1d Range Search
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Prerequisite: Fundamentals of Sweep Line Algorithm
In Line Sweep Algorithms, you would often have to do a Range Search or Range Query. The data structure you'd choose would determine how efficient the range search operation is going to be. Using a Binary Search Tree would guarantee logarithmic time range search operation.
For a Range Search Query you do an Inorder traversal, and at every node traversed you check if the value of the node is in the range you are looking for.
- If the value of the the node we are in is in the range we are looking for, then print the value of the node and traverse its left and right child nodes.
-
If the value of the node is less than the lower limit of the range (say, the range is [lower, upper]),
then traverse only the right child node of the node.
Since the value of the node is less that the lower limit of the range: we are not printing the value of the node we are at because the value is not in the range we are looking for, and also not visiting the left child node of this node because the value of the left child node would be even lesser, so definitely not in the range. -
If the value of the node is greater than the upper limit of the range (say, the range is [lower, upper]),
then traverse only the left child node of the node.
Since the value of the node is greater that the upper limit of the range: we are not printing the value of the node we are at because the value is not in the range we are looking for, and also not visiting the right child node of this node because the value of the right child node would be even greater, so definitely not in the range.
// Returns all keys in the symbol table in the given range an Iterable.
public Iterable< Key > keys(Key lo, Key hi) {
if (lo == null) throw new IllegalArgumentException("first argument to keys() is null");
if (hi == null) throw new IllegalArgumentException("second argument to keys() is null");
Queue< Key > queue = new Queue< Key >();
keys(root, queue, lo, hi);
return queue;
}
private void keys(Node x, Queue< Key > queue, Key lo, Key hi) {
if (x == null) return;
int cmplo = lo.compareTo(x.key);
int cmphi = hi.compareTo(x.key);
if (cmplo < 0) keys(x.left, queue, lo, hi);
if (cmplo <= 0 && cmphi >= 0) queue.enqueue(x.key);
if (cmphi > 0) keys(x.right, queue, lo, hi);
}
For getting the total number of nodes in the tree which has value in the given range, we can use the simple concept as below:
Number of nodes in range =
(total number of nodes with values less than or equal to higher-end limit of range)
- (total number of nodes with values less than lower-end of range)
// Returns the number of keys in the symbol table in the given range.
public int size(Key lo, Key hi) {
if (lo == null) throw new IllegalArgumentException("first argument to size() is null");
if (hi == null) throw new IllegalArgumentException("second argument to size() is null");
if (lo.compareTo(hi) > 0) return 0;
if (contains(hi)) return rank(hi) - rank(lo) + 1;
else return rank(hi) - rank(lo);
}
// Return the number of keys in the symbol table strictly less than given key.
public int rank(Key key) {
if (key == null) throw new IllegalArgumentException("argument to rank() is null");
return rank(key, root);
}
// Number of keys in the subtree less than key.
private int rank(Key key, Node x) {
if (x == null) return 0;
int cmp = key.compareTo(x.key);
if (cmp < 0) return rank(key, x.left);
else if (cmp > 0) return 1 + size(x.left) + rank(key, x.right);
else return size(x.left);
}
Time Complexity:
Range Search Query: O(logN + R) where N = total number of items in the Binary Search Tree and R = number of items returned by the Range Search Query.Below is a O(logN + R) BST Range Search implementation:
Complete code:
public class BST< Key extends Comparable< Key >, Value > {
private Node root; // root of BST
private class Node {
private Key key; // sorted by key
private Value val; // associated data
private Node left, right; // left and right subtrees
private int size; // number of nodes in subtree
public Node(Key key, Value val, int size) {
this.key = key;
this.val = val;
this.size = size;
}
}
// Initializes an empty symbol table.
public BST() {
}
// Returns all keys in the symbol table in the given range an Iterable.
public Iterable< Key > keys(Key lo, Key hi) {
if (lo == null) throw new IllegalArgumentException("first argument to keys() is null");
if (hi == null) throw new IllegalArgumentException("second argument to keys() is null");
Queue< Key > queue = new Queue< Key >();
keys(root, queue, lo, hi);
return queue;
}
// Returns the number of keys in the symbol table in the given range.
public int size(Key lo, Key hi) {
if (lo == null) throw new IllegalArgumentException("first argument to size() is null");
if (hi == null) throw new IllegalArgumentException("second argument to size() is null");
if (lo.compareTo(hi) > 0) return 0;
if (contains(hi)) return rank(hi) - rank(lo) + 1;
else return rank(hi) - rank(lo);
}
private void keys(Node x, Queue< Key > queue, Key lo, Key hi) {
if (x == null) return;
int cmplo = lo.compareTo(x.key);
int cmphi = hi.compareTo(x.key);
if (cmplo < 0) keys(x.left, queue, lo, hi);
if (cmplo <= 0 && cmphi >= 0) queue.enqueue(x.key);
if (cmphi > 0) keys(x.right, queue, lo, hi);
}
// Return key in BST rooted at x of given rank.
