HashSet with len

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

pub fn is_pangram(sentence: &str) -> bool {
    sentence
        .to_lowercase()
        .chars()
        .filter(|c| c.is_ascii_alphabetic())
        .collect::<HashSet<char>>()
        .len()
        == 26
}

This approach chains several functions together to determine the result.

  • It first passes the sentence to_lowercase.
  • The lowercased sentence is then iterated by chars.
  • The chars are filtered in its closure so that only a character that is_ascii_alphabetic makes it through to be collected into a HashSet.
  • The function returns if the len of the HashSet is 26. If the number of unique letters in the HashSet is equal to the 26 letters in the alphabet, then the function 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. 

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)
}
HashSet with is_subset

Use HashSet with is_subset. 

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.