binary heap opereations

/*
A binary heap is a Binary Tree with following properties:
1) It’s a complete tree (All levels are completely filled except possibly the last level and the last level has all keys as left as
possible). This property of Binary Heap makes them suitable to be stored in an array.

2) A Binary Heap is either Min Heap or Max Heap. In a Min Binary Heap, the key at root must be minimum among all keys present in Binary
Heap. The same property must be recursively true for all nodes in Binary Tree. Max Binary Heap is similar to MinHeap.

Given some queries and your task is to implement the three methods insertKey,  deleteKey,  and extractMin on a Binary Min Heap and call
them as per the query given below:
1) 1  x  (a query of this type means to insert an element in the min heap with value x )
2) 2  x  (a query of this type means to remove an element at position x from the min heap)
3) 3  (a query like this removes the min element from the min heap and prints it ).

Input:
The first line of the input contains an integer 'T' denoting the number of test cases. Then T test cases follow. First line of each test
case contains an integer Q denoting the number of queries. The second line of each test case contains Q queries seperated by space.

Output:
The output for each test case will  be space separated integers having -1  if the heap is empty else the min element of the heap.

Your Task:
You are required to complete the 3 methods insertKey() which takes one argument the value to be inserted , deleteKey() which takes one 
argument the position from where element is to be deleted and extractMin() which returns the minimum element in the heap.

Constraints:
1 <= T <= 100
1 <= Q <= 100
1 <= x <= 100

Example:
Input:
2
7
1 4 1 2 3 1 6 2 0 3 3
5
1 8 1 9 2 1 3 3

Output:
2 6 - 1
8 -1
 

Explanation:
Testcase 1:
In the first test case for query 
1 4      the heap will have  {4}  
1 2      the heap will be {2 4}
3         removes min element from heap ie 2 and prints it now heap is {4} 
1 6      inserts 6 to heap now heap is {4 6}
2 0      delete element at position 0 of heap now heap is {6}
3         remove min element from heap ie 6 and prints it  now the heap is empty {}
3         since heap is empty thus no min element exist so -1 is printed .
*/


int MinHeap ::  extractMin()
{
    if(heap_size<=0) return -1;
    if(heap_size==1){
        heap_size--;
        return harr[0];
    }
    int root=harr[0];
    harr[0] = harr[heap_size-1];
    heap_size--;
    MinHeapify(0);
    return root;
}



void MinHeap :: deleteKey(int i)
{
    if(i>=heap_size) return;
    decreaseKey(i, INT_MIN);
    extractMin();
}



void MinHeap ::insertKey(int k)
{
  if(heap_size == capacity) return;
  heap_size++;
  int i=heap_size -1;
  harr[i] = k;
  
  while(i!=0 && harr[parent(i)] > harr[i] ){
      swap(harr[i] , harr[parent(i)]);
      i=parent(i);
  }
  
}



void MinHeap::decreaseKey(int i, int new_val)
{
    harr[i] = new_val;
    while (i != 0 && harr[parent(i)] > harr[i])
    {
       swap(harr[i], harr[parent(i)]);
       i = parent(i);
    }
}



void MinHeap::MinHeapify(int i)
{
    int l = left(i);
    int r = right(i);
    int smallest = i;
    if (l < heap_size && harr[l] < harr[i])
        smallest = l;
    if (r < heap_size && harr[r] < harr[smallest])
        smallest = r;
    if (smallest != i)
    {
        swap(harr[i], harr[smallest]);
        MinHeapify(smallest);
    }
}