Heap sort algorithm is same as selection sort algorithm ,because in selection sort we find the min element from the array and then swap that selected min element with first element of the list and then continue with the remaining n-1 elements, similarly in heap sort we find the min or max element and then we swap the selected element with the last element of the list and then continue with the remaining n-1 elements.

Option A is wrong since ascii value of ‘a’ is 97 and ascii value of ‘O’ is 79 and ascii value of A is 65 so the correct lexographic order in this option will be AND,OR,a,an.
Option B is wrong since ascii since ascii value of 5’’ is 101 and of ‘1’ is 97,so the correct lexographic order for the sequence given in this option is 1,125,15,5.
Option C is correct as it can be seen in Option A explaination.
Option D is wrong as it can be seen in explaination of Option B.

* For first while loop iteration,
j=n, ‘for’ loop will run n times.
* For second while loop iteration,
j=n/2, ‘for’ loop will run n/2 times.
* For third while loop iteration,
j=n/2², ‘for’ loop will run n/22 times.
⋮
For kth while loop iteration,
j=n/2^k-1, for loop will run n/2^k-1 times

Whenever the list is almost sorted then insertion sort will always take O(n) time .But for some sequences even the list is almost sorted then also bubble sort might take O(n^2).And Selection sort will always take O(n^2) time.

Depth first traversal algorithm visits the node depth wise and breadth first traversal algorithm visits the node level wise. So among the given options only in linear chain both Depth first traversal algorithm and
Breadth first traversal algorithm visits the node in the exactly same order because in linear chain nodes either we visit them level wise or depth wise the order of visiting will be same.

The given postfix expression is,
ABC+*DE*/
So the binary tree for above given postfix expression is,

No the prefix expression for above binary tree is nothing but the preorder traversal sequence of above binary tree. So the preorder sequence of above binary tree is,
/*A+BC*DE

Both BFS and DFS can be used to find strongly connected components in linear time O(V+E).

Since key points before solving this question,
Unary operators have higher precedence than arithmetic operators.
If ‘--’ or ‘++’ unary operator is before n then that will be considered in the current equation and not the ‘--’ or ‘++’ which is after n, since it will be calculated after calculating the current equation.
So,
--n*n+1*n+++1
= (n - 1) * n + 1 * n + 1
= n² - n + n + 1
= n² + 1

Whenever a float is calculated with integer we get float value.But when integer is calculated with integer value we will get integer value as a result even if the original value comes as float.
Let’s check option wise,
A)
z * y/x + c
= 10.0 ⅘ + 97
= 40.0/5 + 97
= 8.0 + 97
= 105
B)
z/y * x + c
= 10.0/4 * 5 + 97
= 2.5 * 5 + 97
= 12.5 + 97
= 109.5
C)
z + c/y * x
= 10.0 + 97/4 * 5
= 10.0 + 24 * 5
= 10.0 + 120
= 130
D)
z/c + y * x
= 10.0/97 + 4 * 5
= 0.1031 + 20
= 20.1031

Let’s take example,
x=b=2
k=n=3
z=1
Now,
k!=0 and k is odd, so
z= 1 x 2 = 2
x = 2 x 2 = 4
k = (3/2)=1
Again, k!=0 and k is odd, so
z= 2 x 4 = 8
x= 4 x 4= 16
k = (½) = 0
Since k=0, so while loop will be terminated.
So, finally the values are,
z=8
x=16
k=0
b=2
n=3
Now, let’s check option wise,
A)
x = b ⁿ
x = 2 ³ = 8
Not correct.
B)
z = n ⁿ
z = 2 ³ = 8
Correct.
C)
b = zⁿ = 8 ^{3
Not correct.
D)
b = xn
= 16 3
Not correct.}