Algorithm Heuristics (Rust)

The objective of a heuristic is to produce a solution in a reasonable time frame that is good enough for solving the problem at hand. This solution may not be the best of all the solutions to this problem, or it may simply approximate the exact solution. But it is still valuable because finding it does not require a prohibitively long time.

The list you provided contains several algorithmic heuristics for solving problems that involve reversing strings or words. These heuristics can be useful for solving coding interview challenges that involve manipulating strings.

Here are some examples of how to implement each of these heuristics in Rust:

1. Two-pointer approach: This approach involves using two pointers to traverse the string from both ends, swapping characters as you go.

fn reverse_string(s: &mut [char]) {

    let mut left = 0;

    let mut right = s.len() - 1;

    while left < right {

        s.swap(left, right);

        left += 1;

        right -= 1;

    }

}

2. Reversing the entire string and then reversing each word: This approach involves first reversing the entire string, and then reversing each individual word.

fn reverse_words(s: &mut [char]) {

    s.reverse();

    let mut start = 0;

    for i in 0..=s.len() {

        if i == s.len() || s[i] == ' ' {

            s[start..i].reverse();

            start = i + 1;

        }

    }

}

3. Using a stack to store words in reverse order: This approach involves using a stack to store the words of the string in reverse order, and then popping them off the stack to construct the reversed string.

use std::collections::VecDeque;

fn reverse_words(s: &str) -> String {

    let mut stack = VecDeque::new();

    let mut word = String::new();

    for c in s.chars() {

        if c == ' ' {

            if !word.is_empty() {

                stack.push_front(word);

                word = String::new();

            }

        } else {

            word.push(c);

        }

    }

    if !word.is_empty() {

        stack.push_front(word);

    }

    stack.into_iter().collect::<Vec<_>>().join(" ")

}

4. Splitting the string into an array and then reversing the array: This approach involves splitting the string into an array of words, and then reversing the order of the words in the array.

fn reverse_words(s: &str) -> String {

    let words: Vec<_> = s.split_whitespace().collect();

    words.into_iter().rev().collect::<Vec<_>>().join(" ")

}

5. In-place string reversal: This approach involves reversing a mutable string in place, without using any additional memory.

fn reverse_words(s: &mut String) {

    let s = unsafe { s.as_mut_vec() };

    s.reverse();

    let mut start = 0;

    for i in 0..=s.len() {

        if i == s.len() || s[i] == b' ' {

            s[start..i].reverse();

            start = i + 1;

        }

    }

}

6. Using a list to store the words in reverse order and then joining the list into a string: This approach is similar to using a stack, but it uses a list data structure instead.

fn reverse_words(s: &str) -> String {

    let mut list = Vec::new();

    let mut word = String::new();

    for c in s.chars() {

        if c == ' ' {

            if !word.is_empty() {

                list.insert(0, word);

                word = String::new();

            }

        } else {

            word.push(c);

        }

    }

    if !word.is_empty() {

        list.insert(0, word);

    }

    list.join(" ")

}

Using regular expressions: This approach involves using regular expressions to match and manipulate the words of the string.

use regex::Regex;

fn reverse_words(s: &str) -> String {

    let re = Regex::new(r"\b\w+\b").unwrap();

    let words: Vec<_> = re.find_iter(s).map(|m| m.as_str()).collect();

    words.into_iter().rev().collect::<Vec<_>>().join(" ")

}