package algs33;
import stdlib.*;
import algs13.Queue;
/* ***********************************************************************
 *  Compilation:  javac RedBlackLiteBST.java
 *  Execution:    java RedBlackLiteBST < input.txt
 *  Dependencies: StdIn.java StdOut.java
 *  Data files:   http://algs4.cs.princeton.edu/33balanced/tinyST.txt
 *
 *  A symbol table implemented using a left-leaning red-black BST.
 *  This is the 2-3 version.
 *
 *  This implementation implements only put, get, and contains.
 *  See RedBlackBST.java for a full implementation including delete.
 *
 *
 *  % more tinyST.txt
 *  S E A R C H E X A M P L E
 *
 *  % java RedBlackLiteBST < tinyST.txt
 *  A 8
 *  C 4
 *  E 12
 *  H 5
 *  L 11
 *  M 9
 *  P 10
 *  R 3
 *  S 0
 *  X 7
 *
 *************************************************************************/

public class XRedBlackLiteBST<K extends Comparable<? super K>, V> {

	private static final boolean RED   = true;
	private static final boolean BLACK = false;

	private Node<K,V> root; // root of the BST
	private int N;          // number of key-value pairs in BST

	// BST helper node data type
	private static class Node<K,V> {
		public final K key;         // key
		public V val;         // associated data
		public Node<K,V> left, right;  // links to left and right subtrees
		public boolean color;     // color of parent link

		public Node(K key, V val, boolean color) {
			this.key = key;
			this.val = val;
			this.color = color;
		}
	}

	/* ***********************************************************************
	 *  Standard BST search
	 *************************************************************************/

	// return value associated with the given key, or null if no such key exists
	public V get(K key) { return get(root, key); }
	public V get(Node<K,V> x, K key) {
		while (x != null) {
			int cmp = key.compareTo(x.key);
			if      (cmp < 0) x = x.left;
			else if (cmp > 0) x = x.right;
			else              return x.val;
		}
		return null;
	}

	// is there a key-value pair in the symbol table with the given key?
	public boolean contains(K key) {
		return (get(key) != null);
	}


	/* ***********************************************************************
	 *  Red-black insertion
	 *************************************************************************/

	public void put(K key, V val) {
		root = insert(root, key, val);
		root.color = BLACK;
		assert check();
	}

	private Node<K,V> insert(Node<K,V> h, K key, V val) {
		if (h == null) {
			N++;
			return new Node<>(key, val, RED);
		}

		int cmp = key.compareTo(h.key);
		if      (cmp < 0) h.left  = insert(h.left,  key, val);
		else if (cmp > 0) h.right = insert(h.right, key, val);
		else              h.val   = val;

		// fix-up any right-leaning links
		if (isRed(h.right) && !isRed(h.left))      h = rotateLeft(h);
		if (isRed(h.left)  &&  isRed(h.left.left)) h = rotateRight(h);
		if (isRed(h.left)  &&  isRed(h.right))     flipColors(h);

		return h;
	}

	/* ***********************************************************************
	 *  red-black tree helper functions
	 *************************************************************************/

	// is node x red (and non-null) ?
	private boolean isRed(Node<K,V> x) {
		if (x == null) return false;
		return (x.color == RED);
	}

	// rotate right
	private Node<K,V> rotateRight(Node<K,V> h) {
		assert (h != null) && isRed(h.left);
		Node<K,V> x = h.left;
		h.left = x.right;
		x.right = h;
		x.color = h.color;
		h.color = RED;
		return x;
	}

	// rotate left
	private Node<K,V> rotateLeft(Node<K,V> h) {
		assert (h != null) && isRed(h.right);
		Node<K,V> x = h.right;
		h.right = x.left;
		x.left = h;
		x.color = h.color;
		h.color = RED;
		return x;
	}

	// precondition: two children are red, node is black
	// postcondition: two children are black, node is red
	private void flipColors(Node<K,V> h) {
		assert !isRed(h) && isRed(h.left) && isRed(h.right);
		h.color = RED;
		h.left.color = BLACK;
		h.right.color = BLACK;
	}


	/* ***********************************************************************
	 *  Utility functions
	 *************************************************************************/
	// return number of key-value pairs in symbol table
	public int size() { return N; }

