You are tasked with the following:

Given an integer n, find the number of possible arrays of 1s and 2s, such that the sum of all entries in each array is equal to the integer n.

For example, if you are given as an input n=3, then the arrays of 1s and 2s adding up to 3 could be [1,1,1], [1,2] or [2,1]. So you would return 3.

If you are given as an input n=4, then the arrays of 1s and 2s adding up to 4 could be [1,1,1,1], [1,1,2], [2,1,1], [2,2], [1,2,1], so you would return 5. Etc.

Can you do it?

It's All Fucked Shirt $22.14 |

It's All Fucked Shirt $22.14 |

To solve the problem of finding the number of possible arrays of 1s and 2s that sum up to a given integer nn, we can use a dynamic programming approach.

Here's the Python code to implement this approach:

python

def count_ways(n):

if n == 0:

return 1

if n == 1:

return 1

dp = [0] * (n + 1)

dp[0] = 1

dp[1] = 1

for i in range(2, n + 1):

dp[i] = dp[i-1] + dp[i-2]

return dp[n]

# Test cases

print(count_ways(3)) # Output: 3

print(count_ways(4)) # Output: 5

>brute force solutions

lel

>O(logn)

>brute force

?

There is an O(1) solution.

There isn't though

>t. actually knows what I'm talking about

What do you think is the time complexity of p(n)=sqrt(2)*n ?

>n = 3

>sqrt(2) * 3 = 4.24

Even if you floor it you don't get 3 which is the right answer

I didn't say that's the answer, I'm just asking so when I give him the real solution he can't backtrack and say it's not O(1).

So we agree the expression I gave is O(1)? I ask because you can argue integer multiplication is not O(1) when considering arbitrary precision integers.

it's O(n)

sqrt(2) does not affect the asymptotic complexity

There is. It's just not practical.

Yes it is. It's basically diagonalizing over pascal triangle which generates fibonacci sequence.

Yes but it's not numerically stable. It relies on taking powers of irrational numbers and it loses precision quite fast. My solution is O(logn) but it always gives good results, as long as it doesn't overflow of course.

This isn't nocode, this is NOMATH

truly a new low for IQfy

If you have any more complex problems or need further assistance, feel free to ask!

Dude I'm making fun of (You) for being moronic and thinking O(1) is possible

It's mind-boggling how someone could even consider p(n)=2⋅np(n)=2

⋅n to be O(1)O(1). Seriously, are we even discussing the same function? The utter lack of understanding of basic computational complexity here is astonishing. Let me break it down for you in the simplest terms possible, since apparently, this needs to be spelled out.

When we talk about O(1)O(1), we're referring to constant time complexity. This means that the function's output remains the same regardless of the input size nn. In other words, the function should take a fixed amount of time to compute, whether nn is 1, 10, or 10 million. It's as if the function doesn't give a damn about the value of nn.

Now, look at p(n)=2⋅np(n)=2

⋅n. Are you seeing that nn there? This function scales linearly with nn. For every increase in nn, the output increases proportionally. If nn doubles, the result doubles. How in the world can you even think that a function where the result depends directly on nn could be constant time? The growth of p(n)p(n) is inherently tied to the size of nn.

Dismissing the linear dependency and pretending it's O(1)O(1) is like saying the number of steps you take to cross a room stays the same whether the room is 1 meter wide or 100 meters wide. It's an absurdly incorrect notion and demonstrates a fundamental misunderstanding of what constant time complexity actually means.

In conclusion, calling p(n)=2⋅np(n)=2

⋅n O(1)O(1) is not just wrong; it's an egregious oversight that should infuriate anyone with a basic grasp of algorithm analysis.

If we're talking about normal computer integers and not arbitrary precision integers, then multiplication is O(1). The input is always the same length (8 bytes) as well as the output.

This looks shopped. I can tell from some of the pixels and from seeing quite a few shops in my time.

it is, this is the original

>knees that far up

those shins would fold in instantly if he tried to walk

If he were human, yes

don't trust those trolls, this is the original

Peak human form.

thanks for this, I think I have the full set now

I like the touch of a second chair helping support his massive legs

>all those credentials

fricking hire this man, I don't care what it costs.

it's not going to be cheap to hire Bjarne

why is it feeling suddenly so warm in our ChatGPT computing center?

Not doing your homework Tyrone. Frick off you lazy, stupid fricking Black person.

filtered

fn count(n: u32) -> u32 {

let mut v = (1, 1, 0);

let f = n + 1;

for x in (0..f.ilog2()).rev() {

v = (

v.0 * v.0 + v.1 * v.1,

(v.0 + v.2) * v.1,

v.1 * v.1 + v.2 * v.2,

);

if f / (1 << x) % 2 == 1 {

v = (v.0 + v.1, v.0, v.1);

}

}

return v.1;

}

fn main() {

println!("{}: {}", 3, count(3));

println!("{}: {}", 4, count(4));

}

/thread

it's just the combination of an array of n 1s, and permutations of arrays with 2 1s replaced with a 2. the problem itself is moronic and pointless

>differentiating between [2,1,1] and [1,2,1]

moronic and pointless too

FUNCTION_calcu_Arraynumber.toNUMBERasInt(4)

>5

easy

No

if n = 3 return 3

if n = 4 return 5

else return -1 // todo

I noticed while thinking about this, isn't this basically just the fibonacci series?

mathgays have come up with very efficient algorithms for finding fibonacci numbers, I'm just not going to bother looking them up

just write its code