TopCoder Problem Statement - Cubism |

Single Round Match 172 Round 1 - Division II, Level Three |

Count the number of lines that can be formed in a cube where the line's length equals the length of a side of the cure, and all location along the line match a given color.

1 2 3 public class Cubism { 4 5 public int lines(String[] lattice, String color) { 6 7 int size = lattice.length; 8 char c = "black".equals(color) ? 'B' : 'W'; 9 10 /* 11 * The problem becomes much easier if we transform the 12 * given String[] into a 3-Dimensional array 13 * of chars. 14 */ 15 char[][][] cube = new char[size][size][size]; 16 17 for (int x = 0; x < size; x++) { 18 for (int y = 0; y < size; y++) { 19 for (int z = 0; z < size; z++) { 20 cube[x][y][z] = lattice[x].charAt((y * size) + z); 21 } 22 } 23 } 24 25 int numLines = 0; 26 27 /* 28 * Count the number of lines that begin at each location. 29 */ 30 for (int x = 0; x < size; x++) { 31 for (int y = 0; y < size; y++) { 32 for (int z = 0; z < size; z++) { 33 numLines += countLinesStartingAt(cube, x, y, z, c); 34 } 35 } 36 } 37 38 /* 39 * Each line will be counted twice, once starting from 40 * each end point, so divide the result in half. 41 */ 42 return numLines / 2; 43 } 44 45 /* 46 * Count the number of lines starting from the given coordinate. The line 47 * must be of the given color c. 48 */ 49 private static int countLinesStartingAt(char[][][] cube, 50 int x, int y, int z, 51 char c) { 52 53 int numLines = 0; 54 55 /* 56 * We'll generate the 27 directions that the row could stretch out in 57 * by using deltaX, deltaY, and deltaZ. 58 */ 59 for (int deltaX = -1; deltaX < 2; deltaX++) { 60 for (int deltaY = -1; deltaY < 2; deltaY++) { 61 for (int deltaZ = -1; deltaZ < 2; deltaZ++) { 62 if (isLine(cube, x, y, z, deltaX, deltaY, deltaZ, c)) { 63 numLines++; 64 } 65 } 66 } 67 } 68 return numLines; 69 } 70 71 /* 72 * Returns true if a line originating at (x,y,z) of size cube.length and moving 73 * in the direction given by deltaX, deltaY, and deltaZ can be formed. 74 * All position along the line must match color c. 75 */ 76 private static boolean isLine(char[][][] cube, 77 int x, int y, int z, 78 int deltaX, int deltaY, int deltaZ, 79 char c) { 80 81 // It's not a line if we never move from the origin. 82 if ((deltaX == 0) && (deltaY == 0) && (deltaZ == 0)) return false; 83 84 for (int i = 0; i < cube.length; i++) { 85 86 // Check that we stay within bounds. 87 if ((x < 0) || (x >= cube.length)) return false; 88 if ((y < 0) || (y >= cube.length)) return false; 89 if ((z < 0) || (z >= cube.length)) return false; 90 91 // Check the color matches. 92 if (cube[x][y][z] != c) return false; 93 94 // Move to the next location. 95 x += deltaX; 96 y += deltaY; 97 z += deltaZ; 98 } 99 100 return true; 101 } 102 }

This problem becomes very easy once you break it down into smaller steps. The first task is to take the unwieldly String[] and convert it into an easy to use 3 dimensional array of chars. This is handled on lines 15-23. Nothing trick here, except for a little bit of math to get the right character out of the String.

With our 3-D array, we can now loop through each position and count the number of lines starting at that position. Note that we'll count each row twice. Once starting from each end and going to the other.

The countLinesStartingAt() method does just that. Given a 3-D coordinate, it returns the number of lines starting from that position. Rather than writing out the 27 directions that we could move out from for any given position, I have three nested loops and the variables deltaX, deltaY, and deltaZ. Each of these variables moves from -1 to 1.

With that, we call isLine() providing the staring point, direction, and color. isLine() just checks that we stay within the confines of the cube and stay on the correct color as we move along the line.

Nothing really difficult here. It could probably have been a Division 2 Level 2.

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