/*
Copyright 2014 Workiva, LLC

Licensed under the Apache License, Version 2.0 (the "License");
you may not use this file except in compliance with the License.
You may obtain a copy of the License at

 http://www.apache.org/licenses/LICENSE-2.0

Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.
*/

/*
Package xfast provides access to a sorted tree that treats integers
as if they were words of m bits, where m can be 8, 16, 32, or 64.
The advantage to storing integers as a trie of words is that operations
can be performed in constant time depending on the size of the
universe and not on the number of items in the trie.

The time complexity is as follows:
Space: O(n log M)
Insert: O(log M)
Delete: O(log M)
Search: O(log log M)
Get: O(1)

where n is the number of items in the trie and M is the size of the
universe, ie, 2^63-1 for 64 bit ints.

As you can see, for 64 bit ints, inserts and deletes can be performed
in O(64), constant time which provides very predictable behavior
in the best case.

A get by key can be performed in O(1) time and searches can be performed
in O(6) time for 64 bit integers.

While x-tries have relatively slow insert, deletions, and consume a large
amount of space, they form the top half of a y-fast trie which can
insert and delete in O(log log M) time and consumes O(n) space.
*/
package xfast

import "fmt"

// isInternal returns a bool indicating if the provided
// node is an internal node, that is, non-leaf node.
func isInternal(n *node) bool {
	if n == nil {
		return false
	}
	return n.entry == nil
}

// hasInternal returns a bool indicating if the provided
// node has a child that is an internal node.
func hasInternal(n *node) bool {
	return isInternal(n.children[0]) || isInternal(n.children[1])
}

// isLeaf returns a bool indicating if the provided node
// is a leaf node, that is, has a valid entry value.
func isLeaf(n *node) bool {
	if n == nil {
		return false
	}
	return !isInternal(n)
}

// Entry defines items that can be inserted into the x-fast
// trie.
type Entry interface {
	// Key is the key for this entry.  If the trie has been
	// given bit size n, only the last n bits of this key
	// will matter.  Use a bit size of 64 to enable all
	// 2^64-1 keys.
	Key() uint64
}

// masks are used to determine the prefix of any given key.  The masks
// are stored in a [64] array where each position of the array represents
// a bitmask to the ith bit.  For example, if you wanted to mask the first
// bit of a 64-bit int you'd and it with masks[0].  If you wanted to mask
// the first bit of an 8 bit key, you'd have to shift 56 bits to the right
// and perform the mask operation.  This array is immutable and should not
// be changed after initialization.
var masks = func() [64]uint64 { // we don't technically need the last mask, this is just to be consistent
	masks := [64]uint64{}
	mask := uint64(0)
	for i := uint64(0); i < 64; i++ {
		mask = mask | 1<<(63-i)
		masks[i] = mask
	}
	return masks
}()

// positions are similar to masks and that the positions array allows
// us to determine if a node should go left or right at a specific bit
// position of the key.  Basically, this array stores every 2^n number
// where n is in [0, 64).  This array is immutable and should not be
// changed after initialization.
var positions = func() [64]uint64 {
	positions := [64]uint64{}
	for i := uint64(0); i < 64; i++ {
		positions[i] = uint64(1 << (63 - i))
	}
	return positions
}()

type node struct {
	// entry will store the entry for this node.  Is nil for
	// every internal node and non-nil for all leaves.  It is
	// how the internal/leaf function helpers determine the
	// position of this node.
	entry Entry
	// children stores the left and right child of this node.
	// At any time, and at any layer, it's possible for a pointer
	// to a child to point to a leaf due to threading.
	children [2]*node
	// i hate this, but it is really the best way
	// to walk up successor and predecessor threads
	parent *node
}

// newNode will allocate and initialize a newNode with the provided
// parent and entry.  Parent should never be nil, but entry may be
// if constructing an internal node.
func newNode(parent *node, entry Entry) *node {
	return &node{
		children: [2]*node{},
		entry:    entry,
		parent:   parent,
	}
}

// binarySearchHashMaps will perform a binary search of the provided
// maps to return a node that matches the longest prefix of the provided
// key.  This will return nil if a match could not be found, which would
// also return layer 0.  Layer information is useful when determining the
// distance from the provided node to the leaves.
func binarySearchHashMaps(layers []map[uint64]*node, key uint64) (int, *node) {
	low, high := 0, len(layers)-1
	diff := 64 - len(layers)
	var mid int
	var node *node
	for low <= high {
		mid = (low + high) / 2
		n, ok := layers[mid][key&masks[diff+mid]]
		if ok {
			node = n
			low = mid + 1
		} else {
			high = mid - 1
		}
	}

	return low, node
}

// whichSide returns an int representing the side on which
// the node resides in its parent.  NOTE: this function will panic
// if the child does not within the parent.  This situation should
// should be caught as early as possible as if it happens data
// coming from the x-fast trie cannot be trusted.
func whichSide(n, parent *node) int {
	if parent.children[0] == n {
		return 0
	}

	if parent.children[1] == n {
		return 1
	}

	panic(fmt.Sprintf(`Node: %+v, %p not a child of: %+v, %p`, n, n, parent, parent))
}

