Answer:
No, it can't be verified with a pseudocode.
Explanation:
We can not verify this with a pseudocode because the largest integer that we can store in 32-bit integer goes by the formula 2^32 - 1 = 4, 294, 967,295 and this means that it has 32 ones. Or it may be 2^31 - 1 = 2, 147, 483,647 which is a two complement signed integer.
Despite the fact that it can not be verified by using pseudocode, we can do the Verification by writting programs Through some programming language or in plain English code.
In a 32-bit CPU, the largest integer that we can store is 2147483647 because we can store integer as 2^31 and - (2^31 + 1).
Answer:
Motherboard
Explanation:
Motherboard is the main component of a computer. It connects with other parts of the computer including drive disks, RAM and CPU and any other expansion slots :)
(Hope it helped)
It converts the mic input into a string of raw data, then compares it to hundreds, even thousands of voice samples. The output is a polished string of data in words.
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
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As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
