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For example, for <math>f = \sum</math> and <math>A[1,n]</math> an array of numbers, the range query <math>\sum_{i,j} A</math> computes <math>\sum A[i,j] = (a_i+\ldots + a_j)</math>, for any <math>1 \leq i \leq j \leq n</math>. These queries may be answered in [[constant time]] and using <math>O(n)</math> extra space by calculating the sums of the first {{mvar|i}} elements of {{mvar|A}} and storing them into an auxiliary array {{mvar|B}}, such that <math>B[i]</math> contains the sum of the first {{mvar|i}} elements of {{mvar|A}} for every <math>0\leq i\leq n</math>. Therefore, any query might be answered by doing <math>\sum A[i,j] = B[j] - B[i-1]</math>.
This strategy may be extended for every [[Group (mathematics)|group]] operator {{mvar|f}} where the notion of <math>f^{-1}</math> is well defined and easily computable.<ref name="morin">{{cite journal|first1=Danny|last1=Krizanc|first2=Pat|last2=Morin|author2-link= Pat Morin |first3=Michiel H. M.|last3=Smid|title=Range Mode and Range Median Queries on Lists and Trees|journal=ISAAC|year=2003|pages=517–526|url=http://cg.scs.carleton.ca/~morin/publications/|arxiv=cs/0307034|bibcode=2003cs........7034K }}</ref> Finally, this solution can be extended to two-dimensional arrays with a similar preprocessing.<ref name=menhe>{{cite journal|last1=Meng|first1=He|first2=J. Ian|last2=Munro|first3=Patrick K.|last3=Nicholson|title=Dynamic Range Selection in Linear Space|journal=ISAAC|year=2011|pages=160–169|arxiv=1106.5076 }}</ref>
==Examples==
There have been studied two variants of this problem, the [[offline algorithm|offline]] version, where all the ''k'' queries of interest are given in a batch, and a version where all the preprocessing is done up front. The offline version can be solved with <math>O(n\log k + k \log n)</math> time and <math>O(n\log k)</math> space.
The following pseudocode of the [[Quickselect|quickselect algorithm]] shows how to find the element of rank {{mvar|r}} in <math>A[i,j]</math> an unsorted array of distinct elements, to find the range medians we set <math>r=\frac{j-i}{2}</math>.<ref name="ethpaper">{{cite journal|last1=Beat|first1=Gfeller|first2=Peter|last2=Sanders|author2-link=Peter Sanders (computer scientist)|title=Towards Optimal Range Medians|journal=Icalp (1)|year=2009|pages=475–486|arxiv=0901.1761 }}</ref>
rangeMedian(A, i, j, r) {
===Majority===
Finding frequent elements in a given set of items is one of the most important tasks in data mining. Finding frequent elements might be a difficult task to achieve when most items have similar frequencies. Therefore, it might be more beneficial if some threshold of significance was used for detecting such items. One of the most famous algorithms for finding the majority of an array was proposed by Boyer and Moore <ref>{{Citation|lastlast1=Boyer|firstfirst1=Robert S.|title=MJRTY—A Fast Majority Vote Algorithm|date=1991|url=http://dx.doi.org/10.1007/978-94-011-3488-0_5|work=Automated Reasoning Series|pages=105–117|place=Dordrecht|publisher=Springer Netherlands|access-date=2021-12-18|last2=Moore|first2=J. Strother|series=Automated Reasoning Series |volume=1 |doi=10.1007/978-94-011-3488-0_5 |isbn=978-94-010-5542-0 }}</ref> which is also known as the [[Boyer–Moore majority vote algorithm]]. Boyer and Moore proposed an algorithm to find the majority element of a string (if it has one) in <math>O(n)</math> time and using <math>O(1)</math> space. In the context of Boyer and Moore’s work and generally speaking, a majority element in a set of items (for example string or an array) is one whose number of instances is more than half of the size of that set. Few years later, Misra and Gries <ref>{{Cite journal|lastlast1=Misra|firstfirst1=J.