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Consider any array representation of an $n$ element binary heap where the elements are stored from index $1$ to index $n$ of the array. For the element stored at index $i$ of the array $(i \leq n)$, the index of the parent is

1. $i-1$
2. $\lfloor \frac{i}{2} \rfloor$
3. $\lceil \frac{i}{2} \rceil$
4. $\frac{(i+1)}{2}$

for node at index $i$

left $child(L)$ at $2$i

right $child(R)$ at $2i+1$

for node at index $i$

parent will be at floor $i/2$

Correct Answer: $B$

moved by
Ans is floor(i/2) ..

If index starts with 0 than Ceil(i/2)-1
@papesh What would be the left child and right child if index starts with 0?
left child: 2i+1

right child: 2i+2
ans b)

### 1 comment

ans is B.

just draw the heap and number the elements.