Generate all combinations from multiple lists

Given an unknown amount of lists, each with an unknown length, I need to generate a singular list with all possible unique combinations. For example, given the following lists:

X: [A, B, C] 
Y: [W, X, Y, Z]

Then I should be able to generate 12 combinations:

[AW, AX, AY, AZ, BW, BX, BY, BZ, CW, CX, CY, CZ]

If a third list of 3 elements were added, I'd have 36 combinations, and so forth.

Any ideas on how I can do this in Java?
(pseudo code would be fine too)

You need recursion:

Let's say all your lists are in lists, which is a list of lists. Let result be the list of your required permutations. You could implement it like this:

void generatePermutations(List<List<Character>> lists, List<String> result, int depth, String current) {
    if (depth == lists.size()) {
        result.add(current);
        return;
    }

    for (int i = 0; i < lists.get(depth).size(); i++) {
        generatePermutations(lists, result, depth + 1, current + lists.get(depth).get(i));
    }
}

The ultimate call will be like this:

generatePermutations(lists, result, 0, "");

This operation is called cartesian product. Guava provides an utility function for that: Lists.cartesianProduct

This topic came in handy. I've rewritten the previous solution fully in Java and more user friendly. Furthermore, I use collections and generics for more flexibility:

/**
 * Combines several collections of elements and create permutations of all of them, taking one element from each
 * collection, and keeping the same order in resultant lists as the one in original list of collections.
 * 
 * <ul>Example
 * <li>Input  = { {a,b,c} , {1,2,3,4} }</li>
 * <li>Output = { {a,1} , {a,2} , {a,3} , {a,4} , {b,1} , {b,2} , {b,3} , {b,4} , {c,1} , {c,2} , {c,3} , {c,4} }</li>
 * </ul>
 * 
 * @param collections Original list of collections which elements have to be combined.
 * @return Resultant collection of lists with all permutations of original list.
 */
public static <T> Collection<List<T>> permutations(List<Collection<T>> collections) {
  if (collections == null || collections.isEmpty()) {
    return Collections.emptyList();
  } else {
    Collection<List<T>> res = Lists.newLinkedList();
    permutationsImpl(collections, res, 0, new LinkedList<T>());
    return res;
  }
}

/** Recursive implementation for {@link #permutations(List, Collection)} */
private static <T> void permutationsImpl(List<Collection<T>> ori, Collection<List<T>> res, int d, List<T> current) {
  // if depth equals number of original collections, final reached, add and return
  if (d == ori.size()) {
    res.add(current);
    return;
  }

  // iterate from current collection and copy 'current' element N times, one for each element
  Collection<T> currentCollection = ori.get(d);
  for (T element : currentCollection) {
    List<T> copy = Lists.newLinkedList(current);
    copy.add(element);
    permutationsImpl(ori, res, d + 1, copy);
  }
}

I'm using guava library for collections creation.

Adding an iterator based answer to work for generic list of lists List<List<T>>, extending the idea from Ruslan Ostafiichuk's answer. The idea I followed was:

* List 1: [1 2]
* List 2: [4 5]
* List 3: [6 7]
*
* Take each element from list 1 and put each element
* in a separate list.
* combinations -> [ [1] [2] ]
*
* Set up something called newCombinations that will contains a list
* of list of integers
* Consider [1], then [2]
*
* Now, take the next list [4 5] and iterate over integers
* [1]
*  add 4   -> [1 4]
*      add to newCombinations -> [ [1 4] ]
*  add 5   -> [1 5]
*      add to newCombinations -> [ [1 4] [1 5] ]
*
* [2]
*  add 4   -> [2 4]
*      add to newCombinations -> [ [1 4] [1 5] [2 4] ]
*  add 5   -> [2 5]
*      add to newCombinations -> [ [1 4] [1 5] [2 4] [2 5] ]
*
* point combinations to newCombinations
* combinations now looks like -> [ [1 4] [1 5] [2 4] [2 5] ]
* Now, take the next list [6 7] and iterate over integers
*  ....
*  6 will go into each of the lists
*      [ [1 4 6] [1 5 6] [2 4 6] [2 5 6] ]
*  7 will go into each of the lists
*      [ [1 4 6] [1 5 6] [2 4 6] [2 5 6] [1 4 7] [1 5 7] [2 4 7] [2 5 7]]

Now the code. I used a Set simply to get rid of any duplicates. Can be replaced with a List. Everything should work seamlessly. :)

public static <T> Set<List<T>> getCombinations(List<List<T>> lists) {
    Set<List<T>> combinations = new HashSet<List<T>>();
    Set<List<T>> newCombinations;

    int index = 0;

    // extract each of the integers in the first list
    // and add each to ints as a new list
    for (T i : lists.get(0)) {
        List<T> newList = new ArrayList<T>();
        newList.add(i);
        combinations.add(newList);
    }
    index++;
    while (index < lists.size()) {
        List<T> nextList = lists.get(index);
        newCombinations = new HashSet<List<T>>();
        for (List<T> first : combinations) {
            for (T second : nextList) {
                List<T> newList = new ArrayList<T>();
                newList.addAll(first);
                newList.add(second);
                newCombinations.add(newList);
            }
        }
        combinations = newCombinations;
        index++;
    }
    return combinations;
}

