Discussion on: Daily Challenge #101 - Parentheses Generator

[Idan Arye]

This is an O(n!) because the output is O(n!) - there is no way around it. Still - it can be greatly optimized if we generate the valid combinations in order instead of generating all the combinations and filtering for the valid ones - which both increases the number of combinations we need to generate and forces us to pay for validating each configuration.

First, we need to represent the combinations. It is enough to represent the positions of the opening parenthesis, as we can fill the closing parenthesis when we generate the string. So our data structure will be an n sized vector of positions. The first position is always 0, but we'll keep it in the vector anyway for the sake of simplicity.

What makes a representation valid? The first position - p[0] - must be 0. Technically this means we can just use an n-1 sized vector, but we'll keep it in the vector anyway for the sake of simplicity.

What about the other position? For each k in [1,n), what are the valid values for p[k]? Obviously p[k] > p[k-1] so we have a minimum. The maximum is less obvious but also easy - k * 2. This is the maximum number of characters before it - k opening parenthesis and k closing parenthesis. There cannot be more opening parenthesis because k is the kth, and there cannot be more closing parenthesis because then there won't be enough opening parenthesis to match them.

So, if we have p[0], p[1], ... p[k] we know the valid ranges of p[k+1] - and that's enough to advance p:

struct ValidParenthesesGenerator {
    num_pairs: usize,
    open_paren_locations: Vec<usize>,
}

impl ValidParenthesesGenerator {
    pub fn new(num_pairs: usize) -> Self {
        Self {
            num_pairs,
            open_paren_locations: Vec::new(),
        }
    }

    fn generate_placement(&self) -> String {
        let mut result = String::with_capacity(self.num_pairs * 2);
        for &open_paren_index in self.open_paren_locations.iter() {
            while result.len() < open_paren_index {
                result.push(')');
            }
            result.push('(');
        }
        while result.len() < self.num_pairs * 2 {
            result.push(')');
        }
        result
    }
}

impl Iterator for ValidParenthesesGenerator {
    type Item = String;

    fn next(&mut self) -> Option<Self::Item> {
        loop {
            let last = if let Some(last) = self.open_paren_locations.pop() {
                last
            } else {
                // Special case guard. We put it here as an optimization because we only get here once anyway.
                if self.num_pairs == 0 {
                    return None;
                }
                // This is the first iteration, so we put 0 as the first index and let the
                // following `while` loop fill up the rest.
                self.open_paren_locations.push(0);
                break;
            };
            if self.open_paren_locations.is_empty() {
                return None; // Cannot move the first paren
            }
            let max_position = self.open_paren_locations.len() * 2; // when all previous opening parens are closed
            if last < max_position {
                self.open_paren_locations.push(last + 1);
                break;
            }
        }
        while self.open_paren_locations.len() < self.num_pairs {
            let &last = self
                .open_paren_locations
                .last()
                .expect("After the loop there should be at least one");
            self.open_paren_locations.push(last + 1);
        }
        Some(self.generate_placement())
    }
}

You can see it in action at play.rust-lang.org/?version=stable...

On my machine, when built with release, it can generate all 35,357,670 combinations for n=16 in between 4 and 5 seconds. Printing them takes much much much longer - even when I redirect the output to /dev/null it takes half a minute, and when I don't it takes forever (didn't let it finish). I suspect other implementations will have similar problems, so you should try to time them without printing the combinations (just print how many you got)