[Solved] Modulus of a very large number

Question

BigInteger has the divideAndRemainder(…) method, one that returns a BigInteger array, the first item the division result, and the second, the remainder (which is what mod really does in Java).

Update

Including comment by Mark Dickinson to other answer:

There’s a much simpler linear-time algorithm: set acc to 0, then for each digit d in x in turn (left to right), convert d to an integer and set acc = (acc * 10 + d) % y. Once the digits are consumed, acc is your result.

Implemented all 3 algorithms as described:

private static int testMarkDickinson(String x, int y) {
    int acc = 0;
    for (int i = 0; i < x.length(); i++)
        acc = (acc * 10 + x.charAt(i) - '0') % y;
    return acc;
}

private static int testHovercraftFullOfEels(String x, int y) {
    return new BigInteger(x).divideAndRemainder(BigInteger.valueOf(y))[1].intValue();
}

private static int testMrM(String x, int y) {
    String s = x;
    while (s.length() >= 7) {
        int len = Math.min(9, s.length());
        s = Integer.parseInt(s.substring(0, len)) % y + s.substring(len);
    }
    return Integer.parseInt(s) % y;
}

Testing with seeded random number for verifiable test (didn’t want to hardcode a 98765 digit number):

public static void main(String[] args) {
    Random r = new Random(98765);
    char[] buf = new char[98765];
    for (int i = 0; i < buf.length; i++)
        buf[i] = (char)('0' + r.nextInt(10));

    String x = new String(buf);
    int y = 98765;

    System.out.println(testMarkDickinson(x, y));
    System.out.println(testHovercraftFullOfEels(x, y));
    System.out.println(testMrM(x, y));

    long test1nano = 0, test2nano = 0, test3nano = 0;
    for (int i = 0; i < 10; i++) {
        long nano1 = System.nanoTime();
        testMarkDickinson(x, y);
        long nano2 = System.nanoTime();
        testHovercraftFullOfEels(x, y);
        long nano3 = System.nanoTime();
        testMrM(x, y);
        long nano4 = System.nanoTime();
        test1nano += nano2 - nano1;
        test2nano += nano3 - nano2;
        test3nano += nano4 - nano3;
    }
    System.out.printf("%11d%n%11d%n%11d%n", test1nano, test2nano, test3nano);
}

Output:

All 3 produced same result, but there’s a clear difference in performance, and Mr. M’s “better solution” is the worst of them all.

Accepted Answer