A little test block..

public static void main(String[] args) {
    List<Integer> l1 = Arrays.asList(1, 2, 3);
    List<Integer> l2 = Arrays.asList(4, 5);
    List<Integer> l3 = Arrays.asList(6, 7);

    List<List<Integer>> lists = new ArrayList<List<Integer>>();
    lists.add(l1);
    lists.add(l2);
    lists.add(l3);

    Set<List<Integer>> combs = getCombinations(lists);
    for (List<Integer> list : combs) {
        System.out.println(list.toString());
    }
}

Without recursion unique combinations:

String sArray[] = new String[]{"A", "A", "B", "C"};
//convert array to list
List<String> list1 = Arrays.asList(sArray);
List<String> list2 = Arrays.asList(sArray);
List<String> list3 = Arrays.asList(sArray);

LinkedList<List<String>> lists = new LinkedList<List<String>>();

lists.add(list1);
lists.add(list2);
lists.add(list3);

Set<String> combinations = new TreeSet<String>();
Set<String> newCombinations;

for (String s : lists.removeFirst())
    combinations.add(s);

while (!lists.isEmpty()) {
    List<String> next = lists.removeFirst();
    newCombinations = new TreeSet<String>();
    for (String s1 : combinations)
        for (String s2 : next)
            newCombinations.add(s1 + s2);

    combinations = newCombinations;
}
for (String s : combinations)
    System.out.print(s + " ");

The operation that you need to implement called Cartesian Product. For more details see https://en.wikipedia.org/wiki/Cartesian_product

I recommend to use my open source library that can do exactly what you need: https://github.com/SurpSG/Kombi

There is example how to use it: https://github.com/SurpSG/Kombi#usage-for-lists-1

Note: The library was designed for high performance purposes. You can observe banchmarks results here

The library gives you pretty good throughput and constant memory usage

Use the nested loop solution provided by some other answers here to combine two lists.

When you have more than two lists,

1. Combine the first two into one new list.
2. Combine the resulting list with the next input list.
3. Repeat.

• No recursion
• Ordered
• Possible to get a specific combination by its index (without building all other permutations)
• Supports parallel iteration

Class and main() method in the end:

public class TwoDimensionalCounter<T> {
    private final List<List<T>> elements;
    private final int size;

    public TwoDimensionalCounter(List<List<T>> elements) {
        //Need to reverse the collection if you need the original order in the answers
        this.elements = Collections.unmodifiableList(elements);
        int size = 1;
        for(List<T> next: elements)
            size *= next.size();
        this.size = size;
    }

    public List<T> get(int index) {
        List<T> result = new ArrayList<>();
        for(int i = elements.size() - 1; i >= 0; i--) {
            List<T> counter = elements.get(i);
            int counterSize = counter.size();
            result.add(counter.get(index % counterSize));
            index /= counterSize;
        }
        return result;
    }

    public int size() {
        return size;
    }

    public static void main(String[] args) {
        TwoDimensionalCounter<Integer> counter = new TwoDimensionalCounter<>(
                Arrays.asList(
                        Arrays.asList(1, 2, 3),
                        Arrays.asList(1, 2),
                        Arrays.asList(1, 2, 3)
                ));
        for(int i = 0; i < counter.size(); i++)
            System.out.println(counter.get(i));
    }
}

PS: as it turned out Guava's Cartessian Product uses the same algorithm. But they also created special sub-classes to List to make it several times more efficient.

Late to the party as usual, but here's a nicely explained example using arrays. It can easily be altered for lists. I needed all unique combinations of multiple arrays for my use case in a lexicographical order.

I posted it as none of the answers here give a clear algorithm, and I can't stand recursion. Are we not on stackoverflow after all?

String[][] combinations = new String[][] {
                 new String[] { "0", "1" },
			     new String[] { "0", "1" },
                 new String[] { "0", "1" },
                 new String[] { "0", "1" } };

	int[] indices = new int[combinations.length];
	
	int currentIndex = indices.length - 1;
	outerProcess: while (true) {

		for (int i = 0; i < combinations.length; i++) {
			System.out.print(combinations[i][indices[i]]);
		}
		System.out.println();

		while (true) {
			// Increase current index
			indices[currentIndex]++;
			// If index too big, set itself and everything right of it to 0 and move left
			if (indices[currentIndex] >= combinations[currentIndex].length) {
				for (int j = currentIndex; j < indices.length; j++) {
					indices[j] = 0;
				}
				currentIndex--;
			} else {
				// If index is allowed, move as far right as possible and process next
				// combination
				while (currentIndex < indices.length - 1) {
					currentIndex++;
				}
				break;
			}
			// If we cannot move left anymore, we're finished
			if (currentIndex == -1) {
				break outerProcess;
			}
		}
	}

The output;

0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

Generating combinations with Java 8 Stream map and reduce methods.

