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Java 8 Functional Programming with jOOλ Example

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Java 8 Functional Programming with jOOλ Example

The Java 8 Stream API does not offer enough functionality for this task, so this quick tutorial will use jOOλ.

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I’ve stumbled upon an interesting Stack Overflow question by user “mip”. The question was:

I’m looking for a way of generating an alphabetic sequence:

A, B, C, ..., Z, AA, AB, AC, ..., ZZ.

This can be quickly recognized as the headings of an Excel spreadsheet, which does precisely that:


So far, none of the answers employed any Java 8 functional programming, which I accepted as a challenge. We’re going to use jOOλ, because the Java 8 Stream API does not offer enough functionality for this task.

But first, let’s decompose the algorithm in a functional way. What we need are these components:

  1. A (reproducible) representation of the alphabet
  2. An upper bound, i.e. how many letters we want to produce. The requested sequence goes to ZZ, which means the upper bound would be 2
  3. A way to combine each letter of the alphabet with the previously generated combined letters in a cartesian product

Let’s look into some code:

1. Generating the Alphabet

We could be writing the alphabet like this:

List<String> alphabet = Arrays.asList("A", "B", ..., "Z");

but that would be lame. Let’s generate it instead, using jOOλ:

List<String> alphabet = Seq
    .rangeClosed('A', 'Z')

The above generates a “closed” range (Java-8-Stream-speak for a range with inclusive upper bound) of characters between A and Z, maps characters to strings and collects them into a list.

So far so good. Now:

2. Using an Upper Bound

The requested sequence of characters includes:

A .. Z, AA, AB, .. ZZ

But we could easily imagine extending this requirement generally to produce the following, or even more.

A .. Z, AA, AB, .. ZZ, AAA, AAB, .. ZZZ

For this, we’ll use rangeClosed() again:

// 1 = A .. Z, 2 = AA .. ZZ, 3 = AAA .. ZZZ
Seq.rangeClosed(1, 2)
   .flatMap(length -> ...)

The idea here is to produce a new stream for each individual length in the range [1 .. 2], and to flatten those streams into one single stream. flatMap() is essentially the same as a nested loop in imperative programming.

3. Combine Letters in a Cartesian Product

This is the trickiest part: We need to combine each letter with each letter length times. For this, we’ll use the following stream:

Seq.rangeClosed(1, length - 1)
   .foldLeft(Seq.seq(alphabet), (s, i) -> 
        .map(t -> t.v1 + t.v2))

We’re using again rangeClosed() to produce values in the range [1 .. length-1]. foldLeft() is the same as reduce(), except that foldLeft() is guaranteed to go from “left to right” in a stream, without requiring the folding function to be associative. Whew.

In other, more understandable terms: foldLeft() is nothing else but an imperative loop. The “seed” of the loop, i.e. the loop’s initial value, is a complete alphabet (Seq.seq(alphabet)). Now, for every value in the range [1 .. length-1], we produce a cartesian product (crossJoin()) between the letters “folded” so far and a new alphabet, and we concatenate each combination into a single new string (t.v1 and t.v2).

That’s it!

Combining Everything

The following simple program prints all the values from A .. Z, AA .. ZZ, AAA .. ZZZ to the console:

import java.util.List;

import org.jooq.lambda.Seq;

public class Test {
    public static void main(String[] args) {
        int max = 3;

        List<String> alphabet = Seq
            .rangeClosed('A', 'Z')

        Seq.rangeClosed(1, max)
           .flatMap(length ->
               Seq.rangeClosed(1, length - 1)
                  .foldLeft(Seq.seq(alphabet), (s, i) -> 
                       .map(t -> t.v1 + t.v2)))


This is certainly not the most optimal algorithm for this particular case. One of the best implementations has been given by an unnamed user on Stack Overflow:

import static java.lang.Math.*;

private static String getString(int n) {
    char[] buf = new char[(int) floor(log(25 * (n + 1)) / log(26))];
    for (int i = buf.length - 1; i >= 0; i--) {
        buf[i] = (char) ('A' + n % 26);
        n /= 26;
    return new String(buf);

Needless to say that the latter runs much much faster than the previous functional algorithm.

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