Count operation Lucky twos Lucky Twos determines and displays the number of digits that are 2s in a whole number. For example, the number of 2s in 3487 is 0, while the number of 2s in 272521 is 3. Note: whole numbers are non-negative integers starting at zero 0, 1, 2, 3, 4.
Assume that the fractional part is discarded in the division:
10 / 4 = 2
8 / 5 = 1
20 / 3 = 6
Read number
set count to 0
while number > 0
If the number module 10 is Then add
1 to count
End If
Compute number as number / 10
End while
Display count.
Let n be the number of digits of the whole number. What is the number of operations that are executed in the code in terms of n?

Following are the solution to this question:

Explanation:

Its complexity of both the pseudo-code described is indeed proportional to the number of digits. So, how often number there are in this specific number is the query. Whenever a number is considered, the d digit would be between inclusive exclusive That would be as, let d become the number of digits at N, and the inequalities, They can tell

We get, we take a logarithm,

The increase of 1 to the left inequality, , and Combining the previous outcome, we got, . That's would be to say, that number of number d by is higher and lower. Consequently, the number of transactions in the code is

this seems to be in the wrong category. but i'll take some free points my dude

posting this in wrong category will result in people seeing it who won't know the answer.