Project Euler > Problem 67 > Maximum path sum II (Java Solution)

Problem:

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom in triangle.txt (right click and 'Save Link/Target As...'), a 15K text file containing a triangle with one-hundred rows.

NOTE: This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are 299 altogether! If you could check one trillion (1012) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;o)

Solution:

21124

Code:
The solution may include methods that will be found here: Library.java .

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 17
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p017 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p017().run());
	}
	
	
	public String run() {
		int sum = 0;
		for (int i = 1; i <= 1000; i++)
			sum += toEnglish(i).length();
		return Integer.toString(sum);
	}
	
	
	private static String[] ONES = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};  // Requires 0 <= n <= 9
	private static String[] TEENS = {"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};
	private static String[] TENS = {"twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
	
	
	// Requires 0 <= n <= 99999
	private static String toEnglish(int n) {
		if (n < 0 || n > 99999)
			throw new IllegalArgumentException();
		
		if (n < 100)
			return tens(n);
		else {
			String big = "";
			if (n >= 1000)
				big += tens(n / 1000) + "thousand";
			if (n / 100 % 10 != 0)
				big += ONES[n / 100 % 10] + "hundred";
			
			return big + (n % 100 != 0 ? "and" + tens(n % 100) : "");
		}
	}
	
	
	// Requires 0 <= n <= 99
	private static String tens(int n) {
		if (n < 10)
			return ONES[n];
		else if (n < 20)  // Teens
			return TEENS[n - 10];
		else
			return TENS[n / 10 - 2] + (n % 10 != 0 ? ONES[n % 10] : "");
	}
	
}