introduction.md

Introduction

There are several idiomatic ways to solve Raindrops. One is to use a series of if statements. Another way is to use a Map.

General guidance

The key to solving Raindrops is to know if the input is evenly divisible by 3, 5 and/or 7. For determining that, you will use the remainder operator, also known as the modulo operator.

Approach: if statements

class RaindropConverter {
    String convert(int number) {
        StringBuilder stringBuilder = new StringBuilder();
        if (number % 3 == 0) {
            stringBuilder.append("Pling");
        }
        if (number % 5 == 0) {
            stringBuilder.append("Plang");
        }
        if (number % 7 == 0) {
            stringBuilder.append("Plong");
        }
        return stringBuilder.length() != 0 ? stringBuilder.toString() : Integer.toString(number);
    }
}

For more information, check the if statements approach.

Approach: Map

import java.util.Map;
import java.util.TreeMap;

class RaindropConverter {
    private static final TreeMap < Integer, String > lookup = new TreeMap < Integer, String > (
        Map.of(3, "Pling", 5, "Plang", 7, "Plong"));
        
    String convert(int number) {
        var output = new StringBuilder("");
        lookup.forEach((divisor, drop) -> {
            if (number % divisor == 0)
                output.append(drop);
        });
        return output.length() != 0 ? output.toString() : Integer.toString(number);
    }
}

For more information, check the Map approach.

Approach: Modular Arithmetic

import java.math.BigInteger;
import static java.math.BigInteger.valueOf;

class RaindropConverter {

    String convert (int n) {
        return switch ( valueOf(n).modPow( valueOf(12), valueOf(105) ).intValue() ) {
            case 36 -> "Pling";
            case 85 -> "Plang";
            case 91 -> "Plong";
            case 15 -> "PlingPlang";
            case 21 -> "PlingPlong";
            case 70 -> "PlangPlong";
            case  0 -> "PlingPlangPlong";
            default -> String.valueOf(n); // 1
        };
    }

}

For more information, check the Modular Arithmetic approach.

Which approach to use?

Benchmarking with the Java Microbenchmark Harness is currently outside the scope of this document, but it's likely that the series of if statements is faster than creating and iterating a Map. An advantage for Map is that, if another type of raindrop were to be added, only another entry would be added to the Map, and no other code would need to be added.