// XFastTrie is a datastructure for storing integers in a known
// universe, where universe size is determined by the bit size
// of the desired keys.  This structure should be faster than
// binary search tries for very large datasets and slower for
// smaller datasets.
type XFastTrie struct {
	// layers stores the hashmaps of the individual layers of the trie.
	// The hashmaps store prefixes, allowing use to do a binary search
	// of these maps before visiting the trie for successor/predecessor
	// queries.
	layers []map[uint64]*node
	// root is a pointer to the first node of the trie, which actually
	// adds an additional layer, ie, instead of 64 layers for a
	// uint64, this will cause the number of layers to be 65.
	root *node
	// num is the number of items in the trie.
	num uint64
	// bits represents the number of bits of the keys this trie
	// expects.  Because the time complexity of operations is
	// dependent upon universe size, smaller sized keys will
	// actually cause the trie to be faster.  Diff is the difference
	// between the desired bit size and 64 as we have to offset
	// in the position and mask arrays.
	bits, diff uint8
	// min and max index the lowest and highest seen keys respectively.
	// this immediately allows us to check a desired key against
	// constraints and allows min/max operations to be performed
	// in O(1) time.
	min, max *node
}

// init will initialize the XFastTrie with the provided byte-size.
// I'd prefer generics here, but it is what it is.  We expect uints
// here when ints would perform just as well, but the public methods
// on the XFastTrie all expect uint64, so we expect a uint in the
// constructor for consistency's sake.
func (xft *XFastTrie) init(intType interface{}) {
	bits := uint8(0)
	switch intType.(type) {
	case uint8:
		bits = 8
	case uint16:
		bits = 16
	case uint32:
		bits = 32
	case uint, uint64:
		bits = 64
	default:
		// we'll panic with a bad value to the constructor.
		panic(`Invalid universe size provided.`)
	}

	xft.layers = make([]map[uint64]*node, bits)
	xft.bits = bits
	xft.diff = 64 - bits
	for i := uint8(0); i < bits; i++ {
		xft.layers[i] = make(map[uint64]*node, 50) // we can obviously be more intelligent about this.
	}
	xft.num = 0
	xft.root = newNode(nil, nil)
}

// Exists returns a bool indicating if the provided
// key exists in the trie.  This is an O(1) operation.
func (xft *XFastTrie) Exists(key uint64) bool {
	// the bottom hashmap of the trie has every entry
	// in it.
	_, ok := xft.layers[xft.bits-1][key]
	return ok
}

// Len returns the number of items in this trie.  This is an
// O(1) operation.
func (xft *XFastTrie) Len() uint64 {
	return xft.num
}

// Max will return the highest keyed value in the trie.  This is
// an O(1) operation.
func (xft *XFastTrie) Max() Entry {
	if xft.max == nil {
		return nil
	}

	return xft.max.entry
}

// Min will return the lowest keyed value in the trie.  This is
// an O(1) operation.
func (xft *XFastTrie) Min() Entry {
	if xft.min == nil {
		return nil
	}

	return xft.min.entry
}

// insert will add the provided entry to the trie or overwrite the existing
// entry if it exists.
func (xft *XFastTrie) insert(entry Entry) {
	key := entry.Key() // so we aren't calling this interface method over and over, fucking Go
	n := xft.layers[xft.bits-1][key]
	if n != nil {
		n.entry = entry
		return
	}

	// we need to find a predecessor or successor if it exists
	// to help us set threads later in this method.
	var predecessor, successor *node
	if xft.min != nil && key < xft.min.entry.Key() {
		successor = xft.min
	} else {
		successor = xft.successor(key)
	}

	// only need to find predecessor if successor is nil as otherwise
	// the successor will provide us is the predecessor if it exists.
	if successor == nil {
		if xft.max != nil && key > xft.max.entry.Key() {
			predecessor = xft.max
		} else {
			predecessor = xft.predecessor(key)
		}
	}

