# SPDX-License-Identifier: Apache-2.0
# SPDX-FileCopyrightText: 2021 The Elixir Team
# SPDX-FileCopyrightText: 2012 Plataformatec

defmodule Range do
  @moduledoc """
  Ranges represent a sequence of zero, one or many, ascending
  or descending integers with a common difference called step.

  The most common form of creating and matching on ranges is
  via the [`first..last`](`../2`) and [`first..last//step`](`..///3`)
  notations, auto-imported from `Kernel`:

      iex> 1 in 1..10
      true
      iex> 5 in 1..10
      true
      iex> 10 in 1..10
      true

  Ranges are always inclusive in Elixir. When a step is defined,
  integers will only belong to the range if they match the step:

      iex> 5 in 1..10//2
      true
      iex> 4 in 1..10//2
      false

  When defining a range without a step, the step will be
  defined based on the first and last position of the
  range, If `last >= first`, it will be an increasing range
  with a step of 1. Otherwise, it is a decreasing range.
  Note, however, implicit decreasing ranges are deprecated.
  Therefore, if you need a decreasing range from `3` to `1`,
  prefer to write `3..1//-1` instead.

  `../0` can also be used as a shortcut to create the range `0..-1//1`,
  also known as the full-slice range:

      iex> ..
      0..-1//1

  ## Use cases

  Ranges typically have two uses in Elixir: as a collection or
  to represent a slice of another data structure.

  ### Ranges as collections

  Ranges in Elixir are enumerables and therefore can be used
  with the `Enum` module:

      iex> Enum.to_list(1..3)
      [1, 2, 3]
      iex> Enum.to_list(3..1//-1)
      [3, 2, 1]
      iex> Enum.to_list(1..5//2)
      [1, 3, 5]

  Ranges may also have a single element:

      iex> Enum.to_list(1..1)
      [1]
      iex> Enum.to_list(1..1//2)
      [1]

  Or even no elements at all:

      iex> Enum.to_list(10..0//1)
      []
      iex> Enum.to_list(0..10//-1)
      []

  The full-slice range, returned by `../0`, is an empty collection:

      iex> Enum.to_list(..)
      []

  ### Ranges as slices

  Ranges are also frequently used to slice collections.
  You can slice strings or any enumerable:

      iex> String.slice("elixir", 1..4)
      "lixi"
      iex> Enum.slice([0, 1, 2, 3, 4, 5], 1..4)
      [1, 2, 3, 4]

  In those cases, the first and last values of the range
  are mapped to positions in the collections.

  If a negative number is given, it maps to a position
  from the back:

      iex> String.slice("elixir", 1..-2//1)
      "lixi"
      iex> Enum.slice([0, 1, 2, 3, 4, 5], 1..-2//1)
      [1, 2, 3, 4]

  The range `0..-1//1`, returned by `../0`, returns the
  collection as is, which is why it is called the full-slice
  range:

      iex> String.slice("elixir", ..)
      "elixir"
      iex> Enum.slice([0, 1, 2, 3, 4, 5], ..)
      [0, 1, 2, 3, 4, 5]

  ## Definition

  An increasing range `first..last//step` is a range from `first`
  to `last` increasing by `step` where  `step` must be a positive
  integer and all values `v` must be `first <= v and v <= last`.
  Therefore, a range `10..0//1` is an empty range because there
  is no value `v` that is `10 <= v and v <= 0`.

  Similarly, a decreasing range `first..last//step` is a range
  from `first` to `last` decreasing by `step` where `step` must
  be a negative integer and  values `v` must be `first >= v and v >= last`.
  Therefore, a range `0..10//-1` is an empty range because there
  is no value `v` that is `0 >= v and v >= 10`.

  ## Representation

  Internally, ranges are represented as structs:

      iex> range = 1..9//2
      1..9//2
      iex> first..last//step = range
      iex> first
      1
      iex> last
      9
      iex> step
      2
      iex> range.step
      2

  You can access the range fields (`first`, `last`, and `step`)
  directly but you should not modify nor create ranges by hand.
  Instead use the proper operators or `new/2` and `new/3`.

  Ranges implement the `Enumerable` protocol with memory
  efficient versions of all `Enumerable` callbacks:

      iex> range = 1..10
      1..10
      iex> Enum.reduce(range, 0, fn i, acc -> i * i + acc end)
      385
      iex> Enum.count(range)
      10
      iex> Enum.member?(range, 11)
      false
      iex> Enum.member?(range, 8)
      true

  Such function calls are efficient memory-wise no matter the
  size of the range. The implementation of the `Enumerable`
  protocol uses logic based solely on the endpoints and does
  not materialize the whole list of integers.
  """

  @enforce_keys [:first, :last, :step]
  defstruct first: nil, last: nil, step: nil

  @type limit :: integer
  @type step :: pos_integer | neg_integer
  @type t :: %__MODULE__{first: limit, last: limit, step: step}
  @type t(first, last) :: %__MODULE__{first: first, last: last, step: step}

  @doc """
  Creates a new range.

