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Showing posts from April, 2018

The WeakReference class, monitoring memory leak and garbage collection in a Java application

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 Below is a Stack implementation that uses an internal resizeable array structure.  public class MyStack< T > implements Stack< T > { private static final int CAPACITY = 100 ; private Object[] array ; private int pos = 0 ; public MyStack () { this . array = new Object[ CAPACITY ] ; } @Override public void push ( T item) { if ( pos >= array . length / 2 ) { Object[] newArray = new Object[ pos * 2 ] ; System. arraycopy ( array , 0 , newArray , 0 , array . length ) ; array = newArray ; } array [ pos ++] = item ; } @Override public T pop () { if (isEmpty()) { throw new RuntimeException( "empty stack" ) ; } @SuppressWarnings ( "unchecked" ) T item = ( T ) array [ pos - 1 ] ; pos -= 1 ; return item ; } @Override @SuppressWarnings ( "unchecked" ) public T peek...
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My Crappy Looking Solution to "Binary Tree Common Ancestor" Problem

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I was able to solve this problem and make all the tests pass: https://www.hackerrank.com/challenges/binary-search-tree-lowest-common-ancestor/problem But I'm not happy with the code, so later I need to refactor the code. Here it is: static Node lca(Node root, int v1, int v2) { ArrayList result = new ArrayList<>(); Node result2 = traverse (root, v1, v2, result); if (result.isEmpty()) return result2; return result.get( 0 ); } private static Node traverse(Node root, int v1, int v2, ArrayList result) { if (!result.isEmpty()) { return result.get( 0 ); } if (root. left == null && root. right == null ) { return root; } if (root. data == v1 || root. data == v2) { return root; } Node node1 = null , node2 = null ; if (root. left != null ) node1 = traverse (root. left , v1, v2, result); if (root. right != null ) node2 = traverse (root. right , v1, v...
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Trie Data Structure and Finding Patterns in a Collection of Words

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 I faced a very hard problem recently about finding substrings based on a collection of patterns.  Example For  words = ["Apple", "Melon", "Orange", "Watermelon"]  and  parts = ["a", "mel", "lon", "el", "An"] , the output should be  findSubstrings(words, parts) = ["Apple", "Me[lon]", "Or[a]nge", "Water[mel]on"] . While  "Watermelon"  contains three substrings from the  parts  array,  "a" ,  "mel" , and  "lon" ,  "mel"  is the longest substring that appears first in the string. Note: Number of words and number of parts can be huge! A brute force method or careless usage of Java String methods would make it impossible to handle a large number of words and parts (patterns). Solution Approach Whenever we want to search patterns inside a string, we should think about the Trie data structure:  https://en.wikipedia....
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A Graph Application in Java: Using WordNet to Find Outcast Words

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I'll explain our intent with an example: Let's say we have the following words: horse, zebra, cat, bear, table Which one is the outcast? The table.. But how can a computer system find this? We are talking about the whole English language and any word can be in this list. The number of words in the list can be huge, so the solution should use appropriate data structures and algorithms.  We'll use WordNet which is a graph of English words. It contains the semantic relationships between words and our concern is the "is a" relationship. For our example, we know that horse and cat are both animals. "Table" is not an animal and it is the outcast in this situation. But we even have the ability to tell that horse and zebra is very close. Below is a unit test that shows a high level view of our application in Java: public class OutcastTest {     private WordNet wordNet;     private Outcast outcast;     @Before ...
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My Crappy Looking Solution to "Binary Tree Common Ancestor" Problem

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I was able to solve this problem and make all the tests pass: https://www.hackerrank.com/challenges/binary-search-tree-lowest-common-ancestor/problem But I'm not happy with the code, so later I need to refactor the code. Here it is: static Node lca(Node root, int v1, int v2) { ArrayList result = new ArrayList<>(); Node result2 = traverse (root, v1, v2, result); if (result.isEmpty()) return result2; return result.get( 0 ); } private static Node traverse(Node root, int v1, int v2, ArrayList result) { if (!result.isEmpty()) { return result.get( 0 ); } if (root. left == null && root. right == null ) { return root; } if (root. data == v1 || root. data == v2) { return root; } Node node1 = null , node2 = null ; if (root. left != null ) node1 = traverse (root. left , v1, v2, result); if (root. right != null ) node2 = traverse (root. right , v1, v...
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