3
$\begingroup$

Here is the description of the data structure I am looking for:

Initial Original Data

Index | Frequency |
  1         3
  2         1
  3         7
  4         2
  5         6

Now it should be kept Sort-ed by their frequency inside the data structure like this,

Index | Frequency | Cumulative Frequency |
  2         1               1
  4         2               3
  1         3               6
  5         6               12
  3         7               19

Now, if I Insert a new entry item (index 6) with frequency of 2, the data structure should maintain it like this:

Index | Frequency | Cumulative Frequency |
  2         1               1
  4         2               3
->6         2               5
  1         3               8
  5         6               14
  3         7               21

Now, a frequency increase of 3 at Index 2 should Update the structure as following:

  Index | Frequency | Cumulative Frequency |
|   2           1               1
|   4           2               2
|   6           2               4
|   1           3               7
|-> 2        1+3 = 4            11
    5           6               17
    3           7               24

You can see above that the frequencies can repeat in many indices. (see index 4 and 6 above).

Now my Query to find the index that contains a cumulative frequency should be resulted in the following way

Query(cumulative frequency) | Result (Index at the data structure)
            3                       6
            1                       4
            20                      3
            7                       1

I have already explored Fenwick Tree (Binary Indexed Tree). It has a very efficient, O(log(n)), way to insert and update frequency at an index and to find the index for a specfic cumulative frequency. But I did not find a way to keep the Indexes sorted by their frequency.

Other structures like Treap, Red-Black tree can store the frequency in a sorted way but I could not find a way to search the index with the specific cumulative frequency using those data structures.

$\endgroup$
  • 1
    $\begingroup$ You are looking for dictionaries and/or self-adjusting/-organising data structures. If you want to sort for indices and for frequencies (when does that ever make sense?) you'll need, well, two indices. If you only ever query for frequencies, treat those as keys and indices as elements and you are served by standard dictionaries. Take your pick. $\endgroup$ – Raphael Oct 6 '14 at 16:06
  • $\begingroup$ I do not need it to be sorted by indices, they should be sorted only by frequencies. But I may need to update the frequencies in the indices over time. After the update of frequency, they should again be sorted. $\endgroup$ – Syed Arefinul Haque Oct 6 '14 at 16:27
  • $\begingroup$ Standard dictionaries implement this by delete+insert; it's unlikely to get any better than that. $\endgroup$ – Raphael Oct 6 '14 at 16:41
  • $\begingroup$ @Raphael, In case of same frequency for more than one indices, standard dictionary would not serve properly here, right? Also, I will query on the range of cumulative frequency instead of only frequency. $\endgroup$ – Syed Arefinul Haque Oct 6 '14 at 16:41
  • $\begingroup$ This is information that should go in the question; duplicate keys and range queries do indeed require (minor) changes. The basic properties of many data structures carry over, with notable exceptions; hashtables are bad for range queries, for instance. $\endgroup$ – Raphael Oct 6 '14 at 17:43
1
$\begingroup$

This can be easily handled using a standard balanced tree data structure, augmented with a Fenwick tree and a hashmap:

  • Store the items in a binary tree. Each leaf stores information about a single item. The leaves should be stored in increasing frequency, i.e., sorted by frequency, from least frequent to most frequent.

  • To help compute the cumulative frequency, augment the binary tree to store some additional information in each node. In each node, store the sum of the frequencies of all of the items underneath it (i.e., of all of the leaves under that node). These values can be easily updated as the tree changes. This is basically the idea behind Fenwick trees for computing prefix sums.

  • Also, add a hashmap that maps from the value (what you call "index") to the associated leaf where that index is found in the binary tree.

Now you can support all of the operations you mentioned efficiently. If you use a standard balanced binary tree data structure, the running time for all operations is $O(\log n)$:

  • Sort: All of the items are already stored in sorted order, sorted by increasing frequency, so this happens automatically.

  • Insert: You can insert a new item into the tree in $O(\log n)$ time.

  • Update: You can increase the frequency of a particular item by deleting it from the tree, increasing its frequency, and inserting it into the new location. In a balanced binary tree, this takes $O(\log n)$ time, since deletion can be done in $O(\log n)$ time, as can insertion.

  • Query: You can now find the first index $i$ where the sum of the frequencies of the first $i$ leaves is at least $t$, by using binary search. Note that you can efficiently compute prefix sums of the frequencies by using the extra values stored in each node of the tree. Therefore, binary search suffices, and can be done in $O(\log n)$ time.

There are many balanced binary tree data structures, including red-black trees, AVL trees, 2-3 trees, etc. They all have the property that standard operations are guaranteed to take at most $O(\log n)$ time. You can use any of them for your purposes.

$\endgroup$
0
$\begingroup$

There's 3 sub-problems here:

  1. Maintain a list of items sorted by frequency
  2. Locate an item at a given value of cumulative frequency
  3. Locate an item at a given index

Problem 3 can be solved by adding a hash table to the rest of the answer.

Problems 1 and 2 can be addressed at the same time by using a data structure like the Finger tree or rope (see my answer in another thread) - they are both basically strings with efficient concatenation and splitting, with the twist that you can split at a point where a given monotonic predicate ("measure") becomes true, rather than simply at an index. See Monoids and Finger trees, as well as the rope article, for a discussion of this process.

Then you can use a pair of measures (maximum, sum). Problem 1 is solved by locating the insertion point as the point where maximum-so-far becomes greater than the item being inserted. Problem 2 is the same but for sum-so-far.

$\endgroup$
  • $\begingroup$ Any chance you can give a self-contained explanation of those data structures in this answer? Or at least an explanation of what you mean by "you can split at a point where a given monotonic predicate becomes true, rather that simply at an index"? Surely if we have a monotonic predicate, we can use binary search to find the index where it becomes true, then split at that index, so why is "split at an index" not sufficient? And, actually, in this problem, why do we need to split at all, rather than merely find the index itself? I suspect I might have missed a critical and useful insight... $\endgroup$ – D.W. Oct 6 '14 at 20:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.