	// find the deepest root with a matching prefix, this should
	// save us some time, assuming the hashmap has perfect hashing.
	layer, root := binarySearchHashMaps(xft.layers, key)
	if root == nil {
		n = xft.root
		layer = 0
	} else {
		n = root
	}

	var leftOrRight uint64

	// from the existing node, create new nodes.
	for i := uint8(layer); i < xft.bits; i++ {
		// on 0th, this will be root
		// find out if we need to go left or right
		leftOrRight = (key & positions[xft.diff+i]) >> (xft.bits - 1 - i)
		if n.children[leftOrRight] == nil || isLeaf(n.children[leftOrRight]) {
			var nn *node
			if i < xft.bits-1 {
				nn = newNode(n, nil)
			} else {
				nn = newNode(n, entry)
				xft.num++
			}

			n.children[leftOrRight] = nn
			xft.layers[i][key&masks[xft.diff+i]] = nn // prefix for this layer
		}

		n = n.children[leftOrRight]
	}

	// we need to put the new node where it belongs in the doubly-linked
	// list comprised of all the leaves.
	if successor != nil { // we have to walk predecessor and successor threads
		predecessor = successor.children[0]
		if predecessor != nil {
			predecessor.children[1] = n
			n.children[0] = predecessor
		}
		n.children[1] = successor
		successor.children[0] = n
	} else if predecessor != nil {
		n.children[0] = predecessor
		predecessor.children[1] = n
	}

	// walk up the successor if it exists to set that branch's new
	// predecessor.
	if successor != nil {
		xft.walkUpSuccessor(root, n, successor)
	}

	// walk up the predecessor if it exists to set that branch's
	// new successor.
	if predecessor != nil {
		xft.walkUpPredecessor(root, n, predecessor)
	}

	// finally, walk up our own branch to set both successors
	// and predecessors.
	xft.walkUpNode(root, n, predecessor, successor)

	// and then do a final check against the min/max indicies.
	if xft.max == nil || key > xft.max.entry.Key() {
		xft.max = n
	}

	if xft.min == nil || key < xft.min.entry.Key() {
		xft.min = n
	}
}

// walkUpSuccessor will walk up the successor branch setting
// the predecessor where possible.  This breaks when a common
// ancestor between successor and node is found, ie, the root.
func (xft *XFastTrie) walkUpSuccessor(root, node, successor *node) {
	n := successor.parent
	for n != nil && n != root {
		// we don't really want to overwrite existing internal nodes,
		// or where the child is a leaf that is the successor
		if !isInternal(n.children[0]) && n.children[0] != successor {
			n.children[0] = node
		}
		n = n.parent
	}
}

// walkUpPredecessor will walk up the predecessor branch setting
// the successor where possible.  This breaks when a common
// ancestor between predecessor and node is found, ie, the root.
func (xft *XFastTrie) walkUpPredecessor(root, node, predecessor *node) {
	n := predecessor.parent
	for n != nil && n != root {
		if !isInternal(n.children[1]) && n.children[1] != predecessor {
			n.children[1] = node
		}
		n = n.parent
	}
}

// walkUpNode will walk up the newly created branch and set predecessor
// and successor where possible.  If predecessor or successor are nil,
// this will set nil where possible.
func (xft *XFastTrie) walkUpNode(root, node, predecessor, successor *node) {
	n := node.parent
	for n != nil && n != root {
		if !isInternal(n.children[1]) && n.children[1] != successor && n.children[1] != node {
			n.children[1] = successor
		}
		if !isInternal(n.children[0]) && n.children[0] != predecessor && n.children[0] != node {
			n.children[0] = predecessor
		}
		n = n.parent
	}
}

// Insert will insert the provided entries into the trie.  Any entry
// with an existing key will cause an overwrite.  This is an O(log M)
// operation, for each entry.
func (xft *XFastTrie) Insert(entries ...Entry) {
	for _, e := range entries {
		xft.insert(e)
	}
}

func (xft *XFastTrie) delete(key uint64) {
	n := xft.layers[xft.bits-1][key]
	if n == nil { // there's no matching k, v pair
		return
	}

	successor, predecessor := n.children[1], n.children[0]

	i := uint8(1)
	delete(xft.layers[xft.bits-1], key)
	leftOrRight := whichSide(n, n.parent)
	n.parent.children[leftOrRight] = nil
	n.children[0], n.children[1] = nil, nil
	n = n.parent
	hasImmediateSibling := false
	if successor != nil && successor.parent == n {
		hasImmediateSibling = true
	}
	if predecessor != nil && predecessor.parent == n {
		hasImmediateSibling = true
	}

