76 lines
2.8 KiB
Java
76 lines
2.8 KiB
Java
package DataStructures.Trees;
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import DataStructures.Trees.BinaryTree.Node;
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/**
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* Problem Statement
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* Ceil value for any number x in a collection is a number y which is either equal to x or the least greater number than x.
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*
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* Problem: Given a binary search tree containing positive integer values.
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* Find ceil value for a given key in O(lg(n)) time. In case if it is not present return -1.
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*
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* Ex.1. [30,20,40,10,25,35,50] represents level order traversal of a binary search tree. Find ceil for 10.
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* Answer: 20
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*
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* Ex.2. [30,20,40,10,25,35,50] represents level order traversal of a binary search tree. Find ceil for 22
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* Answer: 25
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*
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* Ex.2. [30,20,40,10,25,35,50] represents level order traversal of a binary search tree. Find ceil for 52
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* Answer: -1
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*/
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/**
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*
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* Solution 1:
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* Brute Force Solution:
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* Do an inorder traversal and save result into an array. Iterate over the array to get an element equal to or greater
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* than current key.
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* Time Complexity: O(n)
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* Space Complexity: O(n) for auxillary array to save inorder representation of tree.
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* <p>
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* <p>
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* Solution 2:
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* Brute Force Solution:
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* Do an inorder traversal and save result into an array.Since array is sorted do a binary search over the array to get an
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* element equal to or greater than current key.
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* Time Complexity: O(n) for traversal of tree and O(lg(n)) for binary search in array. Total = O(n)
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* Space Complexity: O(n) for auxillary array to save inorder representation of tree.
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* <p>
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* <p>
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* Solution 3: Optimal
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* We can do a DFS search on given tree in following fashion.
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* i) if root is null then return null because then ceil doesn't exist
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* ii) If key is lesser than root value than ceil will be in right subtree so call recursively on right subtree
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* iii) if key is greater than current root, then either
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* a) the root is ceil
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* b) ceil is in left subtree: call for left subtree. If left subtree returns a non null value then that will be ceil
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* otherwise the root is ceil
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*/
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public class CeilInBinarySearchTree {
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public static Node getCeil(Node root, int key) {
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if (root == null) {
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return null;
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}
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// if root value is same as key than root is the ceiling
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if (root.data == key) {
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return root;
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}
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// if root value is lesser than key then ceil must be in right subtree
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if (root.data < key) {
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return getCeil(root.right, key);
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}
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// if root value is greater than key then ceil can be in left subtree or if
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// it is not in left subtree then current node will be ceil
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Node result = getCeil(root.left, key);
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// if result is null it means that there is no ceil in children subtrees
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// and the root is the ceil otherwise the returned node is the ceil.
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return result == null ? root : result;
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}
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}
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