I have a very large amount of objects that look like this:

DataDict = {
id1: {"a": true, "b": true, "bc": true, "hgf": true},
id2: {"bcwe": true, "nKNNn": true, "mjj": true, "AAt": true},
id3: {"h": true, "a": true, "mjj": true, "ABwAU": true},
id4: {"wvzy": true, "zzba": true, "abc": true, "a": true},

or just (as a set of sets, note that ids may be excluded)

DataSet = {
{"a", "b", "bc", "hgf"},
{"bcwe", "nKNNn", "mjj", "AAt"},
{"h", "a", "mjj", "ABwAU"},
{"wvzy", "zzba", "abc", "a"},

Let M denote the total number of objects in DataDict (or subsets in DataSet), and let N denote the total number of unique names of properties found in DataDict (or the total number of unique strings found in DataSet).

The question is: given the set of strings {string1, string2, string3, ...}, how to get the Yes/No answer (assuming that there is a way to prepare the “index” for DataSet) to the “Does DataSet contain at least one subset that contains string1 AND string2 AND string3 AND ...?” question as fast as possible?

In another form, the question is: given the array A = [string1, string2, string3, ...], how to build the index (data structure) for DataDict that allows to quickly determine if it contains at least one object obj (I don’t care which object to choose, moreover, I don't want to return this object, all I want is that the function should return true if such an object exists, and false if not) such that DataDict.obj.string1 is true AND DataDict.obj.string2 is true AND DataDict.obj.string3 is true AND ...?

The only way that I see is to build the index like this:

"a": [1, 3, 4], 
"abc": [4],

and, for example, if A = ["a", "abc"], then I need to find the intersection of two arrays ([1, 3, 4] and [4]), but stop as soon as one common element is found and return true (if there are no common elements, return false). But these operations are very costly and time-consuming, which often leads to the unacceptable waiting time. Is there a way that guarantees a significantly better performance for all cases?

It would be extraordinarily nice if there exists a solution that (assuming that the index already exists) finds the answers depending on A.length time. The good solutions may depend on log M or log N or even close to log M multipled by log N, but I cannot imagine how to find such a solution...

  • $\begingroup$ Do you have access to a database? $\endgroup$
    – paparazzo
    Commented Feb 17, 2017 at 14:20
  • $\begingroup$ @Paparazzi: of course, I have full rights to read/modify the data. $\endgroup$ Commented Feb 18, 2017 at 7:34

2 Answers 2


It's possible to answer such queries in time $O(M^{1 - S/N})$, where $S$ is the size of the subset in the query, using a data structure from Ron Rivest:

Partial-Match Retrieval Algorithms. Ronald L. Rivest. SIAM Journal Computing, vol 5 no 1, March 1976.

Basically, you convert each set to a bitvector of length $N$, then use Rivest's partial-match queries. This is ever so slightly faster than the naive algorithm, whose running time will be about $O(M)$ (or a bit more).

Your idea using indices is another approach that might be faster in practice, especially if most sets are much smaller than $N$. Note that it's possible to optimize this a bit. Your index will have, for each string $x$, a list of indices of sets that contain $x$ and the length of that list. Now, given a set $A$, you can check for each $x \in A$ the length of the corresponding list in the index, and find the shortest such list, and enumerate all entries in that list to see if any of them are a superset of $A$. See also https://cstheory.stackexchange.com/q/19526/5038. I think the worst-case running time is no better than the naive algorithm, but in practice it might perform significantly better.


If this does not fit the format let me know and I will delete

I will call the individual elements words


  1. Sort each subset
  2. Sort the search
  3. For each subset
    • Compare count - if search count > subset count move on
    • Check first in search for >= first element in subset
    • Check last in search for <= last element in subset
    • For each word in search do a binary search on subset - quit when you don't get a hit
    • Can start the binary search on last match + 1

The complexity is a little complex as it will depend on the size of the individual subsets.
N is total number of words of all subsets
M is number of subsets
N/M is average subset size
S is number of search terms
I think this would be O(S * log(N/M - 1))
Worse case would be every subset has every word but the last
Since log is not linear taking the average N/M is not precise - should actually sum every log(dataset size)

It might be advantageous to start on the largest subsets as you have better chance of a match but you are also searching a larger subset.

A database does this very efficiently

  1. Put the words in a table (dictionary) with an integer as a key and call it WordIndex. Create a separate index on the Word
  2. Use an integer for the key on the input DataDict and call that DataSetKey
  3. Create a table for the Content of WordIndexKey, DataSetKey as a composite PK
    This table will have one row for each word in each DataSet

In search
1. First search WordIndex to make sure you have them all period
2. Save the WordIndex of the search in a set WordIndexs
3. Get a count of the words in the search S 4. The query is like this

select top (1) DataSetKey  
from Content 
where WordIndexKey in WordIndexs 
group by DataSetKey 
having count(*) = S
  • $\begingroup$ Thank you for the answer. Can you explain more on why exactly is this approach (using the database) efficient? What is the internal scheme of how the search process works? And what is the internal structure of all indexes? I don't understand... Can you draw a few tables (assuming that the database contains only four objects given in the question) step by step, so that I could understand all steps and how the algorithm creates and uses all indexes, and then returns the result? $\endgroup$ Commented Feb 18, 2017 at 7:47
  • $\begingroup$ There are two tables and I define them and indexes in my answer. $\endgroup$
    – paparazzo
    Commented Feb 18, 2017 at 9:54

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