
DSA HOTS MCQ
Authored by Thilagavathy R
Computers
University
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15 questions
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1.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
What is the worst case time complexity of inserting n elements into an empty linked list, if the linked list needs to be maintained in sorted order?
O(n2)
O(2n)
O(n)
O(1)
2.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
The preorder traversal sequence of a binary search tree is 30, 20, 10, 15, 25, 23, 39, 35, 42. Which one of the following is the postorder traversal sequence of the same tree?
10,20,15,23,25,35,42,39,30
15,10,25,23,20,42,35,39,30
15,20,10,23,25,42,35,39,30
15,10,23,25,20,35,42,39,30
3.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
The height of a binary tree is the maximum number of edges in any root to leaf path. The maximum number of nodes in a binary tree of height h is:
2h-1
2h-1-1
2h+1-1
2h+1
4.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
A scheme for storing binary trees in an array X is as follows. Indexing of X starts at 1 instead of 0. the root is stored at X[1]. For a node stored at X[i], the left child, if any, is stored in X[2i] and the right child, if any, in X[2i+1]. To be able to store any binary tree on n vertices the minimum size of X should be ___________
log2n
n
2n+1
2n-1
5.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
The B-trees of the order 4 and height 3 would have a maximum of ________ keys.
188
277
163
255
6.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
A single array A[1..MAXSIZE] is used to implement two stacks, The two stacks grow from opposite ends of the array. Variables top1 and top2 (top1 < top2) point to the location of the topmost element in each of the stacks, If the space is to be used efficiently, the condition for "stack full" is
(top1=MAXSIZE/2) and (top2=MAXSIZE/2 + 1)
top1 + top2 = MAXSIZE
(top1=MAXSIZE/2) and (top2=MAXSIZE)
top1 = top2 - 1
7.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Suppose you are given an array s[1..n] and a procedure reverse (s, i, j) which reverse the order of elements in s between positions i and j (both inclusive). What does the following sequence do, where 1≤k<n:
reverse (s, 1, k);
reverse (s, k+1, k);
reverse (s, 1, n);
Rotates S left by K positions
Reverse all elements of S
Leaves S Unchanged
Rotates S left by K+1 positions
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