Japanese Wordle — Why Duplicate-Character Handling Is Trickier Than It Looks

Wordle's color feedback rule has a subtlety most implementations get wrong: if the answer has one あ and your guess has two, only one of your あs should be highlighted. Doing this correctly requires a two-pass algorithm with a running frequency map of unmatched answer characters.

Wordle is simple to describe but surprisingly easy to implement wrong. The edge case that trips people up: when the guess has more copies of a character than the answer does, how do you decide which ones are "present" and which are "absent"?

🔗 Live demo: https://sen.ltd/portfolio/wordle-jp/
📦 GitHub: https://github.com/sen-ltd/wordle-jp

Features:

5-character hiragana word guessing in 6 attempts
164 curated words
Daily puzzle (same word for everyone per day)
Practice mode with random words
On-screen 50音 hiragana keyboard
Emoji share grid (🟩🟨⬜)
Stats tracking (win rate, streak, distribution)
Japanese / English UI
Zero dependencies, 40 tests

The two-pass duplicate-handling algorithm

Naive implementation:

// WRONG
for (let i = 0; i < 5; i++) {
  if (guess[i] === answer[i]) status[i] = 'correct';
  else if (answer.includes(guess[i])) status[i] = 'present';
  else status[i] = 'absent';
}

This fails when the guess has more copies than the answer. Example: answer あいうえお, guess ああああい. The naive algorithm would mark all four あs as 'present', but only one copy of あ exists in the answer.

The correct algorithm is two passes:

export function checkGuess(guess, answer) {
  const result = new Array(5);
  const answerRemaining = {};

  // Pass 1: mark exact matches, record remaining answer chars
  for (let i = 0; i < 5; i++) {
    if (guess[i] === answer[i]) {
      result[i] = { char: guess[i], status: 'correct' };
    } else {
      answerRemaining[answer[i]] = (answerRemaining[answer[i]] || 0) + 1;
    }
  }

  // Pass 2: mark 'present' only if there's an unmatched answer char
  for (let i = 0; i < 5; i++) {
    if (result[i]) continue;
    if (answerRemaining[guess[i]] > 0) {
      result[i] = { char: guess[i], status: 'present' };
      answerRemaining[guess[i]]--;
    } else {
      result[i] = { char: guess[i], status: 'absent' };
    }
  }

  return result;
}

Worked example with answer あいうあい, guess ああああい:

Pass 1: position 0 (あ=あ) → correct, position 3 (あ=あ) → correct. Positions 1, 2, 4 go to remaining: {い: 2, う: 1}.
Pass 2: position 1 (あ) → not in remaining → absent. Position 2 (あ) → absent. Position 4 (い) → remaining has い → present.

Daily puzzles with a date seed

The daily word uses a simple hash of the date string:

export function getDailyWord(date, wordList) {
  const dateStr = date.toISOString().slice(0, 10);
  let hash = 5381;
  for (let i = 0; i < dateStr.length; i++) {
    hash = ((hash << 5) + hash) + dateStr.charCodeAt(i);
  }
  return wordList[Math.abs(hash) % wordList.length];
}