Qus : 1
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4
Find out the wrong statement based on the characteristics of AVL tree data structure.
1
AVL tree is a binary search tree in nature.
2
AVL tree is known as height-balanced tree. 3
AVL tree has O(log₂ n) search time complexity considering ‘n’ as number of nodes. 4
AVL tree has O(n) search time complexity considering ‘n’ as number of nodes. Go to Discussion
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Solution AVL trees guarantee O(log n) search time, not O(n).
Qus : 16
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2
Which of the following points is/are not true about Linked List data structure when it is compared with an array?
1
Random access is not allowed in a typical implementation of Linked Lists. 2
Access of elements in Linked List takes less time than compared to arrays. 3
Arrays have better cache locality that can make them better in terms of performance. 4
It is easy to insert and delete elements in Linked List. Go to Discussion
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Solution Access in linked list is slower than arrays.
Qus : 19
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2
Five elements P, Q, R, S, T are pushed onto a stack starting from P. The stack is then popped 4 times, and each popped element is inserted into a queue. Two elements are then deleted from the queue and pushed back onto the stack. Finally, one element is popped from the stack. What will be the popped element?
1
P 2
Q 3
R 4
S Go to Discussion
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Solution Final popped element becomes Q after operations.
Qus : 20
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4
Consider the following operations on a singly linked list initially containing the elements: 10 → 20 → 30 → 40
(i) Insert 15 after 10
(ii) Delete node containing 30
(iii) Insert 25 at end
(iv) Delete first node.
What will be the final sequence?
1
15 → 20 → 25 → 40 2
20 → 15 → 40 → 25 3
15 → 20 → 40 → 25 4
20 → 15 → 25 → 40 Go to Discussion
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Solution Final list: 20 → 15 → 25 → 40.
Qus : 24
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3
Which of the given statement is the correct recurrence for the worst case of Binary Search?
1
T(n) = 2T(n/2) + O(1) & T(1)=O(1)=T(0) 2
T(n) = T(n-1) + O(1) & T(1)=O(1)=T(0) 3
T(n) = T(n/2) + O(1) & T(1)=T(0)=O(1) 4
T(n) = T(n-2) + O(1) & T(1)=O(1) Go to Discussion
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Solution
Solution: Binary search recurrence is T(n) = T(n/2) + O(1) . Base case T(1)=O(1).
Answer: (C) T(n) = T(n/2) + O(1)
Qus : 26
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3
Numbers 7, 5, 1, 8, 3, 6, 0, 9, 4, 2 are inserted in that order into an initially empty binary search tree. The tree uses usual ordering. What is the in-order traversal sequence?
1
7, 5, 1, 0, 3, 2, 4, 6, 8, 9 2
0, 2, 4, 3, 1, 6, 5, 9, 8, 7 3
0, 1, 2, 3, 4, 5, 6, 7, 8, 9 4
9, 8, 6, 4, 2, 3, 0, 1, 5, 7 Go to Discussion
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Solution
Solution: In-order traversal of a BST gives the sorted order of keys ⇒ 0 to 9 ascending.
Answer: (C) 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
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