// Precondition: rank is in legal range.
private Key select(Node x, int rank) {
if (x == null) return null;
int leftSize = size(x.left);
if (leftSize > rank) return select(x.left, rank);
else if (leftSize < rank) return select(x.right, rank - leftSize - 1);
else return x.key;
}
// Return the number of keys in the symbol table strictly less than {@code key}.
public int rank(Key key) {
if (key == null) throw new IllegalArgumentException("argument to rank() is null");
return rank(key, root);
}
// Number of keys in the subtree less than key.
private int rank(Key key, Node x) {
if (x == null) return 0;
int cmp = key.compareTo(x.key);
if (cmp < 0) return rank(key, x.left);
else if (cmp > 0) return 1 + size(x.left) + rank(key, x.right);
else return size(x.left);
}
// Returns all keys in the symbol table as an Iterable.
// To iterate over all of the keys in the symbol table
// use the foreach notation: for (Key key : st.keys()).
public Iterable< Key > keys() {
if (isEmpty()) return new Queue();
return keys(min(), max());
}
public boolean contains(Key key) {
if (key == null) throw new IllegalArgumentException("argument to contains() is null");
return get(key) != null;
}
public Value get(Key key) {
return get(root, key);
}
private Value get(Node x, Key key) {
if (key == null) throw new IllegalArgumentException("calls get() with a null key");
if (x == null) return null;
int cmp = key.compareTo(x.key);
if (cmp < 0) return get(x.left, key);
else if (cmp > 0) return get(x.right, key);
else return x.val;
}
// Returns true if this symbol table is empty.
public boolean isEmpty() {
return size() == 0;
}
// Returns the number of key-value pairs in this symbol table
public int size() {
return size(root);
}
// return number of key-value pairs in BST rooted at x
private int size(Node x) {
if (x == null) return 0;
else return x.size;
}
// Inserts the specified key-value pair into the symbol table, overwriting the old
// value with the new value if the symbol table already contains the specified key.
// Deletes the specified key (and its associated value) from this symbol table
// if the specified value is null
public void put(Key key, Value val) {
if (key == null) throw new IllegalArgumentException("calls put() with a null key");
if (val == null) {
delete(key);
return;
}
root = put(root, key, val);
assert check();
}
private Node put(Node x, Key key, Value val) {
if (x == null) return new Node(key, val, 1);
int cmp = key.compareTo(x.key);
if (cmp < 0) x.left = put(x.left, key, val);
else if (cmp > 0) x.right = put(x.right, key, val);
else x.val = val;
x.size = 1 + size(x.left) + size(x.right);
return x;
}
// Removes the smallest key and associated value from the symbol table
public void deleteMin() {
if (isEmpty()) throw new NoSuchElementException("Symbol table underflow");
root = deleteMin(root);
assert check();
}
private Node deleteMin(Node x) {
if (x.left == null) return x.right;
x.left = deleteMin(x.left);
x.size = size(x.left) + size(x.right) + 1;
return x;
}
// Removes the largest key and associated value from the symbol table
public void deleteMax() {
if (isEmpty()) throw new NoSuchElementException("Symbol table underflow");
root = deleteMax(root);
assert check();
}
private Node deleteMax(Node x) {
if (x.right == null) return x.left;
x.right = deleteMax(x.right);
x.size = size(x.left) + size(x.right) + 1;
return x;
}
public void delete(Key key) {
if (key == null) throw new IllegalArgumentException("calls delete() with a null key");
root = delete(root, key);
assert check();
}
private Node delete(Node x, Key key) {
if (x == null) return null;
int cmp = key.compareTo(x.key);
if (cmp < 0) x.left = delete(x.left, key);
else if (cmp > 0) x.right = delete(x.right, key);
else {
if (x.right == null) return x.left;
if (x.left == null) return x.right;
Node t = x;
x = min(t.right);
x.right = deleteMin(t.right);
x.left = t.left;
}
x.size = size(x.left) + size(x.right) + 1;
return x;
}
// Returns the smallest key in the symbol table
public Key min() {
if (isEmpty()) throw new NoSuchElementException("calls min() with empty symbol table");
return min(root).key;
}
private Node min(Node x) {
if (x.left == null) return x;