[Try it online!][1]

public static <T> List<List<T>> combinations(List<List<T>> lists) {
    // incorrect incoming data
    if (lists == null) return Collections.emptyList();
    return lists.stream()
            // non-null and non-empty lists
            .filter(list -> list != null && list.size() > 0)
            // represent each list element as a singleton list
            .map(list -> list.stream().map(Collections::singletonList)
                    // Stream<List<List<T>>>
                    .collect(Collectors.toList()))
            // summation of pairs of inner lists
            .reduce((list1, list2) -> list1.stream()
                    // combinations of inner lists
                    .flatMap(inner1 -> list2.stream()
                            // merge two inner lists into one
                            .map(inner2 -> Stream.of(inner1, inner2)
                                    .flatMap(List::stream)
                                    .collect(Collectors.toList())))
                    // list of combinations
                    .collect(Collectors.toList()))
            // otherwise an empty list
            .orElse(Collections.emptyList());
}

public static void main(String[] args) {
    List<String> list1 = Arrays.asList("A", "B", "C");
    List<String> list2 = Arrays.asList("W", "X", "Y", "Z");
    List<String> list3 = Arrays.asList("L", "M", "K");

    List<List<String>> lists = Arrays.asList(list1, list2, list3);
    List<List<String>> combinations = combinations(lists);

    // column-wise output
    int rows = 6;
    IntStream.range(0, rows).forEach(i -> System.out.println(
            IntStream.range(0, combinations.size())
                    .filter(j -> j % rows == i)
                    .mapToObj(j -> combinations.get(j).toString())
                    .collect(Collectors.joining(" "))));
}

Column-wise output:

[A, W, L] [A, Y, L] [B, W, L] [B, Y, L] [C, W, L] [C, Y, L]
[A, W, M] [A, Y, M] [B, W, M] [B, Y, M] [C, W, M] [C, Y, M]
[A, W, K] [A, Y, K] [B, W, K] [B, Y, K] [C, W, K] [C, Y, K]
[A, X, L] [A, Z, L] [B, X, L] [B, Z, L] [C, X, L] [C, Z, L]
[A, X, M] [A, Z, M] [B, X, M] [B, Z, M] [C, X, M] [C, Z, M]
[A, X, K] [A, Z, K] [B, X, K] [B, Z, K] [C, X, K] [C, Z, K]

---

^{See also: [Cartesian product of an arbitrary number of sets][2]}

[1]: https://tio.run/##rVVbb9owFH7nV5xFWpVMkBYq7aEFpK7aw7SiPQxpN@3BhAN15tiR7RTRqb@d@RLAWbho0vLgOLa/833H55KcPJGeKJHn81@bDS1KITXkZjGtNGXpm9tOa01piaSwW52ymjGaQcaIUjAhlMPvDpinXleaaPMaTsfwQJUeumE6HkMmihnlZlNwFTe3mJmopLZjn8tLoDwTUmKm3aygfAlzosnuCF1A7HAwGgGvGEtAoq4kh3vBmMFZnhSLUq8tT5zc7qD1OYeuPYuT3W6ggQves6aB8Ln7cOY8sHU@XVCmUTpR0PNOwSuvDS4u3Heq6DPGCYzh6iChxFKiQq4BSfboTSDDwq4QBQSUuQeGWnj1bQ0FKRsCdu65neBmbm52puz1tNUEqj47G8NGzMZHAWnmWbZsQqpUCx@D5KDXqioKlxcgFlASKpWdUM5RHrtrifMqw9j52u@6U4Nk63T/eFAD1jAfz/HtY8yInpibdIf7W8LBecKAuEC5RNArEVKauRYgOJ614CLpkAPL74OTikWtqeutDs5LaTllY2Tywhn8B/jJeJ8MgctUc/VhKP5jXgn9iHJFFZr6hX3ptrNJyPdMYXykcdSd46VzoM09CTqHwjTB2MTBFNSPn0DkstHLXMX43To9YQR3UpK1SolyFNFd1IXonR3uo6BRtaCDNvSLRX21wzc7fD@Jv27jHyxqYoePFtrEhgbqNt2yEJagf13/LaFpplF4o@Z/wf8JAhmuUFlV8J4LpKh0We0jaIoGpFhZM2/3nB@4rqtCEr7E@KrrDiXpwgTa9NWYusJZK42mcCqdlkaZZjxuZcYBS6Hcup0nJ/qF/yfkljCH17XYEdATGFPfU/FplntUg2@JOs4Tk/b@Mk9SHyiXXFBuYRFEtjS3ef2y2fwB "Java (OpenJDK 8) – Try It Online" [2]: https://stackoverflow.com/a/67174557/15675591

Here is a sample using bit mask. No recursion and multiple lists

static List<Integer> allComboMatch(List<Integer> numbers, int target) {
	int sz = (int)Math.pow(2, numbers.size());
	for (int i = 1; i < sz; i++) {
		int sum = 0;
		ArrayList<Integer> result = new ArrayList<Integer>();
		for (int j = 0; j < numbers.size(); j++) {
			int x = (i >> j) & 1;
			if (x == 1) {
				sum += numbers.get(j);
				result.add(j);
			}
		}
		if (sum == target) {
			return result;
		}
	}
	return null;
}