segment-trees | Automatically exported from code.google.com/p/segment-trees
kandi X-RAY | segment-trees Summary
kandi X-RAY | segment-trees Summary
Automatically exported from code.google.com/p/segment-trees
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Top functions reviewed by kandi - BETA
- Build a JavaCounter1 from the ranges
- Recursively splits the tree into a tree
- Recursively builds the counter
- Make a binary split
- Computes the number of ranges
- Recursively iterates over the specified node
- Recursively builds the simple counter counts from the given node
- Build a range - set implementation
- Builds a generator for the given node
- Recursively parse the given node
- Searches for a certain range
- Returns true if the given value is within the range inclusive
- Returns counts for the forest
- Determine the counts for a given node
- Record the given value
- Get the counts
segment-trees Key Features
segment-trees Examples and Code Snippets
Community Discussions
Trending Discussions on segment-trees
QUESTION
given N elements of an array compute the sum (min*max) across all the subarrays of the array.
e.g. N = 5
Array: 5 7 2 3 9
output: 346 (5*5 + 7*7 + 2*2 + 3*3 + 9*9 + 5*7 + 2*7 + 2*3 + 3*9 + 2*7+2*7 + 2*9 + 2*7 + 2*9 + 2*9)
i cannot think of anything better than O(n^2). The editorial solution uses segment trees which i couldnt understand.
...ANSWER
Answered 2020-May-18 at 01:29Hint regarding the editorial (the details of which I am uncertain about): if we can solve the problem for all the intervals that include both A[i]
and A[i+1]
, where i
divides A
in half, in O(n)
time, then we can solve the whole problem in O(n log n)
time using divide and conquer, solving left and right separately, and adding to that the intervals that overlap both left and right.
QUESTION
Consider this question. In this segment tree solution, we are updating all nodes of the tree in the given range. Is it possible to apply lazy propagation to this problem?
Edit: Consider that in every update operation arr[i] = (1-(1-arr[i])*a)
, where L<=i<=R
and a
is a constant.
ANSWER
Answered 2018-Apr-13 at 20:26Yes, it is indeed possible, at least in some cases.
Basically, you need a way to efficiently store the lazy operation, and a way to efficiently combine two stored lazy operations into one.
For instance, say the update operation is segment assignment, that is, a[l] = x
, a[l+1] = x
, ...
, a[r-1] = x
, a[r] = x
.
This operation on the whole subtree can be stored as just the value x
, which will mean that the operation was to assign x
to every vertex of this subtree.
For lazy propagation in vertex v
, we just apply it to immediate children of vertex v
, and store the same lazy operation x
there.
Note that any old lazy operation in children is just erased by the assignment.
Such is the nature of assignment.
As for your added example, operation arr[i] = (1 - (1 - arr[i]) * a)
, let us see how the value changes after two such operations with constants a
and b
.
Before operations, let the value be v
.
After the first one, it becomes w = 1 - (1 - v) * a
, which is a*v + (1-a)*1
.
After the second operation, the value becomes 1 - (1 - w) * b
, which is b*w + (1-b)*1
, which in turn is b*a*v + b*(1-a)*1 + (1-b)*1
, and finally becomes (b*a)*v + (1-b*a)*1
.
(I may have mixed up +s and -s, but that hopefully does not change the whole picture.)
We can now see that the value is a linear function of the original value, so we can store the coefficients b*a
and 1-b*a
of the linear and constant terms, respectively.
The problem now is that the coefficients may grow too quickly, and will soon exceed the capacity of the storage type (int
, double
or whatever).
Now, if the problem deals with integer residues modulo some integer instead of just integers or reals, that's not an issue; otherwise, storing the coefficients soon becomes problematic.
QUESTION
I am trying to solve this question. I am making the vector size of Math.ceil(Math.sqrt(arrSize)). I have used the following methods - For constructing sqrt vector
I am taking square root chunks and finding the smallest index in the block and storing them in the vect array.
How can I improve my update query complexity from Sqrt(n).
...ANSWER
Answered 2017-Sep-11 at 09:45If you have to improve complexity you have to use Segment Trees. In this case you cannot directly update the index of the vect array like in case of range sum query. You have to find again the minimum of the block.
QUESTION
I am trying to learn segment tree through https://www.topcoder.com/community/data-science/data-science-tutorials/range-minimum-query-and-lowest-common-ancestor/ After Understanding the basics of segment trees I tried to solve this question. But only one test case is passed and on the second one, I am getting tle. On further inspection comparing the two answers using filediff I found out there are wrong answers. I am unable to find any errors. Please help.
this code is for creating and updating segment tree.
Variables - node = starting index in segment tree which is 1. b = lower limit, e = upper limit
...ANSWER
Answered 2017-Sep-10 at 17:21You are updating the segment tree in case of querying the minimum index. Remove it and the code is working fine. In case of query the segTree should not be changed.
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
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Install segment-trees
You can use segment-trees like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the segment-trees component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .
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