This Answer will explain how to merge two binary max-heaps given as arrays to create a new max-heap.

The most straightforward solution is to first create a new array by copying the elements of both heaps. We then build a max-heap with the newly created array.

We use the max-heapify operation to build max-heaps.

`maxHeapify(idx, heap)`

assumes that the root of the max-heap is at the index`idx`

and both its children are already max-heaps. If the element at`idx`

is not the maximum element, it swaps with the maximum of its left and right children, and then*recursively*calls itself on the affected child sub-heap until the max-heap property is restored.- We build a max-heap by
*iteratively*calling`maxHeapify(idx, heap)`

. This follows a bottom-up approach by building sub-heaps starting at the last element with children and going backward to the root. If`N-1`

is the index of the last element in`heap`

, then its parent's index is`((N-1) - 1)/2`

i.e`N/2 - 1`

. This is the last element with children. All childless nodes are already max-heaps.

int N = heap.size();for(int i = N/2 - 1; i >= 0; i--)maxHeapify(i, heap);

#include <iostream>#include <vector>#include <algorithm>//Merge two binary max-heaps given as arrays to form a new max-heapvoid maxHeapify(int k, std::vector<int> & heap){if(k >= heap.size()) //base conditionreturn;int lc = 2 * k + 1; //left child indexint rc = 2 * k + 2; //right child indexint maxIndex = k;// if left child is greater than parent(at k)if(lc < heap.size() && heap[lc] > heap[maxIndex])maxIndex = lc;// if right child is greater than greatest (at maxIndex)if(rc < heap.size() && heap[rc] > heap[maxIndex])maxIndex = rc;//if greatest is not parent kif(maxIndex != k){std::swap(heap[maxIndex], heap[k]);//recursively heapify the affected sub-heapmaxHeapify(maxIndex, heap);}}std::vector<int> mergeMaxHeaps(std::vector<int> const & arr1, std::vector<int> const & arr2){std::vector<int> heap;//append arr1 and arr2 to heapstd::copy(arr1.begin(), arr1.end(), std::back_inserter(heap));std::copy(arr2.begin(), arr2.end(), std::back_inserter(heap));//build a max-heapint N = heap.size();for(int i = N/2 - 1; i >= 0; i--){maxHeapify(i, heap);}return heap;}void DrawMaxHeap(std::vector<int> const & h);int main(){std::vector<int> const maxHeap1{9, 4, 7, 1, -2}; //input max-heap 1std::vector<int> const maxHeap2{8, 6 , 3, 5}; // input max-heap 2std::cout << "Max-heap 1 : ";DrawMaxHeap(maxHeap1);std::cout << "Max-heap 2 : ";DrawMaxHeap(maxHeap2);auto res = mergeMaxHeaps(maxHeap1, maxHeap2);std::cout << "Merged max-heap: ";DrawMaxHeap(res);return 0;}

The worst-case time complexity of the `maxHeapify`

function is O(log N), where N is the number of elements in the max-heap. In this case, N is the sum of the sizes of the two input max-heaps.

While it appears that we perform N/2 * O(log N) operations to build the max-heap, the actual time complexity is O(N). This is because the number of swap operations is reduced when the children of each node are already max-heapified*.*

Line 11–17: We find the index of the maximum element `maxIndex`

of the max-heap with its root at index `k`

of `heap`

.

Line 20–24: If `heap[k]`

is not the maximum element, we swap `heap[k]`

and `heap[maxIndex]`

. This might affect the heap property in the sub-heap with its root at `maxIndex`

, so we *recursively *call `maxHeapify(maxIndex, heap)`

.

Line 31–32: We copy the elements of `arr1`

and `arr2`

in `heap`

using `copy`

** **function. `std::back_inserter(heap)`

ensures that elements are inserted at the end of `heap`

.` `

Line 36–38: We build the max-heap *iteratively *starting at the last element, with children at the index `N/2 - 1`

and going back to `0`

.

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