Find word in dictionary of unknown size using only a method to get a word by index

A few days ago I had interview in some big company, name is not required :), and interviewer asked me to find solution to the next task:

Predefined: There is dictionary of words with unspecified size, we just know that all words in dictionary are sorted (for example by alphabet). Also we have just a one method

String getWord(int index) throws IndexOutOfBoundsException

Needs: Need to develop algorithm to find some input word in dictionary using java. For this we should implement method

public boolean isWordInTheDictionary(String word)

Limitations: We cannot change the internal structure of dictionary, we have no access to internal structure, we do not know counts of elements in dictionary.

Issues: I have developed modified-binary search, and will publish my variant(works variant) of algorithm, but are there another variants with logarithmic complexity? My variant has complexity O(logN).

My variant of implementation:

public class Dictionary {
    private static final int BIGGEST_TOP_MASK = 0xF00000;
    private static final int LESS_TOP_MASK = 0x0F0000;
    private static final int FULL_MASK = 0xFFFFFF;
    private String[] data;
    private static final int STEP = 100; // for real test step should be Integer.MAX_VALUE
    private int shiftIndex = -1;
    private static final int LESS_MASK = 0x0000FF;
    private static final int BIG_MASK = 0x00FF00;


    public Dictionary() {
        data = getData();
    }

    String getWord(int index) throws IndexOutOfBoundsException {
        return data[index];
    }

    public String[] getData() {
        return new String[]{"a", "aaaa", "asss", "az", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "test", "u", "v", "w", "x", "y", "z"};
    }


    public boolean isWordInTheDictionary(String word) {
        boolean isFound = false;
        int constantIndex = STEP; // predefined step
        int flag = 0;
        int i = 0;
        while (true) {
            i++;
            if (flag == FULL_MASK) {
                System.out.println("Word is not found ... Steps " + i);
                break;
            }
            try {
                String data = getWord(constantIndex);
                if (null != data) {
                    int compareResult = word.compareTo(data);
                    if (compareResult > 0) {
                        if ((flag & LESS_MASK) == LESS_MASK) {
                            constantIndex = prepareIndex(false, constantIndex);
                            if (shiftIndex == 1)
                                flag |= BIGGEST_TOP_MASK;
                        } else {
                            constantIndex = constantIndex * 2;
                        }
                        flag |= BIG_MASK;

                    } else if (compareResult < 0) {
                        if ((flag & BIG_MASK) == BIG_MASK) {
                            constantIndex = prepareIndex(true, constantIndex);
                            if (shiftIndex == 1)
                                flag |= LESS_TOP_MASK;
                        } else {
                            constantIndex = constantIndex / 2;
                        }
                        flag |= LESS_MASK;
                    } else {
// YES!!! We found word.
                        isFound = true;
                        System.out.println("Steps " + i);
                        break;
                    }
                }
            } catch (IndexOutOfBoundsException e) {
                if (flag > 0) {
                    constantIndex = prepareIndex(true, constantIndex);
                    flag |= LESS_MASK;
                } else constantIndex = constantIndex / 2;
            }
        }
        return isFound;
    }

    private int prepareIndex(boolean开发者_运维技巧 isBiggest, int constantIndex) {
        shiftIndex = (int) Math.ceil(getIndex(shiftIndex == -1 ? constantIndex : shiftIndex));
        if (isBiggest)
            constantIndex = constantIndex - shiftIndex;
        else
            constantIndex = constantIndex + shiftIndex;
        return constantIndex;
    }

    private double getIndex(double constantIndex) {
        if (constantIndex <= 1)
            return 1;
        return constantIndex / 2;
    }
}

---

It sounds like the part they really want you to think about is how to handle the fact that you don't know the size of the dictionary. I think they assume that you can give them a binary search. So the real question is how do you manipulate the range of the search as it progresses.

