In the first case the array is contiguoaus while in the second case array is not contiguos because in the second case the declaration is array of ponters to pointers.

In direct mapping only choice is there.
In set associative mapping of set size k,there are k choices.
While in associative mapping the choice is any block of cache.
Hence ascending order of flexibility is ,
Direct mapping , Set associative Mapping, Associative mapping.

When we do min or max heapify in heap sort, we get the min or max element in the root then we exchange it with the last element. And we repeat the procedure for the remaining element .
We do similar type in selection sort,we find min or max element and then we exchange that element with first element and we repeat the element with the remaining procedure .

Lexicographical sorting of numbers is similar to the Lexicographical sorting of words in a dictionary.
So the correct answer is,
109,119,19,2,20,207,27

We have to select such subnet mask by which ANDing the both IP address we get the same network ID. And option D gives same network ID after ANDing .The network ID we get is 186.220.0.0

S ⟶ aS
⟶ aaS (∵ S ⟶ aS)
⟶ aabA (∵ S ⟶ bA)
⟶ aabccA (∵ A ⟶ ccA)
⟶ aabccd (∵ A ⟶ d)

Examples of sorting algorithms that run in linear time are counting sort, radix sort and bucket sort. Counting sort and radix sort assume that the input consists of integers in a small range.

Hexadecimal number consists of 4 bit binary no. An octal number is comprising of 3 bit binary no.
So first we will convert the Hexadecimal number into binary no. and then by taking each 3 bit from right side we will convert it into octal no.

2xyz’ + xy’z’ + x’yz’
= xyz’ + xyz’ + xy’z’ + x’yz’
= xyz’ + x’yz’ + xyz’ + xy’z’
= yz’ (x + x’) + xz’ (y + y’)
= yz’ + xz’
= z’(y + x)

Universal Turing Machine is a machine which can simulate the work of other turing machine.Also universal TM can accept any other machine's input other than itself.
Now lets check option wise,
A) It is the halting problem of turing machine which is undecidable.
B) Universal turing machine cannot recognize its own action,so undecidable.
C) By rice’s theorem it is undecidable.
D) If a TM can simulate the work of any other TM, then it will be UTM. Here definite "yes" and "no" ans possible. So, will be decidable

Queues serve a major role in simulation of limited resource allocation .
Stack serves major role in simulation of recursion and expression evaluation.

The graph K(5) and K(3,3) are two of the most important graphs within the subject of planarity in graph theory.Kuratowski’s theorem tells us that if we can find the subgraph in any graph that is homeomorphic to K(5) or K(3,3) then the graph is not planar,meaning its not possible for the edges to be redrawn such that they are none overlapping. Of course this theorem relies on the fact that K(5) and K(3,3) are themselves are not planar.

Tree with 5 levels of nodes has height 4. So the no. of nodes in a complete binary tree is of height 4 is
2^(4+1) - 1 = 2^5 - 1 = 31