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Linked Lists Questions

NA
SHSTTON
133
Solv. Corr.
29
Solv. In. Corr.
162
Attempted
0 M:0 S
Avg. Time

41 / 48

Choose the correct option.

Are the below statements true about skiplists?
In a sorted set of elements skip lists can implement the below operations

I. given a element find closest element to the given value in the sorted set in O(logn)
II. find the number of elements in the set whose values fall a given range in O(logn)


ATRUE

BFALSE

Answer: Option A

Explanation:

To achieve above operations augment with few additional stuff like partial counts.

Workspace

NA
SHSTTON
110
Solv. Corr.
45
Solv. In. Corr.
155
Attempted
0 M:43 S
Avg. Time

42 / 48

Choose the correct option.

How to maintain multi-level skip list properties when insertions and deletions are done?


Adesign each level of a multi-level skip list with varied probabilities

Bebalancing of lists

Cthat cannot be maintained

Dreconstruction

ENone of these

Answer: Option A

Explanation:

For example consider a 2 level skip list. the level-2 skip list can skip one node on a average and at some places may skip 2 nodes, depending on probabilities. this ensures O(logn).

Workspace

NA
SHSTTON
20
Solv. Corr.
127
Solv. In. Corr.
147
Attempted
0 M:0 S
Avg. Time

43 / 48

Choose the correct option.

What are the disadvantages in implementing buddy system algorithm for free lists ?


Ainternal fragmentation

Bwe no more have the hole lists in order of memory address, so it is difficult to detect if 2 holes remain adjacent in memory and shall be merged into one hole

Cit takes so much space

Dboth a and c are correct

ENone of these

Answer: Option D

Explanation:

Internal fragmentation is an issue to be dealt and it takes so much space.

Workspace

NA
SHSTTON
25
Solv. Corr.
119
Solv. In. Corr.
144
Attempted
0 M:0 S
Avg. Time

44 / 48

Choose the correct option.

Which of the following is true about the Move-To-Front Method for rearranging nodes?


Anode with highest access count is moved to head of the list

Bmay over-reward infrequently accessed nodes

Crequires extra storage

Drequires a counter for each node

ENone of these

Answer: Option B

Explanation:

In Move-To-front Method the element which is searched is moved to the head of the list. And if a node is searched even once, it is moved to the head of the list and given maximum priority even if it is not going to be accessed frequently in the future. Such a situation is referred to as over-rewarding.

Workspace

NA
SHSTTON
49
Solv. Corr.
130
Solv. In. Corr.
179
Attempted
0 M:0 S
Avg. Time

45 / 48

Choose the correct option.

Which of the following data structure is preferred to have lesser search time when the list size is small?


Asearch tree

Bself organizing list

Csorted list

Dlinked list

ENone of these

Answer: Option B

Explanation:

Self-organizing list is easy and simple to implement than search tree and it requires no additional space. So using self organizing list is preferred when list size is small.

Workspace

NA
SHSTTON
81
Solv. Corr.
59
Solv. In. Corr.
140
Attempted
0 M:0 S
Avg. Time

46 / 48

Choose the correct option.

What is buddy memory management of free lists ?


Amodified version of first fit

Bmodified version of best fit

Cbuddy allocation keeps several‭ ‬free lists,‭ ‬each one holds blocks which are of one particular size

Da tree representation of free lists

ENone of these

Answer: Option C

Explanation:

When an allocation request is received,‭ ‬the list that holds blocks that are just large enough to satisfy the request are considered, and an open location is returned.‭ ‬If no‭ ‬free‭ ‬blocks that are smaller than two times the size that are requested are available,‭ ‬a larger block is split in two to satisfy the requirements.

Workspace

NA
SHSTTON
58
Solv. Corr.
60
Solv. In. Corr.
118
Attempted
0 M:0 S
Avg. Time

47 / 48

Choose the correct option.

Assume there is a free list which contains nodes and is filled with a value if it is already assigned and the value will be the size of requested block else will be 0. The below code represents what ?

z = startpoint;
while ((z < end) && (*z <= len))
{
assign this block
}


Acode for first fit

Bcode for worst fit

Ccode for best fit

Dnone of the mentioned

ENone of these

Answer: Option A

Explanation:

As z is start point and now from beginning we are moving and checking if we reached end and then checking size naively assigning the first block which is bigger than required size hence it is first fit.

Workspace

NA
SHSTTON
65
Solv. Corr.
78
Solv. In. Corr.
143
Attempted
0 M:0 S
Avg. Time

48 / 48

Choose the correct option.

How does implicit free lists(garbage collection) works in adding memory to free list ?


Awhichever comes last will be added to free list

Bcertain blocks cannot be used if there are no pointers to them and hence they can be freed

Cwhichever comes first will be added to free list

Dmakes a probabilistic guess

ENone of these

Answer: Option B

Explanation:

When no pointers pointing a block that means it is useless to be in memory.

Workspace

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