Autocomplete using a trie

Question

I am working on an autocompletion script and was thinking about using a trie. My problem is I want everything that matches to be returned. So for example I type in the letter r I want all entries starting with r to be returned. Then all entries starting with re etc. Is this feasible with a trie and how would it work. Also, if there is a better way I am open to suggestions. The reason I ask is it seems like it would be complicated and a whole lot of processing to return all of the nodes off of say the r branch.

And yes I may be reinventing the wheel, but I would like to learn how it works.

Tarango

Answer 1

You can absolutely do it using a trie. Here is some code I threw together that can point you in the right direction:

var tokenTree = function (tokenArray) {
  var createLetterObject = function (l) {
    var pChildren = [];

    var getMatchingWords = function (characterArr, availableWords, children) {
        if (characterArr.length === 0) {
            for (var child in children) {
                if ({}.hasOwnProperty.call(children, child)) {
                    var currentChild = children[child];

                    var words = currentChild.getWords(characterArr);

                    for (var pos in words) {
                        if ({}.hasOwnProperty.call(words, pos)) {
                            availableWords.push(words[pos]);
                        }
                    }

                    if (currentChild.word) {
                        availableWords.push(currentChild.word);
                    }
                }
            }
        } else {
            var currentCharacter = characterArr.pop();
            getMatchingWords(characterArr, availableWords, children[currentCharacter].children);
        }
    };

    function doGetWords(wordPart) {
        var len = wordPart.length;
        var ar = [];
        var wordList = [];

        for (var ii = len - 1; ii >= 0; ii --) {
            ar.push(wordPart[ii].toUpperCase());
        }

        getMatchingWords(ar, wordList, pChildren);

        return wordList;
    }

    return {
        letter: l,
        children: pChildren,
        parent: null,
        word: null,
        getWords: doGetWords
    };
};

var startingPoint = createLetterObject();

function parseWord(wordCharacterArray, parent, fullWord) {
    if (wordCharacterArray.length === 0) {
        parent.word = fullWord;
        return;
    }

    var currentCharacter = wordCharacterArray.pop().toUpperCase();

    if (!parent.children[currentCharacter]) {
        parent.children[currentCharacter] = createLetterObject(currentCharacter);
    }

    parseWord(wordCharacterArray, parent.children[currentCharacter], fullWord);
}

for (var counter in tokenArray) {
    if ({}.hasOwnProperty.call(tokenArray, counter)) {
        var word = tokenArray[counter];

        if (!word) {
            continue;
        }

        var ar = [];

        var wordLength = word.length;

        for (var ii = wordLength - 1; ii >= 0; ii--) {
            ar.push(word[ii]);
        }

        parseWord(ar, startingPoint, word);
    }
}

  return startingPoint;
};

Usage

var tokens = ["Token", "words", "whohaa", "mommy", "test", "wicked"];
var tree = tokenTree(tokens);
var currentTokenSet = 'w'; 
var list = tree.getWords(currentTokenSet);

// it will return words,whohaa,wicked.
console.log(list) 

I'm not saying this is anywhere near the best or most efficient way, but it should at least get you pointed in the right direction.

Here is a jsfiddle showing it working: https://jsfiddle.net/es6xp8h9/

Ciapas

Why to you use CharacterArr and not CharacterArray, which you declared? :) – Stonefish

Answer 2

Regarding the time to discover items at a root note, if you're doing this for an autocomplete, it's likely you won't be returning too many results per 'match'. If you wanted to trade off space for speed, you could store references to the 'top n' items at each of the nodes. This, of course, would require more time on update

Oeo