Feedback: (06) ticket_management - (02) vec could lead to a much more complex problem

[fungiboletus]

Hi, thanks for the exercices, I enjoy them.

In exercice (06) ticket_management - (02) vec, you have to complete the following function:

pub fn fibonacci(n: u32) -> u32 {
    // TODO: implement the `fibonacci` function
    //
    // Hint: use a `Vec` to memoize the results you have already calculated
    // so that you don't have to recalculate them several times.
    todo!()
}

Following the hint, I used a vec to memoize the results in a vec, but my code ended up a tiny bit too complex with some once_cell::sync::Lazy and std::sync::Mutex, which doesn't seem right for such an exercice. I have done memoization in Rust before, and it's not super fun, so that could explain why I used once_cell stuff, but to me the hint was about caching the results between calls.

use once_cell::sync::Lazy;
use std::sync::Mutex;

static FIBONACCI_MEMO: Lazy<Mutex<Vec<u32>>> = Lazy::new(|| Mutex::new(vec![0, 1]));

pub fn fibonacci(n: u32) -> u32 {
    let nusize = n as usize;
    let mut memo = FIBONACCI_MEMO.lock().unwrap();

    if memo.len() <= nusize {
        if memo.capacity() <= nusize {
            memo.reserve(nusize * 2);
        }
        std::mem::drop(memo);
        let result = fibonacci(n - 2) + fibonacci(n - 1);
        let mut memo = FIBONACCI_MEMO.lock().unwrap();
        memo.push(result);
        return result;
    }

    memo[nusize]
}

I looked at the solution and I see that it's much simpler.

pub fn fibonacci(n: u32) -> u32 {
    let n = n as usize;
    let mut memo = vec![0, 1];
    for i in 2..=n {
        memo.push(memo[i - 1] + memo[i - 2]);
    }
    memo[n]
}

But I'm not sure the Vec is necessary at all:

pub fn fibonacci(n: u32) -> u32 {
    if n <= 1 {
        return n;
    }

    let mut a = 0;
    let mut b = 1;

    for _ in 2..=n {
        let c = a + b;
        a = b;
        b = c;
    }

    b
}