	// this loop will kill any nodes that no longer link to internal
	// nodes
	for n != nil && n.parent != nil {
		// if we have an internal node remaining we should abort
		// now as no further node will be removed.  We should also
		// abort if the first parent of a leaf references the pre
		if hasInternal(n) || (i == 1 && hasImmediateSibling) {
			n = n.parent // we had one side deleted, need to set the other
			break
		}

		leftOrRight = whichSide(n, n.parent)
		n.parent.children[leftOrRight] = nil
		n.children[0], n.children[1] = nil, nil
		delete(xft.layers[xft.bits-i-1], key&masks[len(masks)-1-int(i)])
		n = n.parent
		i++
	}

	// we need to check now and update threads, but in the leaves
	// and in their branches
	if predecessor != nil {
		predecessor.children[1] = successor
		xft.walkUpPredecessor(n, successor, predecessor)
	}

	if successor != nil {
		successor.children[0] = predecessor
		xft.walkUpSuccessor(n, predecessor, successor)
	}

	// check max/min indices
	if xft.max.entry.Key() == key {
		xft.max = predecessor
	}

	if xft.min.entry.Key() == key {
		xft.min = successor
	}

	// decrement number of nodes
	xft.num--
}

// Delete will delete the provided keys from the trie.  If an entry
// associated with a provided key cannot be found, that deletion is
// a no-op.  Each deletion is an O(log M) operation.
func (xft *XFastTrie) Delete(keys ...uint64) {
	for _, key := range keys {
		xft.delete(key)
	}
}

// predecessor will find the node equal to or immediately less
// than the provided key.
func (xft *XFastTrie) predecessor(key uint64) *node {
	if xft.root == nil || xft.max == nil { // no successor if no nodes
		return nil
	}

	if key >= xft.max.entry.Key() {
		return xft.max
	}

	if key < xft.min.entry.Key() {
		return nil
	}

	n := xft.layers[xft.bits-1][key]
	if n != nil {
		return n
	}

	layer, n := binarySearchHashMaps(xft.layers, key)
	if n == nil && layer > 1 {
		return nil
	} else if n == nil {
		n = xft.root
	}

	if isInternal(n.children[0]) && isLeaf(n.children[1]) {
		return n.children[1].children[0]
	}
	return n.children[0]
}

// successor will find the node equal to or immediately more
// than the provided key.
func (xft *XFastTrie) successor(key uint64) *node {
	if xft.root == nil || xft.min == nil { // no successor if no nodes
		return nil
	}

	if key <= xft.min.entry.Key() {
		return xft.min
	}

	if key > xft.max.entry.Key() {
		return nil
	}

	n := xft.layers[xft.bits-1][key]
	if n != nil {
		return n
	}

	layer, n := binarySearchHashMaps(xft.layers, key)
	if n == nil && layer > 1 {
		return nil
	} else if n == nil {
		n = xft.root
	}

	if isInternal(n.children[1]) && isLeaf(n.children[0]) {
		return n.children[0].children[1]
	}
	return n.children[1]
}

// Successor will return an Entry which matches the provided
// key or its immediate successor.  Will return nil if a successor
// does not exist.  This is an O(log log M) operation.
func (xft *XFastTrie) Successor(key uint64) Entry {
	n := xft.successor(key)
	if n == nil {
		return nil
	}

	return n.entry
}

// Predecessor will return an Entry which matches the provided
// key or its immediate predecessor.  Will return nil if a predecessor
// does not exist.  This is an O(log log M) operation.
func (xft *XFastTrie) Predecessor(key uint64) Entry {
	n := xft.predecessor(key)
	if n == nil {
		return nil
	}

	return n.entry
}

// Iter will return an iterator that will iterate over all values
// equal to or immediately greater than the provided key.  Iterator
// will iterate successor relationships.
func (xft *XFastTrie) Iter(key uint64) *Iterator {
	return &Iterator{
		n:     xft.successor(key),
		first: true,
	}
}

// Get will return a value in the trie associated with the provided
// key if it exists.  Returns nil if the key does not exist.  This
// is expected to take O(1) time.
func (xft *XFastTrie) Get(key uint64) Entry {
	// only have to check the last hashmap for the provided
	// key.
	n := xft.layers[xft.bits-1][key]
	if n == nil {
		return nil
	}

	return n.entry
}

// New will construct a new X-Fast Trie with the given "size,"
// that is the size of the universe of the trie.  This expects
// a uint of some sort, ie, uint8, uint16, etc.  The size of the
// universe will be 2^n-1 and will affect the speed of all operations.
// IFC MUST be a uint type.
func New(ifc interface{}) *XFastTrie {
	xft := &XFastTrie{}
	xft.init(ifc)
	return xft
}