We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Strings
  4. Pangrams
  5. Discussions

Pangrams

Sort by

recency

|

1942 Discussions

|

  • bahaa_lashin
     + 0 comments
    function pangrams(s) {
    s = Array.from(new Set(s.toLowerCase().split("").filter((x)=> x <= 'z' && x >= 'a').join("")))
    console.log(s)
    if(s.length >= 26)
        return "pangram";
    return "not pangram"
}

  • geovanni_luna
     + 0 comments

    Here is a TypeScrip solution with O(n) time and O(1) space.

    I know the apace can also be consider O(n) because it grows; however, it will never be bigger then 26 (size of alphabet) so it can be consider constant space.

    function pangrams(s: string): string {
    const alpha: string = "abcdefghigklmnopqrstuvwxyz"
    let content = new Set()
    for (let ch of s.toLocaleLowerCase()) {
        if (alpha.includes(ch)) {
            content.add(ch)
        }
    }
    
    if (content.size === 25) 
        return "pangram ";
    else
        return "not pangram ";
}

  • yashdeora98294
     + 0 comments

    Here is problem solution in python java c++ c and javascript - https://programmingoneonone.com/hackerrank-pangrams-problem-solution.html

  • shiaramoon
     + 0 comments

    python

    def pangrams(s):
    l = set(s.lower())
    if " " in l:
        l.remove(" ")
    
    return "pangram" if len(l) == 26 else "not pangram"

  • vaibhav1206
     + 0 comments

    C++ Easy Solution

    string pangrams(string s) {
    unordered_set<char> res;
    for(char c : s){
        if(isalpha(c)){
            res.insert(tolower(c));
        }
    }
    return (res.size()==26)? "pangram":"not pangram";
}