Building a Word Unscrambler with JavaScript (part 2)

bridget_amana

Bridget Amana

Posted on September 30, 2024

Building a Word Unscrambler with JavaScript (part 2)

In my last article, I was stuck on a little bug. I thought I had fixed it, but well... let's just say I had an "oops" moment. All that was missing was the logical NOT operator “!” in the if condition like this:

if (!uniquePermutations.includes(newPermutation)) { 
    uniquePermutations.push(newPermutation);
}
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I added it and now, we move on to the next problem.

The Length Issue

After generating all the permutations, I noticed something else. My word unscrambler was only giving me permutations that were exactly the length of the input word. So if I typed in "wood," I got permutations like "doow" and "odwo" but never shorter words like "wo" or "ow." That's not what I wanted.

I needed to find a way to generate all possible combinations, not just those of the same length as the input word.

So, here’s what I did:

function generatePermutations(inputString) {
    const uniquePermutations = [];

    function generateUniquePermutations(arr, currentIndex, length) {
        if (currentIndex === arr.length - 1) {
            const newPermutation = arr.slice(0, length).join("");
            if (!uniquePermutations.includes(newPermutation)) {
                uniquePermutations.push(newPermutation);
            }
        } else {
            for (let i = currentIndex; i < arr.length; i++) {
                [arr[currentIndex], arr[i]] = [arr[i], arr[currentIndex]];
                generateUniquePermutations([...arr], currentIndex + 1, length);
            }
        }
    }

    for (let length = 2; length <= inputString.length; length++) {
        generateUniquePermutations(inputString.split(""), 0, length);
    }

    return uniquePermutations;
}

console.log(generatePermutations('wood'));
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Breaking It Down:

I modified the generateUniquePermutations function to accept a length argument, which slices the array and ensures it generates permutations of different lengths. I also added a loop that starts at 2 and goes up to the full length of the input word.

Categorizing by Length

Once I had all my word permutations, I thought it would be much nicer (and way easier on the eyes) if I could categorize them by word length. So here’s how I broke them down:

function categorizeByLength(combinations) {
    const categorized = {};
    combinations.forEach(word => {
        const length = word.length;
        if (!categorized[length]) {
            categorized[length] = [];
        }
        categorized[length].push(word);
    });
    return categorized;
}
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This function takes the list of permutations, groups them by their length, and returns an object where each key is a word length (2, 3, 4, etc.), and each value is an array of words of that length.

Displaying the Results

Next up, I wanted these categorized results to actually show up nicely in the browser. I wrote a function that dynamically creates sections in the HTML for each word length, then lists all the words under those sections.

function displayResults(categorizedWords) {
    const resultsContainer = document.getElementById('resultsContainer');
    resultsContainer.innerHTML = '';

    for (const [length, words] of Object.entries(categorizedWords)) {
        const section = document.createElement('div');
        section.className = 'result-section';

        const header = document.createElement('h2');
        header.textContent = `${length}-Letter Words`;
        section.appendChild(header);

        const wordList = document.createElement('ul');
        words.forEach(word => {
            const listItem = document.createElement('li');
            listItem.textContent = word;
            wordList.appendChild(listItem);
        });
        section.appendChild(wordList);
        resultsContainer.appendChild(section);
    }
}
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This is what we have

Image description

Next up? I need to work on filtering out the invalid words. Right now, I’m getting all sorts of gibberish, and I only want real, meaningful words. Still stick around 🙃

💖 💪 🙅 🚩
bridget_amana
Bridget Amana

Posted on September 30, 2024

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