Once you have found a value in the dictionary that is greater than your search target (or out of bounds), the rest looks like standard binary search. The hard part is how do you optimally expand the range when the target value is greater than the dictionary value that you've looked up. It looks like you are expanding by a factor of 1.5. This could be really problematic with a huge dictionary and a small fixed initial step like you have (100). Think if there were 50 million words how many times your algorithm would have to expand the range upwards if you're searching for 'zebra'.

Here's an idea: use the ordered nature of the collection to your advantage by assuming the first letter of each word is evenly distributed amongst the letters of the alphabet (this will never be true, but without knowing more about the collection of words it's probably the best you can do). Then weight the amount of your range expansion by how far from the end you would expect the dictionary word to be.

So if you took your initial step of 100 and looked up the dictionary word at that index and it was 'aardvark', you would expand your range a lot more for the next step than if it was 'walrus.' Still O(log n) but probably much better for most collections of words.

---

Here is an alternative implementation that uses Collections.binarySearch. It fails if one of the words in the list starts with the Character '\uffff' (that is Unicode 0xffff and not a legal not a valid unicode character).

public static class ListProxy extends AbstractList<String> implements RandomAccess
{
    @Override public String get( int index )
    {
        try {
            return getWord( index );
        } catch( IndexOutOfBoundsException ex ) {
            return "\uffff";
        }
    }

    @Override public int size()
    {
        return Integer.MAX_VALUE;
    }
}

public static boolean isWordInTheDictionary( String word )
{
    return Collections.binarySearch( new ListProxy(), word ) >= 0;
}

Update: I modified it so that it implements RandomAccess since the binarySearch in Collections would otherwise use a iterator based search on such a large list which would be extremely slow. This should now however be decently fast since the binary search will need only 31 iterations even though the List pretends to be as large as possible.

Here is a slightly modified version that remembers the smallest failed index to converge its proclaimed size to the actual size of the dictionary en passant and thus avoids almost all exceptions in successive lookups. Although you would need to create a new ListProxy instance whenever the size of the dictionary could have changed.

public static class ListProxy extends AbstractList<String> implements RandomAccess
{
    private int size = Integer.MAX_VALUE;

    @Override public String get( int index )
    {
        try {
            if( index < size )
                return getWord( index );
        } catch( IndexOutOfBoundsException ex ) {
            size = index;
        }
        return "\uffff";
    }

    @Override public int size()
    {
        return size;
    }
}

private static ListProxy listProxy = new ListProxy();

public static boolean isWordInTheDictionary( String word )
{
    return Collections.binarySearch( listProxy , word ) >= 0;
}

---

You have the right idea, but I think your implementation is overly complicated. You want to do a binary search, but you don't know what the upper bound is. So instead of starting at the middle, you start at index 1 (assuming dictionary indexes start at 0).

If the word you're looking for is "less than" the current dictionary word, halve the distance between the current index and your "low" value. ("low" starts at 0, of course).

If the word you're looking for is "greater than" the word at the index you just examined, then either halve the distance between the current index and your "high" value ("high" starts at 2) or, if index and "high" are the same, double the index.

If doubling the index gives you an out of range exception, you halve the distance between the current value and the doubled value. So if going from 16 to 32 throws an exception, try 24. And, of course, keep track of the fact that 32 is more than the max.

So a search sequence might look like 1, 2, 4, 8, 16, 12, 14 - found!

It's the same concept as a binary search, but rather than starting with low = 0, high = n-1, you start with low = 0, high = 2, and double the high value when you need to. It's still O(log N), although the constant is going to be a bit larger than with a "normal" binary search.

---

You can incur a one-time cost of O(n), if you know that the dictionary will not change. You can add all the words in the dictionary to a hashtable, and then any subsequent calls to isWordInDictionary() will be O(1) (in theory).

