Solution for "Add binary" Leetcode problem

main.rs

impl Solution {
    pub fn add_binary(a: String, b: String) -> String {
        enum These {
            Both(u8, u8),
            Single(u8),
        }
        use These::*;

        let len = a.len().max(b.len()) + 1;
        let mut a = a.as_bytes().iter().rev().copied();
        let mut b = b.as_bytes().iter().rev().copied();
        let digits = std::iter::from_fn(move || {
            match (a.next(), b.next()) {
                (Some(a), Some(b)) => Some(Both(a, b)),
                (Some(d), None) | (None, Some(d)) => Some(Single(d)),
                (None, None) => None,
            }
        });
        
        let (mut ret, carry) = digits.fold((Vec::with_capacity(len), false), |(mut acc, carry), digit| {
            let (digit, carry) = match digit {
                Both(a, b) => sum_binary(a, b, carry),
                Single(d) => sum_binary_with_carry(d, carry),
            };
            acc.push(digit);
            (acc, carry)
        });
        if carry {
            ret.push(b'1');
        }
        ret.reverse();
        String::from_utf8(ret).unwrap()
    }
}

fn sum_binary_with_carry(digit: u8, carry: bool) -> (u8, bool) {
    match (digit, carry) {
        (b'0', false) => (b'0', false),
        (_, false) | (b'0', true) => (b'1', false),
        (_, true) => (b'0', true),
    }
}

fn sum_binary(a: u8, b: u8, carry: bool) -> (u8, bool) {
    let sum = (a == b'1') as u8 + (b == b'1') as u8 + carry as u8;
    match sum {
        0 => (b'0', false),
        1 => (b'1', false),
        2 => (b'0', true),
        _ => (b'1', true),
    }
}