HashSet with is_subset

Pangram
Pangram in Rust 
use std::collections::HashSet;

pub fn is_pangram(sentence: &str) -> bool {
    let all: HashSet<char> = HashSet::from_iter("abcdefghijklmnopqrstuvwxyz".chars());
    let used: HashSet<char> = HashSet::from_iter(sentence.to_lowercase().chars());
    all.is_subset(&used)
}

In this approach a HashSet is made of the lowercase alphabet chars using the from_iter method, and another HashSet is made from the to_lowercase sentence chars.

The function returns if the alphabet HashSet is_subset of the sentence HashSet. If all of the letters in the alphabet are a subset of the letters in the sentence, then is_subset will return true.

8th May 2024 · Found it useful?

Other Approaches to Pangram in Rust

Other ways our community solved this exercise
pub fn is_pangram_all_contains(sentence: &str) -> bool {
    let sentence_lowered = sentence.to_lowercase();
    ('a'..='z').all(|ltr| sentence_lowered.contains(ltr))
}
all with contains on lower case

Use all with contains on lowercased letters. 

sentence
    .to_lowercase()
    .chars()
    .filter(|c| c.is_ascii_alphabetic())
    .collect::<HashSet<char>>()
    .len()
    == 26
HashSet with len

Use HashSet with len. 

for letter in sentence.chars() {
    if letter >= 'a' && letter <= 'z' {
        letter_flags |= 1 << (letter as u8 - A_LCASE);
    } else if letter >= 'A' && letter <= 'Z' {
        letter_flags |= 1 << (letter as u8 - A_UCASE);
    }
}
letter_flags == ALL_26_BITS_SET
Bit field

Use a bit field to keep track of used letters.