Heap Data Structure MCQ

1. (i/2) position
2. (i+1)/ position
3. floor(i/2) position
4. ceil(i/2) position

int function(vector arr)

int len=arr.length();

if(len==0)

return;

temp=arr[len-1];

arr.pop_back();

return temp;

1. o(n)
2. O(logn)
3. O(1)
4. O(n logn)

1. O(n)
2. O(1)
3. O(logn)
4. O(nlogn)

int myfun(heap_arr[])

{

	int mini=INF;

	for(int i=0;i

	mini=min(mini,heap_arr)

	return mini;

}

1. Last added element to heap
2. First element added to heap
3. Root of the heap
4. Leftmost node of the heap

1. Priority queue
2. Stack
3. A decreasing order array
4. None of the mentioned

1. Yes
2. No
3. Both 1and2
4. None of these

FIB-INSERT(H, x)

degree[x]= 0

p[x]=  NIL

child[x] =NIL

left[x] =x

right[x] =x

mark[x] =FALSE

concatenate the root list containing x with root list H 

if min[H] = NIL or key[x] > key[min[H]]

then min[H]= x

n[H]= n[H] + 1

1. Line -11
2. Line -3
3. Line -9
4. Line -7

FIB_UNION(H1,H2)

{

	H =MAKE_HEAP()

	min[H]= min[H1]

	concatenate the root list of H2 with the root list of H

	if (min[H1] = NIL) or (min[H2]!= NIL and min[H2] < min[H1])

	then min[H] = min[H2]

	n[H]=  n[H1] + n[H2]

	free the objects H1 and H2

	return H

}

1. n+1
2. n+n/2
3. nlogn
4. 2*n

1. O(log n)
2. O(h)
3. O(log n) & O(h)
4. None of the mentioned

1. 2
2. 100
3. 17
4. 3