What is the worst case time complexity of inserting n
2
2
elements into an AVL-tree with n elements intially ?

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Θ(n
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Θ
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Question

What is the worst case time complexity of inserting n
2
2
  elements into an AVL-tree with n elements intially ?

Θ
(
n
4
)
Θ(n 
4
 )

Θ
(
n
2
)
Θ(n 
2
 )

Θ
(
n
2
log
⁡
n
)
Θ(n 
2
 logn)

Θ
(
n
3
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Θ(n 
3
 )
No, the answer is incorrect.
Score: 0
Accepted Answers:
Θ
(
n
2
log
⁡
n
)
Θ(n 
2
 logn)
Explain like a tutor like i do not know anything about graphs:

Gate Mock Test, Gate Practice Test question, Answer carefully with all clear steps and accuracy with given answer:14) What is the worst case time complexity of inserting n elements nto an AVL-tree with n elements intially? 1 point
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No,the answers incorect,
score 0
Accepted Answers:
B(n*logn)

VIPSolutions · Accepted Answer

Let's analyze the problem step by step:

## Problem:
What is the **worst case time complexity** of inserting $ n^2 $ elements into an **AVL tree** that initially has $ n $ elements?

---

## Key Concepts:
- **AVL Tree**: A self-balancing binary search tree that maintains a height of $ O(\log n) $.
- **Insertion in AVL Tree**: 
  - Insert the element as in a normal binary search tree (BST).
  - Rebalance the tree if needed (via rotations).
- **Time for one insertion**:
  - In the worst case, insertion takes $ O(\log n) $ time because the tree height is $ O(\log n) $.

---

## Step-by-step reasoning:

### 1. Initial size of the tree:
- The tree initially has $ n $ elements.

### 2. Number of insertions:
- We are inserting $ n^2 $ new elements.

### 3. Size of the tree during insertion:
- After inserting 1 element, size becomes $ n + 1 $.
- After inserting $ n^2 $ elements, total size will be approximately $ n + n^2 $.

### 4. Cost of inserting each element:
- Each insertion in an AVL tree costs $ O(\log k) $, where $ k $ is the size of the tree at that moment.
- Since the size of the tree increases with each insertion, the cost varies from $ O(\log n) $ (initially) to $ O(\log (n + n^2)) $.

### 5. Approximate total cost:
- The total time for inserting all $ n^2 $ elements:
  
$$
\sum_{k=1}^{n^2} O(\log (n + k))
$$

- For large $ n $, the dominant term is $ \log (n + n^2) \approx \log n^2 = 2 \log n $.

- The sum over $ n^2 $ insertions becomes:

$$
n^2 \times O(\log n)
$$

which simplifies to:

$$
O(n^2 \log n)
$$

### 6. Final complexity:
- The worst-case time complexity is:

$$
\boxed{\Theta(n^2 \log n)}
$$

---

## **Answer:**
**$\Theta(n^2 \log n)$**

This matches the accepted answer and explains why inserting $ n^2 $ elements into an AVL tree initially with $ n $ elements takes approximately $ n^2 \log n $ time in the worst case.

---

## Summary:
- Each insertion costs $ O(\log n) $.
- Total insertions: $ n^2 $.
- Total time: $ O(n^2 \log n) $.

**Thus, the correct choice is: $\Theta(n^2 \log n)$.**

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