|last2=Gries|first2=David|date=November 1982|title=Finding repeated elements|url=http://dx.doi.org/10.1016/0167-6423(82)90012-0|journal=Science of Computer Programming|volume=2|issue=2|pages=143–152|doi=10.1016/0167-6423(82)90012-0|issn=0167-6423}}</ref> proposed a more general version of Boyer and Moore's algorithm using <math>O \left ( n \log \left ( \frac{1}{\tau} \right ) \right )</math> comparisons to find all items in an array whose relative frequencies are greater than some threshold <math>0<\tau<1</math>. A range <math>\tau</math>-majority query is one that, given a subrange of a data structure (for example an array) of size <math>|R|</math>, returns the set of all distinct items that appear more than (or in some publications equal to) <math>\tau |R|</math> times in that given range. In different structures that support range <math>\tau</math>-majority queries, <math>\tau </math> can be either static (specified during preprocessing) or dynamic (specified at query time). Many of such approaches are based on the fact that, regardless of the size of the range, for a given <math>\tau</math> there could be at most <math>O(1/\tau)</math> distinct ''candidates'' with relative frequencies at least <math>\tau</math>. By verifying each of these candidates in constant time, <math>O(1/\tau)</math> query time is achieved. A range <math>\tau</math>-majority query is decomposable <ref name=":1">{{Cite book|last=Verfasser|first=Karpiński, Marek 1948-|url=http://worldcat.org/oclc/277046650|title=Searching for frequent colors in rectangles|oclc=277046650}}</ref> in the sense that a <math>\tau</math>-majority in a range <math>R</math> with partitions <math>R_1</math> and <math>R_2</math> must be a <math>\tau</math>-majority in either <math>R_1</math>or <math>R_2</math>. Due to this decomposability, some data structures answer <math>\tau</math>-majority queries on one-dimensional arrays by finding the [[Lowest common ancestor]] (LCA) of the endpoints of the query range in a [[Range tree]] and validating two sets of candidates (of size <math>O(1/\tau)</math>) from each endpoint to the lowest common ancestor in constant time resulting in <math>O(1/\tau)</math> query time.
==== Two-dimensional arrays ====
Gagie et al.<ref>{{Citation|lastlast1=Gagie|firstfirst1=Travis|title=Finding Frequent Elements in Compressed 2D Arrays and Strings|date=2011|url=http://dx.doi.org/10.1007/978-3-642-24583-1_29|work=String Processing and Information Retrieval|pages=295–300|place=Berlin, Heidelberg|publisher=Springer Berlin Heidelberg|isbn=978-3-642-24582-4|access-date=2021-12-18|last2=He|first2=Meng|last3=Munro|first3=J. Ian|last4=Nicholson|first4=Patrick K.|doi=10.1007/978-3-642-24583-1_29 }}</ref> proposed a data structure that supports range <math>\tau</math>-majority queries on an <math>m\times n</math> array <math>A</math>. For each query <math>\operatorname{Q}=(\operatorname{R}, \tau)</math> in this data structure a threshold <math>0<\tau<1</math> and a rectangular range <math>\operatorname{R}</math> are specified, and the set of all elements that have relative frequencies (inside that rectangular range) greater than or equal to <math>\tau</math> are returned as the output. This data structure supports dynamic thresholds (specified at query time) and a preprocessing threshold <math>\alpha</math> based on which it is constructed. During the preprocessing, a set of ''vertical'' and ''horizontal'' intervals are built on the <math>m \times n</math> array. Together, a vertical and a horizontal interval form a ''block.'' Each block is part of a ''superblock'' nine times bigger than itself (three times the size of the block's horizontal interval and three times the size of its vertical one). For each block a set of candidates (with <math>\frac{9}{\alpha}</math> elements at most) is stored which consists of elements that have relative frequencies at least <math>\frac{\alpha}{9}</math> (the preprocessing threshold as mentioned above) in its respective superblock. These elements are stored in non-increasing order according to their frequencies and it is easy to see that, any element that has a relative frequency at least <math>\alpha</math> in a block must appear its set of candidates. Each <math>\tau</math>-majority query is first answered by finding the ''query block,'' or the biggest block that is contained in the provided query rectangle in <math>O(1)</math> time. For the obtained query block, the first <math>\frac{9}{\tau}</math> candidates are returned (without being verified) in <math>O(1/\tau)</math> time, so this process might return some false positives. Many other data structures (as discussed below) have proposed methods for verifying each candidate in constant time and thus maintaining the <math>O(1/\tau)</math> query time while returning no false positives. The cases in which the query block is smaller than <math>1/\alpha</math> are handled by storing <math>\log \left ( \frac{1}{\alpha} \right )</math> different instances of this data structure of the following form:
