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Problem solution · Java

CCC 2004 S4 - Space Turtle

CCC 2004 S4 - Space Turtle: a Java solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Sliding window or two pointers
Source
CCCSolutions
Length
105 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

Sliding window or two pointers

For CCC 2004 S4 - Space Turtle, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 105 lines of Java from the credited upstream file ccc04s4.java.
  • The implementation visibly relies on sequence storage.
  • 1 loop block detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 2004 S4 - Space Turtle · JavaJava
Use this to learn the idea, then write your own version.
// CCC 2004// Problem S4 Space Turtle//// This algoritm is due to Richard Peng.//// The idea is: given the position of a space turtle and a// golden shell, determine the closest the turtle comes to the shell.// (The turtle moves forward than turns left, right up and down.)//// Peng's insight is to ALWAYS think of the turtle at 0,0,0.// This puts the shell at x,y,z.//// As an example lets think only in 2d: assume the shell is at (5,2)// moving forward 2 is the same as moving the shell to (3,2)// that is, each movement is: x = x - d//// Now here's Peng's second great insight: the closest the turtle comes// to the shell will be at the ends points of the movement, OR if he zooms by// so to speak, its at x = 0. EG: if the shell is at (5,2) and the turtle moves 8// and the new shell location is (-3,2) the closest the turtle was, was// 2. At one point the shell was at (0,2)// in 3D terms if the new x and x are different signs, the closest distance// is sqrt(y^2 + z^2) versus the normal sqrt(x^2+ y^2 + z^2)//// Turning is a bit of a mind bending exercise, but again not so bad// if you imagine it in only 2D: if the shell is at (5,2) and// the turtle turns left the shell will be at (2,-5)// that is (x,y,z) becomes (y,-x,z) (z is unaffected by a left or right turn)// DRAW a grid, plot the point, and you'll see!// similarly right means (x,y,z) becomes (-y,x,z)//           down  means (x,y,z) becomes (-z,y,x) (y is unaffected by down/up)//       and up    means (x,y,z) becomes (z,y,-x) import hsa.*; public class S4SpaceTurtlePeng{    static Console c;    static double tx, ty, tz;    static double sx, sy, sz;    static double x, y, z, newX, t;     public static void main (String[] args)    {	c = new Console ();	double closest, distance, d;	char turn;	String file; 	c.print ("file name: ");	file = c.readString ();	TextInputFile fi = new TextInputFile (file); 	tx = fi.readDouble (); // turtle coordinates	ty = fi.readDouble ();	tz = fi.readDouble (); 	sx = fi.readDouble (); // shell coordinates	sy = fi.readDouble ();	sz = fi.readDouble (); 	x = sx - tx;	y = sy - ty;	z = sz - tz; 	closest = x * x + y * y + z * z;	do	{	    distance = fi.readDouble ();	    turn = fi.readChar (); 	    newX = x - distance; 	    if (newX * x < 0)		closest = Math.min (closest, y * y + z * z);	    else		closest = Math.min (closest, newX * newX + y * y + z * z);	    x = newX;	    t = x;	    if (turn == 'L')	    {		x = y;		y = -t;	    }	    else if (turn == 'R')	    {		x = -y;		y = t;	    }	    else if (turn == 'U')	    {		x = z;		z = -t;	    }	    else	    {		x = -z;		z = t;	    }	}	while (turn != 'E');	c.println ((int) ((Math.sqrt (closest) * 100) + 0.5) / 100.0);    }}

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