Discussion on: Challenge: find 'Kaprekar numbers'

[Thomas Much]

Ok, here's a Java 8 version:

import java.util.function.IntPredicate;
import java.util.stream.IntStream;

public class FirstEightKaprekarNumbers {

  public static void main(String... args) {

    IntPredicate isKaprekar = i -> {
      String square = String.valueOf(i*i);
      String padded = (square.length() % 2 == 0) ? square : "0" + square;
      int halfLen = padded.length() / 2;
      int left = Integer.parseInt(padded.substring(0, halfLen));
      int right = Integer.parseInt(padded.substring(halfLen));
      return i == left + right;
    };

    IntStream.iterate(1, i -> i + 1).filter(isKaprekar).limit(8).forEach(System.out::println);
  }
}

[Thomas Much]

As Heiko points out, the solution was incorrect, because the split was always done in the middle of the string...

So, here's a (hopefully correct) Java 9 solution (Java 9 because I use takeWhile):

import java.util.function.IntPredicate;
import java.util.stream.IntStream;

public class FirstSixteenKaprekarNumbers {

  public static void main(String... args) {

    IntPredicate isKaprekar = i -> {
      long square = i * i;
      // first, generate a stream of possible 10-base divisors (if the left side is zero, we're done):
      return IntStream.iterate(10, div -> div * 10).takeWhile(div -> square / div > 0)
          // then, filter out zero right sides:
          .filter(div -> square % div > 0)
          // finally, see if the sum of the parts match the original number:
          .anyMatch(div -> i == square / div + square % div);
    };

    IntStream.concat(IntStream.of(1), IntStream.iterate(2, i -> i + 1).filter(isKaprekar))
      .limit(16)
      .forEach(System.out::println);

    // 1, 9, 45, 55, 99, 297, 703, 999, 2223, 2728, 4879, 4950, 5050, 5292, 7272, 7777
  }
}

I chose to output the first 16 numbers so we can see the output includes numbers 4879 and 5292.

[Heiko Dudzus]

I really like to see your use of Streams and lambdas solving this problem. Nice occasion for me to learn a little bit more about them.