<math>\beta=2^{-i}, \;\; i\in \left \{ 1,\dots,\log \left (\frac{1}{\alpha} \right ) \right \}
==== One-dimensional arrays ====
Chan et al.<ref name=":0">{{Citation|lastlast1=Chan|firstfirst1=Timothy M.|title=Linear-Space Data Structures for Range Minority Query in Arrays|date=2012|url=http://dx.doi.org/10.1007/978-3-642-31155-0_26|work=Algorithm Theory – SWAT 2012|pages=295–306|place=Berlin, Heidelberg|publisher=Springer Berlin Heidelberg|isbn=978-3-642-31154-3|access-date=2021-12-20|last2=Durocher|first2=Stephane|last3=Skala|first3=Matthew|last4=Wilkinson|first4=Bryan T.|doi=10.1007/978-3-642-31155-0_26 }}</ref> proposed a data structure that given a one-dimensional array<math>A</math>, a subrange <math>R</math> of <math>A</math> (specified at query time) and a threshold <math>\tau</math> (specified at query time), is able to return the list of all <math>\tau</math>-majorities in <math>O(1/\tau)</math> time requiring <math>O(n \log n)</math> words of space. To answer such queries, Chan et al.<ref name=":0" /> begin by noting that there exists a data structure capable of returning the ''top-k'' most frequent items in a range in <math>O(k)</math> time requiring <math>O(n)</math> words of space. For a one-dimensional array <math>A[0,..,n-1]</math>, let a one-sided top-k range query to be of form <math>A[0..i] \text { for } 0 \leq i \leq n-1</math>. For a maximal range of ranges <math>A[0..i] \text { through } A[0..j]</math> in which the frequency of a distinct element <math>e</math> in <math>A</math> remains unchanged (and equal to <math>f</math>), a horizontal line segment is constructed. The <math>x</math>-interval of this line segment corresponds to <math>[i,j]</math> and it has a <math>y</math>-value equal to <math>f</math>. Since adding each element to <math>A</math> changes the frequency of exactly one distinct element, the aforementioned process creates <math>O(n)</math> line segments. Moreover, for a vertical line <math>x=i</math> all horizonal line segments intersecting it are sorted according to their frequencies. Note that, each horizontal line segment with <math>x</math>-interval <math>[\ell,r]</math> corresponds to exactly one distinct element <math>e</math> in <math>A</math>, such that <math>A[\ell]=e</math>. A top-k query can then be answered by shooting a vertical ray <math>x=i</math> and reporting the first <math>k</math> horizontal line segments that intersect it (remember from above that these line line segments are already sorted according to their frequencies) in <math>O(k)</math> time.
Chan et al.<ref name=":0" /> first construct a [[range tree]] in which each branching node stores one copy of the data structure described above for one-sided range top-k queries and each leaf represents an element from <math>A</math>. The top-k data structure at each node is constructed based on the values existing in the subtrees of that node and is meant to answer one-sided range top-k queries. Please note that for a one-dimensional array <math>A</math>, a range tree can be constructed by dividing <math>A</math> into two halves and recursing on both halves; therefore, each node of the resulting range tree represents a range. It can also be seen that this range tree requires <math>O(n \log n)</math> words of space, because there are <math>O(\log n)</math> levels and each level <math>\ell</math> has <math>2^{\ell}</math> nodes. Moreover, since at each level <math>\ell</math> of a range tree all nodes have a total of <math>n</math> elements of <math>A</math> at their subtrees and since there are <math>O(\log n)</math> levels, the space complexity of this range tree is <math>O(n \log n)</math>.
