How to check if a string contains all the letters of the alphabet?

All these solutions seem to do a lot of work for a relatively simple check, especially given Java 8's stream API:

/* Your lowercase string */.chars()
    .filter(i -> i >= 'a' && i <= 'z')
    .distinct().count() == 26;

Edit: For speed

If you want to end the string iteration as soon as the entire alphabet is found while still using streams, then you can keep track with a HashSet internally:

Set<Integer> chars = new HashSet<>();
String s = /* Your lowercase string */;
s.length() > 25 && s.chars()
    .filter(i -> i >= 'a' && i <= 'z') //only alphabet
    .filter(chars::add)                //add to our tracking set if we reach this point
    .filter(i -> chars.size() == 26)   //filter the 26th letter found
    .findAny().isPresent();            //if the 26th is found, return

This way, the stream will cease as soon as the Set is filled with the 26 required characters.

There are some (even still) more efficient solutions in terms of performance below, but as a personal note I will say to not bog yourself in premature optimization too much, where you could have readability and less effort in writing the actual code.

Adding to @Leon answer, creating a List and removing from it seems quite unnecessary. You could simply loop over 'a' - 'z' and do a check with each char. Additionally you are looping over the whole String to find out, if each letter is present. But the better version would be to loop over each letter itself. This can potentionally safe you a few iterations.

In the end a simple example could look like this:

// This is the string- I've just put a random example
String str = "a dog is running crazily on the ground who doesn't care about the world";
str = str.toLowerCase();

boolean success = true;
for(char c = 'a';c <= 'z'; ++c) {
    if(!str.contains(String.valueOf(c))) {
        success = false;
        break;
    }
}

if (success)
    System.out.println("String contains all alphabets");
else
    System.out.println("String DOESN'T contains all alphabets");

List.remove removes by index. Since a char can be cast to an int you are effectively removing index values that do not exist, ie char 'a' is equal to int 97. As you can see your list does not have 97 entries.

You can do alphabet.remove(alphabets.indexOf(inp));

As pointed out by @Scary Wombat(https://stackoverflow.com/a/39263836/1226744) and @Kevin Esche (https://stackoverflow.com/a/39263917/1226744), there are better alternative to your algorithm

O(n) solution

static Set<Integer> alphabet = new HashSet<>(26);

public static void main(String[] args) {

    int cnt = 0;

    String str = "a dog is running crazily on the ground who doesn't care about the world";

    for (char c : str.toCharArray()) {
        int n = c - 'a';
        if (n >= 0 && n < 26) {
            if (alphabet.add(n)) {
                cnt += 1;
                if (cnt == 26) {
                    System.out.println("found all letters");
                    break;
                }
            }
        }
    }
}