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GATE | GATE CS 2020 | Question 16

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What is the worst case time complexity of inserting n2 elements into an AVL-tree with n elements initially ? (A) Θ(n4) (B) Θ(n2) (C) Θ(n2 log n) (D) Θ(n3)

Answer: (C)

Explanation: Since AVL tree is balanced tree, the height is O(log n). So, time complexity to insert an element in an AVL tree is O(log n) in worst case. Note:
Every insertion of element:
Finding place to insert = O(log n)
If property not satisfied (after insertion) do rotation = O(log n)

So, an AVL insertion take = O(log n) + O(log n) = O(log n) in worst case. 
Now, given n2 element need to insert into given AVL tree, therefore, total time complexity will be O(n2 log n). Alternative method: Time complexity in worst case,
1st insertion time complexity = O(log n)
2nd insertion time complexity = O(log(n+1))
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n2th insertion time complexity = O(log(n + n2)) 
So, total time complexity will be,
= O(log n) +  O(log n+1)) + .... +  O(log(n + n2))
= O(log n*(n+1)*(n+2)*...(n+n2))
= O(log nn2)
= O(n2 log n) 
Option (C) is correct.

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Last Updated : 11 Aug, 2021
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