# 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.

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;

}

}

}

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;

}

}

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:

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;

}

}

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;

}

}

}

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(" ")

}

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

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

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

}

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;

}

}

}

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(" ")

}