Using this structure, a range <math>\tau</math>-majority query <math>A[i..j]</math> on <math>A[0..n-1]</math> with <math>0\leq i\leq j \leq n</math> is answered as follows. First, the [[lowest common ancestor]] (LCA) of leaf nodes <math>i</math> and <math>j</math> is found in constant time. Note that there exists a data structure requiring <math>O(n)</math> bits of space that is capable of answering the LCA queries in <math>O(1)</math> time.<ref>{{Cite journal|lastlast1=Sadakane|firstfirst1=Kunihiko|last2=Navarro|first2=Gonzalo|date=2010-01-17|title=Fully-Functional Succinct Trees|url=http://dx.doi.org/10.1137/1.9781611973075.13|journal=Proceedings of the Twenty-First Annual ACM-SIAM Symposium on Discrete Algorithms|pages=134–149 |___location=Philadelphia, PA|publisher=Society for Industrial and Applied Mathematics|doi=10.1137/1.9781611973075.13|isbn=978-0-89871-701-3 |s2cid=3189222 }}</ref> Let <math>z</math> denote the LCA of <math>i </math> and <math>j</math>, using <math>z</math> and according to the decomposability of range <math>\tau</math>-majority queries (as described above and in <ref name=":1" />), the two-sided range query <math>A[i..j]</math> can be converted into two one-sided range top-k queries (from <math>z</math> to <math>i</math> and <math>j</math>). These two one-sided range top-k queries return the top-(<math>1/\tau</math>) most frequent elements in each of their respective ranges in <math>O(1/\tau)</math> time. These frequent elements make up the set of ''candidates'' for <math>\tau</math>-majorities in <math>A[i..j]</math> in which there are <math>O(1/\tau)</math> candidates some of which might be false positives. Each candidate is then assessed in constant time using a linear-space data structure (as described in Lemma 3 in <ref>{{Cite journal|lastlast1=Chan|firstfirst1=Timothy M.|last2=Durocher|first2=Stephane|last3=Larsen|first3=Kasper Green|last4=Morrison|first4=Jason|last5=Wilkinson|first5=Bryan T.|date=2013-03-08|title=Linear-Space Data Structures for Range Mode Query in Arrays|url=http://dx.doi.org/10.1007/s00224-013-9455-2|journal=Theory of Computing Systems|volume=55|issue=4|pages=719–741|doi=10.1007/s00224-013-9455-2|s2cid=253747004 |issn=1432-4350}}</ref>) that is able to determine in <math>O(1)</math> time whether or not a given subrange of an array <math>A</math> contains at least <math>q</math> instances of a particular element <math>e</math>.
==== Tree paths ====
Gagie et al.<ref name=":2">{{Cite journal|lastlast1=Gagie|firstfirst1=Travis|last2=He|first2=Meng|last3=Navarro|first3=Gonzalo|last4=Ochoa|first4=Carlos|date=September 2020|title=Tree path majority data structures|url=http://dx.doi.org/10.1016/j.tcs.2020.05.039|journal=Theoretical Computer Science|volume=833|pages=107–119|doi=10.1016/j.tcs.2020.05.039|issn=0304-3975}}</ref> proposed a data structure which supports queries such that, given two nodes <math>u</math> and <math>v</math> in a tree, are able to report the list of elements that have a greater relative frequency than <math>\tau</math> on the path from <math>u</math> to <math>v</math>. More formally, let <math>T</math> be a labelled tree in which each node has a label from an alphabet of size <math>\sigma</math>. Let <math>label(u)\in [1,\dots,\sigma]</math> denote the label of node <math>u</math> in <math>T</math>. Let <math>P_{uv}</math> denote the unique path from <math>u</math> to <math>v</math> in <math>T</math> in which middle nodes are listed in the order they are visited. Given <math>T</math>, and a fixed (specified during preprocessing) threshold <math>0<\tau<1</math>, a query <math>Q(u,v)</math> must return the set of all labels that appear more than <math>\tau|P_{uv}|</math> times in <math>P_{uv}</math>.
