It is great but that's really it.
Don't get me wrong I adore Python, no complications, pure simplicity, wonderful community. But for any larger project that will be scaled I'd never use it. It's slow (mostly because of GIL) and gets pretty hard to organise once you have thousands of .py files but it's still a great language (my first one) when doing quick prototyping, personal projects, learning and it's also AI de facto programming language because of its readability works as a glue with AI.
It's related to flowchart in a way we write algorithms, for eg. in python we rarely use counter in for loop the inverse is thus C++ where most for loops are for loops not for each loops.
Hope this helps.
The answer is D, because it's called a back-up.
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.
The problem with the swap function is that it loses the value at the first index, as soon as it gets overwritten by the value at the second index. This happens in the first statement. To fix it, you need a helper variable.
First you're going to "park" the index at the first index in that helper variable, then you can safely overwrite it with the value at the second index. Then finally you can write the parked value to the second index:
var swap = function(array, firstIndex, secondIndex) {
let helper = array[firstIndex];
array[firstIndex] = array[secondIndex];
array[secondIndex] = helper;
};
I hope this makes sense to you.
Answer:
In the presence of packets transfer between client and server in a TCP session, the attack will be successful if the number of the sequence is approximately X+100. Otherwise, the attack will not be successful (i.e. fail).
Explanation:
Generally, it is important to ensure that there is a successful operation in the operation of a TCP session. In the presence of packets transfer between client and server in a TCP session, the attack will be successful if the number of the sequence is approximately X+100. Otherwise, the attack will not be successful (i.e. fail).
