JavaAlgorithms/ProjectEuler/Problem12.java

package ProjectEuler;
/**
 * The sequence of triangle numbers is generated by adding the natural numbers.
 * So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
 * The first ten terms would be:
 * <p>
 * 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
 * <p>
 * Let us list the factors of the first seven triangle numbers:
 * <p>
 * 1: 1
 * 3: 1,3
 * 6: 1,2,3,6
 * 10: 1,2,5,10
 * 15: 1,3,5,15
 * 21: 1,3,7,21
 * 28: 1,2,4,7,14,28
 * We can see that 28 is the first triangle number to have over five divisors.
 * <p>
 * What is the value of the first triangle number to have over five hundred divisors?
 * <p>
 * link: https://projecteuler.net/problem=12
 */
public class Problem12 {

    /**
     * Driver Code
     */
    public static void main(String[] args) {
        assert solution1(500) == 76576500;
    }

    /* returns the nth triangle number; that is, the sum of all the natural numbers less than, or equal to, n */
    public static int triangleNumber(int n) {
        int sum = 0;
        for (int i = 0; i <= n; i++)
            sum += i;
        return sum;
    }

    public static int solution1(int number) {
        int j = 0; // j represents the jth triangle number
        int n = 0; // n represents the triangle number corresponding to j
        int numberOfDivisors = 0; // number of divisors for triangle number n

        while (numberOfDivisors <= number) {

            // resets numberOfDivisors because it's now checking a new triangle number
            // and also sets n to be the next triangle number
            numberOfDivisors = 0;
            j++;
            n = triangleNumber(j);

            // for every number from 1 to the square root of this triangle number,
            // count the number of divisors
            for (int i = 1; i <= Math.sqrt(n); i++)
                if (n % i == 0)
                    numberOfDivisors++;

            // 1 to the square root of the number holds exactly half of the divisors
            // so multiply it by 2 to include the other corresponding half
            numberOfDivisors *= 2;
        }
        return n;
    }
}