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Problem solution · Java

CCC 2010 S2 - Huffman Encoding

CCC 2010 S2 - Huffman Encoding: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Direct simulation
Source
CCCSolutions
Length
50 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For CCC 2010 S2 - Huffman Encoding, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 50 lines of Java from the credited upstream file ccc10s2.java.
  • The implementation visibly relies on sequence storage.
  • 3 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2010 S2 - Huffman Encoding · JavaJava
Use this to learn the idea, then write your own version.
// S2 2010: Huffman Encoding//// arrays, searching and string handling// quite straight forward.// import java.awt.*;import hsa.*; public class S22010{     public static void main (String[] args)    {	TextInputFile c;	c = new TextInputFile ("s2.6.in"); 	int k, i;	char[] letter = new char [20];	String[] code = new String [20];	String binary;	String answer; 	// input section	k = c.readInt ();	for (i = 0 ; i < k ; i++)	{	    letter [i] = c.readChar ();	    code [i] = c.readString ();	}	binary = c.readString (); 	//Translate	answer = "";	while (binary.length () > 0)	{	    // simple search as one code MUST be found	    i = 0;	    while (!binary.startsWith (code [i]))		i++; 	    // add letter to answer and remove that code from binary	    answer = answer + letter [i];	    binary = binary.substring (code [i].length ());	} 	System.out.println (answer);    }}

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