else return min(x.left);
}
// Returns the largest key in the symbol table
public Key max() {
if (isEmpty()) throw new NoSuchElementException("calls max() with empty symbol table");
return max(root).key;
}
private Node max(Node x) {
if (x.right == null) return x;
else return max(x.right);
}
public Key floor(Key key) {
if (key == null) throw new IllegalArgumentException("argument to floor() is null");
if (isEmpty()) throw new NoSuchElementException("calls floor() with empty symbol table");
Node x = floor(root, key);
if (x == null) throw new NoSuchElementException("argument to floor() is too small");
else return x.key;
}
private Node floor(Node x, Key key) {
if (x == null) return null;
int cmp = key.compareTo(x.key);
if (cmp == 0) return x;
if (cmp < 0) return floor(x.left, key);
Node t = floor(x.right, key);
if (t != null) return t;
else return x;
}
public Key floor2(Key key) {
Key x = floor2(root, key, null);
if (x == null) throw new NoSuchElementException("argument to floor() is too small");
else return x;
}
private Key floor2(Node x, Key key, Key best) {
if (x == null) return best;
int cmp = key.compareTo(x.key);
if (cmp < 0) return floor2(x.left, key, best);
else if (cmp > 0) return floor2(x.right, key, x.key);
else return x.key;
}
public Key ceiling(Key key) {
if (key == null) throw new IllegalArgumentException("argument to ceiling() is null");
if (isEmpty()) throw new NoSuchElementException("calls ceiling() with empty symbol table");
Node x = ceiling(root, key);
if (x == null) throw new NoSuchElementException("argument to floor() is too large");
else return x.key;
}
private Node ceiling(Node x, Key key) {
if (x == null) return null;
int cmp = key.compareTo(x.key);
if (cmp == 0) return x;
if (cmp < 0) {
Node t = ceiling(x.left, key);
if (t != null) return t;
else return x;
}
return ceiling(x.right, key);
}
public Key select(int rank) {
if (rank < 0 || rank >= size()) {
throw new IllegalArgumentException("argument to select() is invalid: " + rank);
}
return select(root, rank);
}
public int height() {
return height(root);
}
private int height(Node x) {
if (x == null) return -1;
return 1 + Math.max(height(x.left), height(x.right));
}
public Iterable levelOrder() {
Queue keys = new Queue();
Queue queue = new Queue();
queue.enqueue(root);
while (!queue.isEmpty()) {
Node x = queue.dequeue();
if (x == null) continue;
keys.enqueue(x.key);
queue.enqueue(x.left);
queue.enqueue(x.right);
}
return keys;
}
/*************************************************************************
* Check integrity of BST data structure.
***************************************************************************/
private boolean check() {
if (!isBST()) StdOut.println("Not in symmetric order");
if (!isSizeConsistent()) StdOut.println("Subtree counts not consistent");
if (!isRankConsistent()) StdOut.println("Ranks not consistent");
return isBST() && isSizeConsistent() && isRankConsistent();
}
// 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 x, Key min, Key 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);
}
// are the size fields correct?
private boolean isSizeConsistent() { return isSizeConsistent(root); }
private boolean isSizeConsistent(Node x) {
if (x == null) return true;
if (x.size != size(x.left) + size(x.right) + 1) return false;
return isSizeConsistent(x.left) && isSizeConsistent(x.right);
}
// check that ranks are consistent
private boolean isRankConsistent() {
for (int i = 0; i < size(); i++)
if (i != rank(select(i))) return false;
for (Key key : keys())
if (key.compareTo(select(rank(key))) != 0) return false;
return true;
}
public static void main(String[] args) {
BST< String, Integer > st = new BST< String, Integer >();
for (int i = 0; !StdIn.isEmpty(); i++) {
String key = StdIn.readString();
st.put(key, i);
}
for (String s : st.levelOrder())
System..out.println(s + " " + st.get(s));
StdOut.println();
for (String s : st.keys())
System.out.println(s + " " + st.get(s));
}
}
Related Must-Read Topics:
- Closest Pair
- Convex Hull
-
Dual Line Sweep
& Rectangles Union -
2D Intervals Union
& Skyline Problem - Overlapping 1D Intervals
- Merging Overlapping 1D Intervals
- Separating Overlapping 1D Intervals
