Data.Sequence.Ordered

newtype OrdSeq a

An ordered sequence. The Semigroup instance uses the merge function.

Instances

• (Eq a) => Eq (OrdSeq a)
• (Show a) => Show (OrdSeq a)
• (Ord a) => Semigroup (OrdSeq a)
• (Ord a) => Monoid (OrdSeq a)
• Foldable OrdSeq

empty :: forall a. OrdSeq a

fromFoldable :: forall f a. Foldable f => Ord a => f a -> OrdSeq a

Probably O(n*log(n)), but depends on the Foldable instance. Consruct an ordered sequence from any any Foldable.

insert :: forall a. Ord a => a -> OrdSeq a -> OrdSeq a

O(log(n)). Insert the given value into the correct place in the sequence.

null :: forall a. OrdSeq a -> Boolean

O(1). Returns true if the given sequence is empty, false otherwise.

length :: forall a. OrdSeq a -> Int

O(n). Return the length of the sequence.

least :: forall a. Ord a => OrdSeq a -> Maybe a

O(1). Access the least element of the sequence, or Nothing if the sequence is empty.

greatest :: forall a. Ord a => OrdSeq a -> Maybe a

O(1). Access the greatest element of the sequence, or Nothing if the sequence is empty.

popLeast :: forall a. Ord a => OrdSeq a -> Maybe (Tuple a (OrdSeq a))

O(1). Remove the least element of the sequence, returning that element and the remainder of the sequence. If the sequence is empty, return Nothing.

popGreatest :: forall a. Ord a => OrdSeq a -> Maybe (Tuple a (OrdSeq a))

O(1). Remove the greatest element of the sequence, returning that element and the remainder of the sequence. If the sequence is empty, return Nothing.

partition :: forall a. Ord a => a -> OrdSeq a -> Tuple (OrdSeq a) (OrdSeq a)

O(log(n)). Split an ordered sequence into two halves. The first element of the returned tuple contains all elements which compared less than or equal to the argument; the second element contains the rest.

merge :: forall a. Ord a => OrdSeq a -> OrdSeq a -> OrdSeq a

O(m*log(n/m)), where m and n are the lengths of the longer and shorter sequences respectively. Create a new sequence containing every element in both of the given sequences.

intersection :: forall a. Ord a => OrdSeq a -> OrdSeq a -> OrdSeq a

O(n*log(n)), where n is the length of the longer sequence. Create a new sequence containing only elements which are common to both sequences.

deleteAll :: forall a. Ord a => a -> OrdSeq a -> OrdSeq a

O(log(n)). Delete all elements from the sequence which compare EQ to the given value.

toUnfoldable :: forall f. Functor f => Unfoldable f => OrdSeq ~> f

Probably O(n), but depends on the Unfoldable instance. Unfold an ordered sequence in ascending order.

toUnfoldableDescending :: forall f a. Functor f => Unfoldable f => OrdSeq a -> f a

Probably O(n), but depends on the Unfoldable instance. Unfold an ordered sequence in descending order.

sort :: forall f a. Functor f => Foldable f => Unfoldable f => Ord a => f a -> f a

Sort any structure (which has Foldable, Unfoldable, and Functor instances) by converting to an OrdSeq and back again. I am fairly sure this is usually O(n*log(n)), although of course this depends on the Unfoldable and Foldable instances.