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  1. Pangrams
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Pangrams

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228 Discussions

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  • jay_garala
     + 0 comments
    	Set<Character> chars = new HashSet<>();
        for (int i='a'; i<='z'; i++) {
            chars.add((char)i);
        }
        for (char c : s.toCharArray()) {
            if (c == ' ') continue;
            if (c >= 'A' && c <= 'Z') {
                chars.remove((char)(c+32));
            } else {
                chars.remove(c);
            }
        }
        return chars.size() == 0 ? "pangram" : "not pangram";

  • nelarakipaj
     + 0 comments

    Python3 Solution: def pangrams(s): string = s.lower() for j in [chr(i) for i in range(ord('a'), ord('z') + 1)]: if j not in string: return('not pangram')

    return('pangram')

  • victor_reial
     + 0 comments

    Rust best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    pub fn pangrams(s: &str) -> String {
    //Time complexity: O(n)
    //Space complexity (ignoring input): O(1)
    let mut bit_mask = 0;
    for letter in s.to_lowercase().chars() {
        if ('a'..='z').contains(&letter) {
        let bit_pos = letter as u32 - 'a' as u32;
        bit_mask |= 1 << bit_pos;
        };
    }
    if bit_mask == (1 << 26) - 1 {
        return "pangram".to_string();
    };

    "not pangram".to_string()
}

  • victor_reial
     + 0 comments

    Python best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    def pangrams(s: str):
    # Time complexity: O(n)
    # Space complexity (ignoring input): O(1)
    bit_mask = 0
    for letter in s.lower():
        if (letter <= "z") and (letter >= "a"):
            bit_pos = ord(letter) - ord("a")
            bit_mask |= 1 << bit_pos

    if bit_mask == (1 << 26) - 1:
        return "pangram"
    return "not pangram"


def pangrams_not_elegant(s: str):
    # Time complexity: O(n)
    # Space complexity (ignoring input): O(1)
    dict_letters = {}
    for letter in s.lower():
        if (letter <= "z") and (letter >= "a") and (letter not in dict_letters):
            dict_letters[letter] = 1

    if len(dict_letters) == 26:
        return "pangram"

    return "not pangram"

  • shabloel
     + 0 comments

    //java 17 solution

     public static String pangrams(String s) {
 // Write your code here
 List<Character> chars = s.toLowerCase().chars().mapToObj(c -> (char)c)
    .distinct()
    .filter(c -> c != ' ')
    .collect(Collectors.toList());

 if(chars.size() == 26) {
    return "pangram";
 }
 return "not pangram";
}