Programming Challenge

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?

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  1. 6 days ago
    Anonymous

    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

    • 6 days ago
      Anonymous

      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

      >brute force solutions
      lel

      • 6 days ago
        Anonymous

        >O(logn)
        >brute force
        ?

        • 6 days ago
          Anonymous

          There is an O(1) solution.

          • 6 days ago
            Anonymous

            There isn't though
            >t. actually knows what I'm talking about

          • 6 days ago
            Anonymous

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

          • 6 days ago
            Anonymous

            >n = 3
            >sqrt(2) * 3 = 4.24
            Even if you floor it you don't get 3 which is the right answer

          • 6 days ago
            Anonymous

            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).

            sqrt(2) does not affect the asymptotic complexity

            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.

          • 6 days ago
            Anonymous

            it's O(n)

          • 6 days ago
            Anonymous

            sqrt(2) does not affect the asymptotic complexity

          • 6 days ago
            Anonymous

            There is. It's just not practical.

            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

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

          • 6 days ago
            Anonymous

            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.

      • 6 days ago
        Anonymous

        There is an O(1) solution.

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

        This isn't nocode, this is NOMATH
        truly a new low for IQfy

        • 6 days ago
          Anonymous

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

          • 6 days ago
            Anonymous

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

          • 6 days ago
            Anonymous

            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.

          • 6 days ago
            Anonymous
          • 6 days ago
            Anonymous

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

            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.

  2. 6 days ago
    Anonymous

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

    • 6 days ago
      Anonymous

      it is, this is the original

      • 6 days ago
        Anonymous

        >knees that far up
        those shins would fold in instantly if he tried to walk

        • 6 days ago
          Anonymous

          If he were human, yes

        • 6 days ago
          Anonymous

          don't trust those trolls, this is the original

          • 6 days ago
            Anonymous

            Peak human form.

          • 6 days ago
            Anonymous

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

          • 6 days ago
            Anonymous

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

          • 6 days ago
            Anonymous

            >all those credentials
            fricking hire this man, I don't care what it costs.

          • 6 days ago
            Anonymous

            it's not going to be cheap to hire Bjarne

  3. 6 days ago
    Anonymous

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

  4. 6 days ago
    Anonymous

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

    • 6 days ago
      Anonymous

      filtered

  5. 6 days ago
    Anonymous

    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

  6. 6 days ago
    Anonymous

    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

  7. 6 days ago
    Anonymous

    FUNCTION_calcu_Arraynumber.toNUMBERasInt(4)
    >5
    easy

  8. 6 days ago
    Anonymous

    No

  9. 6 days ago
    Anonymous

    if n = 3 return 3
    if n = 4 return 5
    else return -1 // todo

  10. 6 days ago
    Anonymous

    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

  11. 6 days ago
    Anonymous

    just write its code

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