To construct this data structure, first <math>{O}(\tau n)</math> nodes are ''marked''. This can be done by marking any node that has distance at least <math>\lceil 1 / \tau\rceil</math> from the bottom of the three (height) and whose depth is divisible by <math>\lceil 1 / \tau\rceil</math>. After doing this, it can be observed that the distance between each node and its nearest marked ancestor is less than <math>2\lceil 1 / \tau\rceil</math>. For a marked node <math>x</math>, <math>\log(depth(x))</math> different sequences (paths towards the root) <math>P_i(x)</math> are stored,
for <math>0\leq i \leq \log(depth(x))</math> where <math>\operatorname{par}(x)</math> returns the label of the direct parent of node <math>x</math>. Put another way, for each marked node, the set of all paths with a power of two length (plus one for the node itself) towards the root is stored. Moreover, for each <math>P_i(x)</math>, the set of all majority ''candidates'' <math>C_i(x)</math> are stored. More specifically, <math>C_i(x)</math> contains the set of all <math>(\tau/2)</math>-majorities in <math>P_i(x)</math> or labels that appear more than <math>(\tau/2).(2^i+1)</math> times in <math>P_i(x)</math>. It is easy to see that the set of candidates <math>C_i(x)</math> can have at most <math>2/\tau</math> distinct labels for each <math>i</math>. Gagie et al.<ref name=":2"/> then note that the set of all <math>\tau</math>-majorities in the path from any marked node <math>x</math> to one of its ancestors <math>z</math> is included in some <math>C_i(x)</math> (Lemma 2 in <ref name=":2"/>) since the length of <math>P_i(x)</math> is equal to <math>(2^i+1)</math> thus there exists a <math>P_i(x)</math> for <math>0\leq i \leq \log(depth(x))</math> whose length is between <math>d_{xz} \text{ and } 2 d_{xz}</math> where <math>d_{xz}</math> is the distance between x and z. The existence of such <math>P_i(x)</math> implies that a <math>\tau</math>-majority in the path from <math>x</math> to <math>z</math> must be a <math>(\tau/2)</math>-majority in <math>P_i(x)</math>, and thus must appear in <math>C_i(x)</math>. It is easy to see that this data structure require <math>O(n \log n)</math> words of space, because as mentioned above in the construction phase <math>O(\tau n)</math> nodes are marked and for each marked node some candidate sets are stored. By definition, for each marked node <math>O(\log n)</math> of such sets are stores, each of which contains <math>O(1/\tau)</math> candidates. Therefore, this data structure requires <math>O(\log n \times (1/\tau) \times \tau n)=O(n \log n)</math> words of space. Please note that each node <math>x</math> also stores <math>count(x)</math> which is equal to the number of instances of <math>label(x)</math> on the path from <math>x</math> to the root of <math>T</math>, this does not increase the space complexity since it only adds a constant number of words per node.
Each query between two nodes <math>u</math> and <math>v</math> can be answered by using the decomposability property (as explained above) of range <math>\tau</math>-majority queries and by breaking the query path between <math>u</math> and <math>v</math> into four subpaths. Let <math>z</math> be the lowest common ancestor of <math>u</math> and <math>v</math>, with <math>x</math> and <math>y</math> being the nearest marked ancestors of <math>u</math> and <math>v</math> respectively. The path from <math>u</math> to <math>v</math> is decomposed into the paths from <math>u</math> and <math>v</math> to <math>x</math> and <math>y</math> respectively (the size of these paths are smaller than <math>2\lceil 1 / \tau\rceil</math> by definition, all of which are considered as candidates), and the paths from <math>x</math> and <math>y</math> to <math>z</math> (by finding the suitable <math>C_i(x)</math> as explained above and considering all of its labels as candidates). Please note that, boundary nodes have to be handled accordingly so that all of these subpaths are disjoint and from all of them a set of <math>O(1/\tau)</math> candidates is derived. Each of these candidates is then verified using a combination of the <math>labelanc (x, \ell)</math> query which returns the lowest ancestor of node <math>x</math> that has label <math>\ell</math> and the <math>count(x)</math> fields of each node. On a <math>w</math>-bit RAM and an alphabet of size <math>\sigma</math>, the <math>labelanc (x, \ell)</math> query can be answered in <math>O\left(\log \log _{w} \sigma\right) </math> time whilst having linear space requirements.<ref>{{Cite journal|lastlast1=He|firstfirst1=Meng|last2=Munro|first2=J. Ian|last3=Zhou|first3=Gelin|date=2014-07-08|title=A Framework for Succinct Labeled Ordinal Trees over Large Alphabets|url=http://dx.doi.org/10.1007/s00453-014-9894-4|journal=Algorithmica|volume=70|issue=4|pages=696–717|doi=10.1007/s00453-014-9894-4|s2cid=253977813 |issn=0178-4617}}</ref> Therefore, verifying each of the <math>O(1/\tau)</math> candidates in <math>O\left(\log \log _{w} \sigma\right) </math> time results in <math>O\left((1/\tau)\log \log _{w} \sigma\right) </math> total query time for returning the set of all <math>\tau </math>-majorities on the path from <math>u </math> to <math>v </math>.
==Related problems==
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