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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