For the Quick sort algorithm, what is the time complexity of the best/worst case?

Algorithm - Quiz(MCQ)

best case: O(log(n)) worst case: O(n2)

best case: O(n2), worst case: O(n log(n))

best case: O(n log(n) ) worst case: O(n2)

best case: O(n2), worst case: O(n2log(n) )

Correct Answer : best case: O(n log(n) ) worst case: O(n2)

Recently Updated in Algorithm Questions

External entities in DFD are shown by_________

Circle

Ellipse

Hexagon

Rectangle

Correct Answer : Rectangle

Assume that we use Bubble Sort to sort n distinct elements in ascending order. When does the best case of Bubble Sort occur?

When elements are sorted in ascending order

When elements are not sorted by any order

When elements are sorted in descending order

There is no best case for Bubble Sort. It always takes O(n*n) time

Correct Answer : When elements are sorted in ascending order

Consider the following pseudo code, where x and y are positive integers.

begin q := 0 r := x while r >= y do begin r := r – y q := q + 1 end end

The post condition that needs to be satisfied after the program terminates is

{r = qx + y ∧ r < y}

{ q + 1 < r–y ∧ y > 0}

{x = qy + r ∧ r < y}

{y = qx + r ∧ 0 < r < y}

Correct Answer : {x = qy + r ∧ r < y}

Explaination :

The given pseudo code does following for given x and y which positive integers.

1) It initializes r as x.

2) It repeatedly subtracts y from r until r becomes

smaller than y. For every subtraction, it

increments count q.

3) Finally r contains remainder, i.e., x%y and q contains

⌊x/y⌋

See below pseudo code with comments.

begin
   q := 0  // q is going to contain floor(x/y)
   r := x  // r is going to contain x % y

// Repeatedly subtract y from x.
while r >= y do
   begin
      r := r – y
      q := q + 1
   end
end