { maxHeapify(arr, n, i); } } // Driver Code public static void Main() { int [] a = {10, 5, 6, 2}; int [] b = {12, 7, 9}; int n = a.Length; int m = b.Length; int [] merged = new int [m + n]; mergeHeaps(merged, a, b, n, m); for ( int i = 0; i < m + n; i++) Console.Write(merged[i] + " " ); Console.WriteLine(); } } // This code is contributed by nitin mittal |
Javascript
<script> // javascript program to merge two max heaps. // Standard heapify function to heapify a // subtree rooted under idx. It assumes // that subtrees of node are already heapified. function maxHeapify(arr , n , i) { // Find largest of node and its children if (i >= n) { return ; } var l = i * 2 + 1; var r = i * 2 + 2; var max; if (l < n && arr[l] > arr[i]) { max = l; } else max = i; if (r < n && arr[r] > arr[max]) { max = r; } // Put maximum value at root and // recur for the child with the // maximum value if (max != i) { var temp = arr[max]; arr[max] = arr[i]; arr[i] = temp; maxHeapify(arr, n, max); } } // Merges max heaps a and b into merged function mergeHeaps(arr, a, b , n , m) { for ( var i = 0; i < n; i++) { arr[i] = a[i]; } for ( var i = 0; i < m; i++) { arr[n + i] = b[i]; } n = n + m; // Builds a max heap of given arr[0..n-1] for ( var i = parseInt(n / 2 - 1); i >= 0; i--) { maxHeapify(arr, n, i); } } // Driver Code var a = [ 10, 5, 6, 2 ]; var b = [ 12, 7, 9 ]; var n = a.length; var m = b.length; var merged = Array(m + n).fill(0); mergeHeaps(merged, a, b, n, m); for ( var i = 0; i < m + n; i++) document.write(merged[i] + " " ); // This code is contributed by umadevi9616 </script> |
Output:
12 10 9 2 5 7 6
Since time complexity for building the heap from array of n elements is O(n). The complexity of merging the heaps is equal to O(n + m).
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