Data.IntSet.NonEmpty

A non-empty (by construction) set of integers. At least one value exists in an NEIntSet a at all times.

Functions that take an NEIntSet can safely operate on it with the assumption that it has at least one item.

Functions that return an NEIntSet provide an assurance that the result has at least one item.

Data.IntSet.NonEmpty re-exports the API of Data.IntSet, faithfully reproducing asymptotics, typeclass constraints, and semantics. Functions that ensure that input and output sets are both non-empty (like insert) return NEIntSet, but functions that might potentially return an empty map (like delete) return a IntSet instead.

You can directly construct an NEIntSet with the API from Data.IntSet.NonEmpty; it's more or less the same as constructing a normal IntSet, except you don't have access to empty. There are also a few ways to construct an NEIntSet from a IntSet:

1. The nonEmptySet smart constructor will convert a IntSet a into a Maybe (NEIntSet a), returning Nothing if the original IntSet was empty.

2. You can use the Data.IntSet.NonEmpty.insertIntSet family of functions to insert a value into a IntSet to create a guaranteed NEIntSet.

3. You can use the IsNonEmpty and IsEmpty patterns to "pattern match" on a IntSet to reveal it as either containing a NEIntSet or an empty map.

4. withNonEmpty offers a continuation-based interface for deconstructing a IntSet and treating it as if it were an NEIntSet.

You can convert an NEIntSet into a IntSet with toSet or IsNonEmpty, essentially "obscuring" the non-empty property from the type.

O(1). Create a singleton set.

O(n*log n). Create a set from a list of elements.

O(n). Build a set from an ascending list in linear time. /The precondition (input list is ascending) is not checked./

O(n). Build a set from an ascending list of distinct elements in linear time. The precondition (input list is strictly ascending) is not checked.

O(log n). Insert an element in a set. If the set already contains an element equal to the given value, it is replaced with the new value.

O(log n). Delete an element from a set.

O(log n). Delete an element from a set, returning Nothing if the result would be empty.

This is more efficient than nonEmptySet . delete x because it avoids converting the known-minimum representation back through IntSet when the deleted element is not the minimum.

O(log n). Is the element in the set?

O(log n). Is the element not in the set?

O(log n). Find largest element smaller than the given one.

lookupLT 3 (fromList (3 :| [5])) == Nothing
lookupLT 5 (fromList (3 :| [5])) == Just 3

O(log n). Find smallest element greater than the given one.

lookupLT 4 (fromList (3 :| [5])) == Just 5
lookupLT 5 (fromList (3 :| [5])) == Nothing

O(log n). Find largest element smaller or equal to the given one.

lookupLT 2 (fromList (3 :| [5])) == Nothing
lookupLT 4 (fromList (3 :| [5])) == Just 3
lookupLT 5 (fromList (3 :| [5])) == Just 5

O(log n). Find smallest element greater or equal to the given one.

lookupLT 3 (fromList (3 :| [5])) == Just 3
lookupLT 4 (fromList (3 :| [5])) == Just 5
lookupLT 6 (fromList (3 :| [5])) == Nothing

O(1). The number of elements in the set. Guaranteed to be greater than zero.

O(n+m). Is this a subset? (s1 `isSubsetOf` s2) tells whether s1 is a subset of s2.

O(n+m). Is this a proper subset? (ie. a subset but not equal).

O(n+m). Check whether two sets are disjoint (i.e. their intersection is empty).

disjoint (fromList (2:|[4,6]))   (fromList (1:|[3]))     == True
disjoint (fromList (2:|[4,6,8])) (fromList (2:|[3,5,7])) == False
disjoint (fromList (1:|[2]))     (fromList (1:|[2,3,4])) == False

O(m*log(n/m + 1)), m <= n. The union of two sets, preferring the first set when equal elements are encountered.

O(m*log(n/m + 1)), m <= n. Union of a possibly-empty IntSet and a non-empty set.

O(m*log(n/m + 1)), m <= n. Union of a non-empty set and a possibly-empty IntSet.

The union of a non-empty list of sets

O(m*log(n/m + 1)), m <= n. Difference of two sets.

Returns a potentially empty set (IntSet) because the first set might be a subset of the second set, and therefore have all of its elements removed.

Same as difference.

O(m*log(n/m + 1)), m <= n. The intersection of two sets.

Returns a potentially empty set (IntSet), because the two sets might have an empty intersection.

Elements of the result come from the first set, so for example

import qualified Data.IntSet.NonEmpty as NES
data AB = A | B deriving Show
instance Ord AB where compare _ _ = EQ
instance Eq AB where _ == _ = True
main = print (NES.singleton A `NES.intersection` NES.singleton B,
              NES.singleton B `NES.intersection` NES.singleton A)

prints (fromList (A:|[]),fromList (B:|[])).

O(n). Filter all elements that satisfy the predicate.

Returns a potentially empty set (IntSet) because the predicate might filter out all items in the original non-empty set.

O(n). Partition the map according to a predicate.

Returns a These with potentially two non-empty sets:

• This n1 means that the predicate was true for all items.

• That n2 means that the predicate was false for all items.

