CODEWITHSUNDEEP
javascriptwordle-game
rohan1122
rohan1122

Reputation: 130

Duplicate verification in Wordle clone

I'm working on a Wordle clone, and I'm having an issue with comparing the user input with the correct word.

For anybody unfamiliar with the verification process of Wordle; the user inputs a 5 letter word to be compared to the actual answer. If the letter is in the same position as the correct word, it returns a green value. If the letter position is incorrect, but the letter is in the word, return a yellow value. If the letter is not in the word, return a grey value.

My problem arises when I try to verify a word with a double letter, and the second letter is the correct word. Because I'm iterating through the word index-by-index, the letter is being categorised as yellow incorrectly.

The correct answer for this is CODES.

The first E gets verified as wrong position, correct letter. The second E is in the correct position, therefore the first E should not return yellow, but grey instead.

This is the code I've written below to verify the letter.

let x = input.value().toUpperCase();
let y = correctWord.toUpperCase();
for (let i = 0; i < 5; i++) {
    if (y.includes(x[i])) {
        if (y[i] == x[i]) {
            return green;
        } else {
            return yellow;
        }
    } else {
        return grey;
    }
}

Is there an efficient way to fix this?

Try the demo below: https://wordleclone.butterchoke.repl.co/

The correct word is printed onto the console. You can change the word with

correctWord = "WORD OF YOUR CHOICE";

in the console.

Upvotes: 0

Views: 502

Answers (1)

Nina Scholz
Nina Scholz

Reputation: 386736

You could take two loops and replace found letters of the word to omit finding the same, but already consumed letter.

function check(input, wordle) {
    const
        letters = Array.from(input.toUpperCase()),
        word =  Array.from(wordle.toUpperCase()),
        result = Array(5).fill('');

    for (let i = 0; i < 5; i++) {
        if (letters[i] === word[i]) {
            result[i] = 'green',
            word[i] = '';
            letters[i] = '';
        }
    }    
    for (let i = 0; i < 5; i++) {
        if (!letters[i]) continue;
        const index = word.indexOf(letters[i]);
        if (index !== -1) {
            result[i] = 'yellow',
            word[index] = '';
        }
    }    
    return result;
}

console.log(check('eeege', 'green'));

Upvotes: 0

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