Find groups of anagrams in a list of strings

main.rs

use std::collections::HashMap;

/// Given an array of strings, group the anagrams together in separate arrays.
/// An anagram is a word or phrase formed by rearranging the letters of another
/// word or phrase, using all the original letters exactly once.
///
/// Example:
/// ```
/// $ group_anagrams(["eat","tea","ten","poop","net","ate"])
/// [["poop"],["net","ten"],["eat","tea","ate"]]
/// ```
fn main() {
    println!(
        "{:?}",
        group_anagrams(vec!["eat", "tea", "ten", "poop", "net", "ate"])
    );
}

fn group_anagrams(words: Vec<&'static str>) -> Vec<Vec<String>> {
    let mut anagram_map: HashMap<String, Vec<String>> = HashMap::new();
    for word in words.into_iter() {
        let mut sorted: Vec<char> = word.clone().chars().collect();
        sorted.sort_by(|a, b| b.cmp(a));
        let sorted = sorted.iter().collect::<String>();
        let val = String::from(word);

        match anagram_map.contains_key(&sorted.clone()) {
            true => {
                if let Some(v) = anagram_map.get_mut(&sorted) {
                    v.push(val);
                }
            }
            false => {
                anagram_map.insert(sorted, vec![val]);
            }
        }
    }

    anagram_map.into_values().collect()
}

#[cfg(test)]
mod test {
    use super::group_anagrams;

    #[test]
    fn test_group_anagrams() {
        let words = vec!["foo", "bar", "baz"];
        let anagrams = group_anagrams(words);

        // since there are no anagram, each string is alone in its own group.
        assert!(anagrams.len() == 3);

        let words = vec!["eat", "ate", "tea"];
        let anagrams = group_anagrams(words.clone());

        assert!(anagrams.len() == 1);
        assert_eq!(
            anagrams[0],
            words
                .iter()
                .map(|&s| { String::from(s) })
                .collect::<Vec<String>>()
        );
    }
}