	// is the symbol table empty?
	public boolean isEmpty() { return N == 0; }

	// height of tree (empty tree height = 0)
	public int height() { return height(root); }
	private int height(Node<K,V> x) {
		if (x == null) return 0;
		return 1 + Math.max(height(x.left), height(x.right));
	}

	// return the smallest key; null if no such key
	public K min() { return min(root); }
	private K min(Node<K,V> x) {
		K key = null;
		while (x != null) {
			key = x.key;
			x = x.left;
		}
		return key;
	}

	// return the largest key; null if no such key
	public K max() { return max(root); }
	private K max(Node<K,V> x) {
		K key = null;
		while (x != null) {
			key = x.key;
			x = x.right;
		}
		return key;
	}


	/* *********************************************************************
	 *  Iterate using an inorder traversal.
	 *  Iterating through N elements takes O(N) time.
	 ***********************************************************************/
	public Iterable<K> keys() {
		Queue<K> queue = new Queue<>();
		keys(root, queue);
		return queue;
	}

	private void keys(Node<K,V> x, Queue<K> queue) {
		if (x == null) return;
		keys(x.left, queue);
		queue.enqueue(x.key);
		keys(x.right, queue);
	}


	/* ***********************************************************************
	 *  Check integrity of red-black BST data structure
	 *************************************************************************/
	private boolean check() {
		if (!isBST())            StdOut.println("Not in symmetric order");
		if (!is23())             StdOut.println("Not a 2-3 tree");
		if (!isBalanced())       StdOut.println("Not balanced");
		return isBST() && is23() && isBalanced();
	}

	// does this binary tree satisfy symmetric order?
	// Note: this test also ensures that data structure is a binary tree since order is strict
	private boolean isBST() {
		return isBST(root, null, null);
	}

	// is the tree rooted at x a BST with all keys strictly between min and max
	// (if min or max is null, treat as empty constraint)
	// Credit: Bob Dondero's elegant solution
	private boolean isBST(Node<K,V> x, K min, K max) {
		if (x == null) return true;
		if (min != null && x.key.compareTo(min) <= 0) return false;
		if (max != null && x.key.compareTo(max) >= 0) return false;
		return isBST(x.left, min, x.key) && isBST(x.right, x.key, max);
	}

	// Does the tree have no red right links, and at most one (left)
	// red links in a row on any path?
	private boolean is23() { return is23(root); }
	private boolean is23(Node<K,V> x) {
		if (x == null) return true;
		if (isRed(x.right)) return false;
		if (x != root && isRed(x) && isRed(x.left))
			return false;
		return is23(x.left) && is23(x.right);
	}

	// do all paths from root to leaf have same number of black edges?
	private boolean isBalanced() {
		int black = 0;     // number of black links on path from root to min
		Node<K,V> x = root;
		while (x != null) {
			if (!isRed(x)) black++;
			x = x.left;
		}
		return isBalanced(root, black);
	}

	// does every path from the root to a leaf have the given number of black links?
	private boolean isBalanced(Node<K,V> x, int black) {
		if (x == null) return black == 0;
		if (!isRed(x)) black--;
		return isBalanced(x.left, black) && isBalanced(x.right, black);
	}


	/* ***********************************************************************
	 *  Test client
	 *************************************************************************/
	public static void main(String[] args) {

		String test = "S E A R C H E X A M P L E";
		String[] keys = test.split(" ");
		XRedBlackLiteBST<String, Integer> st = new XRedBlackLiteBST<>();
		for (int i = 0; i < keys.length; i++)
			st.put(keys[i], i);

		StdOut.println("size = " + st.size());
		StdOut.println("min  = " + st.min());
		StdOut.println("max  = " + st.max());
		StdOut.println();


		// print keys in order using allKeys()
		StdOut.println("Testing keys()");
		StdOut.println("--------------------------------");
		for (String s : st.keys())
			StdOut.println(s + " " + st.get(s));
		StdOut.println();

		// insert N elements in order if one command-line argument supplied
		if (args.length == 0) return;
		int N = Integer.parseInt(args[0]);
		XRedBlackLiteBST<Integer, Integer> st2 = new XRedBlackLiteBST<>();
		for (int i = 0; i < N; i++) {
			st2.put(i, i);
			int h = st2.height();
			StdOut.println("i = " + i + ", height = " + h + ", size = " + st2.size());
		}


		StdOut.println("size = " + st2.size());
	}
}