---

Use the getWord() API to copy the entire contents of the dictionary into a more sensible data structure (e.g. hash table, trie, perhaps even augmented by a Bloom filter). ;-)

---

In a different language:

#!/usr/bin/perl

$t=0;
$cur=1;
$under=0;
$EOL=int(rand(1000000))+1;
$TARGET=int(rand(1000000))+1;
if ($TARGET>$EOL)
{
  $x=$EOL;
  $EOL=$TARGET;
  $TARGET=$x;
}
print "Looking for $TARGET with EOL $EOL\n";

sub testWord($)
{
  my($a)=@_;
  ++$t;
 return 0 if ($a eq $TARGET);
 return -2 if ($a > $EOL);
 return 1 if ($a > $TARGET);
 return -1;
}

while ($r = testWord($cur))
{
  print "Tested $cur, got $r\n";
  if ($r == 1) { $over=$cur; }
  if ($r == -1) { $under=$cur; }
  if ($r == -2) { $over = $cur; }
  if ($over)
  {
    $cur = int(($over-$under)/2)+$under;
    $cur++ if ($cur <= $under);
    $cur-- if ($cur >= $over);
  }
  else
  {
    $cur *= 2;
  }
}
print "Found $TARGET at $r in $t tests\n";

The main benefit of this one is it is a bit simpler to understand. I think it may be more efficient if your first guesses are below the target since I don't think you are taking advantage of the space you have already "searched", but that is just with a quick glance at your code. Since it is looking for numbers for simplicity, it doesn't have to deal with not finding the target, but that is an easy extension.

---

@Sergii Zagriichuk hope the interview went well. Good luck with that.

I think just as @alexcoco said Binary Search is the answer.

Other options I see are only available if you could extend the dictionary. You could make it slightly better. E.g. You could count the words on each letter, and keep their track this way you would effectively had to work only on a subset of words.

Or yea as guys are saying to entirely implement your own dictionary structure.

I know this doesn't answer you question properly. But I cannot see other possibilities.

BTW would be nice to see your algorithm.

EDIT: Expanding on my comment under answer of bshields...

@Sergii Zagriichuk even better it would be to remember the last index where we had null (no word), I think. Then at each run you could check if it is still true. If not then expand the range to a 'previous index' obtained by reversing the binary search behaviour, so we have null again. This way you would always adjust the size of the range of your search algorithm, thus adapting to the current state of the dictionary as needed. Plus the changes would have to be significant in order to cause your range adjustment so the adjustment wouldn't have any real negative impact on the algorithm. Also dictionaries tend to be static in nature so this should work :)

---

On one hand yes you are right with binary search implementation. But on the other hand in case dictionary is static and is not changed between lookups - we could suggest different algorithm. Here we have common problem - string sorting/search is different comparing to sorting/searching int array, so getWord(int i).compareTo(string) is O(min(length0, length1)).

Suppose we have request to find words w0, w1, ... wN, during lookup we could build up a tree with indicies (probably some suffix tree will good enough for this task). During next lookup request we have following set a1, a2, ... aM, so to decrease average time we could first decrease range by searching position in the tree. The problem with this implementation is concurrency and memory usage, so next step is implementing strategy to make search tree smaller.

PS: main aim was to check ideas and problems you suggest.

---

Well i think the info that dictionary is sorted can be utilized in a better way. Say you are looking for a word "Zebra" , whereas the first guess search resulted in "abcg". So we can use this info in chossing the second guess index . like in my case the resulted word is starting with a , whereas i am looking for something starting with z. So rather than making a static jump , i can make some calculated jump based on the current result and desired result. So in this way suppose if my next jump takes me to the word "yvu" , i now i am very near , so i will make a rather slow small jump than in the prev case.