  If `first` is less than `last`, the range will be increasing from
  `first` to `last`. If `first` is equal to `last`, the range will contain
  one element, which is the number itself.

  If `first` is greater than `last`, the range will be decreasing from `first`
  to `last`, albeit this behavior is deprecated. Therefore, it is advised to
  explicitly list the step with `new/3`.

  ## Examples

      iex> Range.new(-100, 100)
      -100..100

  """

  @spec new(limit, limit) :: t
  def new(first, last) when is_integer(first) and is_integer(last) do
    step =
      if first <= last do
        1
      else
        # TODO: Remove me on v2.0
        IO.warn_once(
          {__MODULE__, :new},
          fn ->
            "Range.new/2 and first..last default to a step of -1 when last < first. Use Range.new(first, last, -1) or first..last//-1, or pass 1 if that was your intention"
          end,
          3
        )

        -1
      end

    %Range{first: first, last: last, step: step}
  end

  def new(first, last) do
    raise ArgumentError,
          "ranges (first..last) expect both sides to be integers, " <>
            "got: #{inspect(first)}..#{inspect(last)}"
  end

  @doc """
  Creates a new range with `step`.

  ## Examples

      iex> Range.new(-100, 100, 2)
      -100..100//2

  """
  @doc since: "1.12.0"
  @spec new(limit, limit, step) :: t
  def new(first, last, step)
      when is_integer(first) and is_integer(last) and is_integer(step) and step != 0 do
    %Range{first: first, last: last, step: step}
  end

  def new(first, last, step) do
    raise ArgumentError,
          "ranges (first..last//step) expect both sides to be integers and the step to be a " <>
            "non-zero integer, got: #{inspect(first)}..#{inspect(last)}//#{inspect(step)}"
  end

  @doc """
  Returns the size of `range`.

  ## Examples

      iex> Range.size(1..10)
      10
      iex> Range.size(1..10//2)
      5
      iex> Range.size(1..10//3)
      4
      iex> Range.size(1..10//-1)
      0

      iex> Range.size(10..1//-1)
      10
      iex> Range.size(10..1//-2)
      5
      iex> Range.size(10..1//-3)
      4
      iex> Range.size(10..1//1)
      0

  """
  @doc since: "1.12.0"
  @spec size(t) :: non_neg_integer
  def size(range)
  def size(first..last//step) when step > 0 and first > last, do: 0
  def size(first..last//step) when step < 0 and first < last, do: 0
  def size(first..last//step), do: abs(div(last - first, step)) + 1

  # TODO: Remove me on v2.0
  def size(%{__struct__: Range, first: first, last: last} = range) do
    step = if first <= last, do: 1, else: -1
    size(Map.put(range, :step, step))
  end

  @doc """
  Shifts a range by the given number of steps.

  ## Examples

      iex> Range.shift(0..10, 1)
      1..11
      iex> Range.shift(0..10, 2)
      2..12

      iex> Range.shift(0..10//2, 2)
      4..14//2
      iex> Range.shift(10..0//-2, 2)
      6..-4//-2

  """
  @doc since: "1.14.0"
  @spec shift(t, integer) :: t
  def shift(%Range{} = range, 0), do: range

  def shift(first..last//step, steps_to_shift) when is_integer(steps_to_shift) do
    shift = steps_to_shift * step

    new(first + shift, last + shift, step)
  end

  @doc """
  Splits a range in two.

  It returns a tuple of two elements.

  If `split` is less than the number of elements in the range, the first
  element in the range will have `split` entries and the second will have
  all remaining entries.

  If `split` is more than the number of elements in the range, the second
  range in the tuple will emit zero elements.

  ## Examples

  Increasing ranges:

      iex> Range.split(1..5, 2)
      {1..2, 3..5}

      iex> Range.split(1..5//2, 2)
      {1..3//2, 5..5//2}

      iex> Range.split(1..5//2, 0)
      {1..-1//2, 1..5//2}

      iex> Range.split(1..5//2, 10)
      {1..5//2, 7..5//2}

  Decreasing ranges can also be split:

      iex> Range.split(5..1//-1, 2)
      {5..4//-1, 3..1//-1}

      iex> Range.split(5..1//-2, 2)
      {5..3//-2, 1..1//-2}

      iex> Range.split(5..1//-2, 0)
      {5..7//-2, 5..1//-2}

      iex> Range.split(5..1//-2, 10)
      {5..1//-2, -1..1//-2}

  Empty ranges preserve their property but still return empty ranges:

      iex> Range.split(2..5//-1, 2)
      {2..3//-1, 4..5//-1}

      iex> Range.split(2..5//-1, 10)
      {2..3//-1, 4..5//-1}

      iex> Range.split(5..2//1, 2)
      {5..4//1, 3..2//1}

      iex> Range.split(5..2//1, 10)
      {5..4//1, 3..2//1}

  If the number to split is negative, it splits from the back:

      iex> Range.split(1..5, -2)
      {1..3, 4..5}

      iex> Range.split(5..1//-1, -2)
      {5..3//-1, 2..1//-1}

  If it is negative and greater than the elements in the range,
  the first element of the tuple will be an empty range:

      iex> Range.split(1..5, -10)
      {1..0//1, 1..5}

      iex> Range.split(5..1//-1, -10)
      {5..6//-1, 5..1//-1}

  ## Properties

  When a range is split, the following properties are observed.
  Given `split(input)` returns `{left, right}`, we have:

      assert input.first == left.first
      assert input.last == right.last
      assert input.step == left.step
      assert input.step == right.step
      assert Range.size(input) == Range.size(left) + Range.size(right)

