Check if string has all the letters of the alphabet

Question

What would be the best logic to check all the letters in a given string.

If all the 26 letters are available in the provided string, I want to check that and perform

Answer 1

try this method, is very simple and works fast

public static String checkSentence(String mySentence)
{
    String result = "";
    char[] alphabet = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

    String missingLetters = "";
    int countMissLetters = 0;

    for (int i = 0; i < alphabet.length; i++)
    {
        if (!mySentence.toLowerCase().contains(String.valueOf(alphabet[i]).toLowerCase()))
            {
                missingLetters = missingLetters + String.valueOf(alphabet[i]) + " ";
                countMissLetters++;
            }
    }

    result = "Letters who are missing: " + missingLetters + "Counted: " + countMissLetters;
    return result;
}

with this exemple

System.out.println(checkSentence("Your own sentence to test."));
System.out.println(checkSentence("abcdefghijklmnoprstuvwxyz"));

will return this results

Letters who are missing: a b d f g h i j k l m p q v x z Counted: 16
Letters who are missing: q Counted: 1

Answer 2

Just another solution, using Array instead of Set

public static boolean isPanagram(String input) {
    if(input == null || input.trim().equals(""))
        return false;

    boolean[] chars = new boolean[26];
    int count = 0;
    for(char c : input.toLowerCase().toCharArray()) {
        if(Character.isAlphabetic(c)) {
            int val = (int) c - 'a';
            if(val >= 0 || val <= 26) {
                if(!chars[val]) {
                    count++;
                    chars[val] = true;
                }
            }
            // checking if all 26 alphabets are set, for every new alphabet found. 
            // so can return true immediately instead of processing entire string
            if(count==26)
                return true;
        }
    }

    return false;
}

Answer 3

Try this one out it's easy and simple to understand

import java.util.*;

public class Pnagram{
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        String Str=sc.nextLine();
        Set<Character> set = new HashSet<Character>();

        for(char c:Str.toUpperCase().toCharArray()){
            if(Character.isLetter(c))
            set.add(c);
        }
        if(set.size()==26)
            System.out.println("pnagram");
        else
            System.out.println("not pnagram");
    }
}