---

Here is my solution.. uses O(logn) operations. First part of the code tries to find a estimate of the length and then the second part takes advantage of the fact that the dictionary is sorted and performs a binary search.

boolean isWordInTheDictionary(String word){
    if (word == null){
        return false;
    }
    // estimate the length of the dictionary array
    long len=2;
    String temp= getWord(len);

    while(true){
        len = len * 2;
        try{
          temp = getWord(len);
        }catch(IndexOutOfBoundsException e){
           // found upped bound break from loop
           break;
        }
    }

    // Do a modified binary search using the estimated length
    long beg = 0 ;
    long end = len;
    String tempWrd;
    while(true){
        System.out.println(String.format("beg: %s, end=%s, (beg+end)/2=%s ", beg,end,(beg+end)/2));
        if(end - beg <= 1){
            return false;
        }
        long idx = (beg+end)/2;
        tempWrd = getWord(idx);
        if(tempWrd == null){
            end=idx;
            continue;
        }
        if ( word.compareTo(tempWrd) > 0){
            beg = idx;
        }
        else if(word.compareTo(tempWrd) < 0){
            end= idx;
        }else{
            // found the word..
            System.out.println(String.format("getword at index: %s, =%s", idx,getWord(idx)));
            return true;
        }
    }
}

---

Assuming the dictionary is 0-based, I would decompose the search in two parts.

First, given that the index to parameter to getWord() is an integer, and assuming that the index must be a number between 0 and the maximum positive integer, perform a binary search over that range in order to find the maximum valid index (irrespective of the word values). This operation is O(log N), since is a simple binary search.

Once obtained the size of the dictionary, a second ordinary binary search (again of complexity O(log N)) will bring on the desired answer.

Since O(log N)+O(log N) is O(log N), this algorithm complies with your requirement.

---

I'm in a hiring proccess which asked me this same problem... My approach was a bit different, and considering the dictionary (webservice) I have, it's about 30% more efficient (for the words I've tested).

Here is the solution: https://github.com/gustavompo/wordfinder

I'll not post the whole solution here because it's decoupled through classes and methods, but the core algorithm is this:

public WordFindingResult FindWord(string word)
    {
        var callsCount = 0;
        var lowerLimit = new WordFindingLimit(0, null);
        var upperLimit = new WordFindingLimit(int.MaxValue, null);
        var wordToFind = new Word(word);
        var wordIndex = _initialIndex;

        while (callsCount <= _maximumCallsCount)
        {
            if (CouldNotFindWord(lowerLimit, upperLimit))
                return new WordFindingResult(callsCount, -1, string.Empty, WordFindingResult.ErrorCodes.NOT_FOUND);

            var wordFound = RetrieveWordAt(wordIndex);
            callsCount++;

            if (wordToFind.Equals(wordFound))
                return new WordFindingResult(callsCount, wordIndex, wordFound.OriginalWordString);

            else if (IsIndexTooHigh(wordToFind, wordFound))
            {
                upperLimit = new WordFindingLimit(wordIndex, wordFound);
                wordIndex = IndexConsideringTooHighPreviousResult(lowerLimit, wordIndex);
            }
            else
            {
                lowerLimit = new WordFindingLimit(wordIndex, wordFound);
                wordIndex = IndexConsideringTooLowPreviousResult(lowerLimit, upperLimit, wordToFind);
            }

        }
        return new WordFindingResult(callsCount, -1, string.Empty, WordFindingResult.ErrorCodes.CALLS_LIMIT_EXCEEDED);
    }

    private int IndexConsideringTooHighPreviousResult(WordFindingLimit maxLowerLimit, int current)
    {
        return BinarySearch(maxLowerLimit.Index, current);
    }

    private int IndexConsideringTooLowPreviousResult(WordFindingLimit maxLowerLimit, WordFindingLimit minUpperLimit, Word target)
    {
        if (AreLowerAndUpperLimitsDefined(maxLowerLimit, minUpperLimit))
            return BinarySearch(maxLowerLimit.Index, minUpperLimit.Index);

        var scoreByIndexPosition = maxLowerLimit.Index / maxLowerLimit.Word.Score;
        var indexOfTargetBasedInScore = (int)(target.Score * scoreByIndexPosition);
        return indexOfTargetBasedInScore;
    }