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  1. Prepare
  2. Algorithms
  3. Strings
  4. Pangrams
  5. Discussions

Pangrams

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1943 Discussions

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  • animeshtyagi2
     + 0 comments
        public static String pangrams(String s) {
    // Write your code here
        int count[] = new int[26];
        
        for(int i = 0; i < s.length(); i++) {
            char ch = Character.toLowerCase(s.charAt(i));
            if(ch >= 'a' && ch <= 'z') { //small letters
            count[ch - 'a']++;   
            }
        }
        
        for(int i = 0; i < count.length; i++) {
            if(count[i] == 0) {
                return "not pangram";
            }
        }
        return "pangram";
    }

  • 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"