## Wednesday, January 14, 2015

### Target

Problem:

 TopCoder Problem Statement - Target Single Round Match 633 Round 1 - Division II, Level One

Overview:

Draw a series of concentric squares.

Java Source:
```01: public class Target {
02:
03:     private static final char OFF = ' ';
04:     private static final char ON = '#';
05:
06:     public String[] draw(int n) {
07:
08:         // Create a char[][] and initialize it.
09:         char[][] target = new char[n][n];
10:
11:         for (int y = 0; y < n; y++) {
12:             for (int x = 0; x < n; x++) {
13:                 target[y][x] = OFF;
14:             }
15:         }
16:
17:         // Start the recursive calls.
18:         drawSquare(target, 0);
19:
20:         // Convert the char[][] into a String[] and return it.
21:         String[] result = new String[n];
22:
23:         for (int y = 0; y < n; y++) {
24:             result[y] = new String(target[y]);
25:         }
26:
27:         return result;
28:     }
29:
30:     private void drawSquare(char[][] target, int topOrLeft) {
31:
32:         int bottomOrRight = target.length - topOrLeft - 1;
33:
34:         if (topOrLeft < bottomOrRight) {
35:             drawSquare(target, topOrLeft + 2);
36:         }
37:
38:         for (int i = topOrLeft; i <= bottomOrRight; i++) {
39:             target[topOrLeft][i] = ON;      // Top Row
40:             target[bottomOrRight][i] = ON;  // Bottom Row
41:             target[i][topOrLeft] = ON;      // Left Edge
42:             target[i][bottomOrRight] = ON;  // Right Edge
43:         }
44:
45:     }
46: }
```
Notes:

The solution uses recursion to draw all of the inner squares, and then completes the outer square. The draw() method simply initializes a new 2-dimensional char array and then calls drawSquare(). When drawSquare() completes, it just converstion the char[][] back to a String[] and returns it. The call to drawSquare() passes the char[][] and the index 0, which indicates the top-left corner of the outermost square.

The drawSquare() method handles all the work on turning on the appropriate locations. It is passed the top-left corner of the current square, and uses that and the size of the char[][] to determine the location of the bottom-right corner. If these values are not the same, recursively call drawSquare moving the top-left corner in and down by 2.

As the recursion unwinds, we have a loop to draw each of the sides. Note the index representing the top edge will always equal the index of the left edge. So one variable (topOrLeft) will suffice. Similarly, the index for the bottom edge will be the same as the index of the right edge - thus bottomOrRight.