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  • + 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";
        }
    
  • + 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"
    }
    
  • + 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 ";
    }
    
  • + 0 comments

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

  • + 0 comments

    python

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