• These n1 n2 gives n1 (all of the items that were true for the predicate) and n2 (all of the items that were false for the predicate).

See also split.

partition (> 3) (fromList (5 :| [3])) == These (singleton 5) (singleton 3)
partition (< 7) (fromList (5 :| [3])) == This  (fromList (3 :| [5]))
partition (> 7) (fromList (5 :| [3])) == That  (fromList (3 :| [5]))

O(log n). The expression (split x set) is potentially a These containing up to two NEIntSets based on splitting the set into sets containing items before and after the value x. It will never return a set that contains x itself.

• Nothing means that x was the only value in the the original set, and so there are no items before or after it.

• Just (This n1) means x was larger than or equal to all items in the set, and n1 is the entire original set (minus x, if it was present)

• Just (That n2) means x was smaller than or equal to all items in the set, and n2 is the entire original set (minus x, if it was present)

• Just (These n1 n2) gives n1 (the set of all values from the original set less than x) and n2 (the set of all values from the original set greater than x).

split 2 (fromList (5 :| [3])) == Just (That  (fromList (3 :| [5]))      )
split 3 (fromList (5 :| [3])) == Just (That  (singleton 5)              )
split 4 (fromList (5 :| [3])) == Just (These (singleton 3) (singleton 5))
split 5 (fromList (5 :| [3])) == Just (This  (singleton 3)              )
split 6 (fromList (5 :| [3])) == Just (This  (fromList (3 :| [5]))      )
split 5 (singleton 5)         == Nothing

O(log n). The expression (splitMember x set) splits a set just like split but also returns member x set (whether or not x was in set)

splitMember 2 (fromList (5 :| [3])) == (False, Just (That  (fromList (3 :| [5)]))))
splitMember 3 (fromList (5 :| [3])) == (True , Just (That  (singleton 5)))
splitMember 4 (fromList (5 :| [3])) == (False, Just (These (singleton 3) (singleton 5)))
splitMember 5 (fromList (5 :| [3])) == (True , Just (This  (singleton 3))
splitMember 6 (fromList (5 :| [3])) == (False, Just (This  (fromList (3 :| [5])))
splitMember 5 (singleton 5)         == (True , Nothing)

O(1). Decompose a set into pieces based on the structure of the underlying tree. This function is useful for consuming a set in parallel.

No guarantee is made as to the sizes of the pieces; an internal, but deterministic process determines this. However, it is guaranteed that the pieces returned will be in ascending order (all elements in the first subset less than all elements in the second, and so on).

Note that the current implementation does not return more than four subsets, but you should not depend on this behaviour because it can change in the future without notice.

O(n*log n). map f s is the set obtained by applying f to each element of s.

It's worth noting that the size of the result may be smaller if, for some (x,y), x /= y && f x == f y

O(n). Fold the elements in the set using the given right-associative binary operator, such that foldr f z == foldr f z . toAscList.

For example,

elemsList set = foldr (:) [] set

O(n). Fold the elements in the set using the given left-associative binary operator, such that foldl f z == foldl f z . toAscList.

For example,

descElemsList set = foldl (flip (:)) [] set

O(n). A version of foldr that uses the value at the maximal value in the set as the starting value.

Note that, unlike foldr1 for IntSet, this function is total if the input function is total.

O(n). A version of foldl that uses the value at the minimal value in the set as the starting value.

Note that, unlike foldl1 for IntSet, this function is total if the input function is total.

O(1). The minimal element of a set. Note that this is total, making lookupMin obsolete. It is constant-time, so has better asymptotics than Data.IntSet.lookupMin and Data.Map.findMin as well.

findMin (fromList (5 :| [3])) == 3

O(log n). The maximal key of a set Note that this is total, making lookupMin obsolete.

findMax (fromList (5 :| [3])) == 5

O(1). Delete the minimal element. Returns a potentially empty set (IntSet), because we might delete the final item in a singleton set. It is constant-time, so has better asymptotics than Data.IntSet.deleteMin.

deleteMin (fromList (5 :| [3, 7])) == Data.IntSet.fromList [5, 7]
deleteMin (singleton 5) == Data.IntSet.empty

O(log n). Delete the maximal element. Returns a potentially empty set (IntSet), because we might delete the final item in a singleton set.

deleteMax (fromList (5 :| [3, 7])) == Data.IntSet.fromList [3, 5]
deleteMax (singleton 5) == Data.IntSet.empty

O(1). Delete and find the minimal element. It is constant-time, so has better asymptotics that Data.IntSet.minView for IntSet.

Note that unlike Data.IntSet.deleteFindMin for IntSet, this cannot ever fail, and so is a total function. However, the result IntSet is potentially empty, since the original set might have contained just a single item.

deleteFindMin (fromList (5 :| [3, 10])) == (3, Data.IntSet.fromList [5, 10])

O(log n). Delete and find the minimal element.

Note that unlike Data.IntSet.deleteFindMax for IntSet, this cannot ever fail, and so is a total function. However, the result IntSet is potentially empty, since the original set might have contained just a single item.

deleteFindMax (fromList (5 :| [3, 10])) == (10, Data.IntSet.fromList [3, 5])

O(n). Test if the internal set structure is valid.