  """
  @doc since: "1.15.0"
  @spec split(t, integer) :: {t, t}
  def split(first..last//step = range, split) when is_integer(split) do
    if split >= 0 do
      split(first, last, step, split)
    else
      split(first, last, step, size(range) + split)
    end
  end

  defp split(first, last, step, split) when first < last or (first == last and step > 0) do
    if step > 0 do
      mid = max(min(first + step * (split - 1), last), first - step)
      {first..mid//step, (mid + step)..last//step}
    else
      {first..(first - step)//step, (last + step)..last//step}
    end
  end

  defp split(last, first, step, split) do
    if step < 0 do
      mid = min(max(last + step * (split - 1), first), last - step)
      {last..mid//step, (mid + step)..first//step}
    else
      {last..(last - step)//step, (first + step)..first//step}
    end
  end

  @doc """
  Converts a range to a list.

  ## Examples

      iex> Range.to_list(0..5)
      [0, 1, 2, 3, 4, 5]
      iex> Range.to_list(-3..0)
      [-3, -2, -1, 0]

  """
  @doc since: "1.15.0"
  @spec to_list(t) :: list(integer)
  def to_list(first..last//step)
      when step > 0 and first <= last
      when step < 0 and first >= last do
    :lists.seq(first, last, step)
  end

  def to_list(_first.._last//_step) do
    []
  end

  # TODO: Remove me on v2.0
  def to_list(%{__struct__: Range, first: first, last: last}) do
    step = if first <= last, do: 1, else: -1
    :lists.seq(first, last, step)
  end

  @doc """
  Checks if two ranges are disjoint.

  ## Examples

      iex> Range.disjoint?(1..5, 6..9)
      true
      iex> Range.disjoint?(5..1//-1, 6..9)
      true
      iex> Range.disjoint?(1..5, 5..9)
      false
      iex> Range.disjoint?(1..5, 2..7)
      false

  Steps are also considered when computing the ranges to be disjoint:

      iex> Range.disjoint?(1..10//2, 2..10//2)
      true

      # First element in common is 29
      iex> Range.disjoint?(1..100//14, 8..100//21)
      false
      iex> Range.disjoint?(57..-1//-14, 8..100//21)
      false
      iex> Range.disjoint?(1..100//14, 50..8//-21)
      false
      iex> Range.disjoint?(1..28//14, 8..28//21)
      true

      # First element in common is 14
      iex> Range.disjoint?(2..28//3, 9..28//5)
      false
      iex> Range.disjoint?(26..2//-3, 29..9//-5)
      false

      # Starting from the back without alignment
      iex> Range.disjoint?(27..11//-3, 30..0//-7)
      true

  """
  @doc since: "1.8.0"
  @spec disjoint?(t, t) :: boolean
  def disjoint?(first1..last1//step1 = range1, first2..last2//step2 = range2) do
    if size(range1) == 0 or size(range2) == 0 do
      true
    else
      {first1, last1, step1} = normalize(first1, last1, step1)
      {first2, last2, step2} = normalize(first2, last2, step2)

      cond do
        last2 < first1 or last1 < first2 ->
          true

        abs(step1) == 1 and abs(step2) == 1 ->
          false

        true ->
          # We need to find the first intersection of two arithmetic
          # progressions and see if they belong within the ranges
          # https://math.stackexchange.com/questions/1656120/formula-to-find-the-first-intersection-of-two-arithmetic-progressions
          {gcd, u, v} = Integer.extended_gcd(-step1, step2)

          if rem(first2 - first1, gcd) == 0 do
            c = first1 - first2 + step2 - step1
            t1 = -c / step2 * u
            t2 = -c / step1 * v
            t = max(floor(t1) + 1, floor(t2) + 1)
            x = div(c * u + t * step2, gcd) - 1
            y = div(c * v + t * step1, gcd) - 1

            x < 0 or first1 + x * step1 > last1 or
              y < 0 or first2 + y * step2 > last2
          else
            true
          end
      end
    end
  end

  @compile inline: [normalize: 3]
  defp normalize(first, last, step) when first > last,
    do: {first - abs(div(first - last, step) * step), first, -step}

  defp normalize(first, last, step), do: {first, last, step}

  @doc false
  @deprecated "Pattern match on first..last//step instead"
  def range?(%{__struct__: Range, first: first, last: last})
      when is_integer(first) and is_integer(last),
      do: true

  def range?(_), do: false
end
