How to get every permutation of elements in an array of arrays - arrays

Question

I have an array of arrays in TypeScript. I would like to take action on every permutation of elements between the inner arrays.
Here is an example:
const arrOfArrays = [
[1],
[2, 3, 4],
[5, 6],
[7, 8]
];
recursePermutations(arrOfArrays, permutation => {
console.log(permutation);
});
And when I run this, I would like to have the following output:
[ 1, 2, 5, 7 ]
[ 1, 2, 5, 8 ]
[ 1, 2, 6, 7 ]
[ 1, 2, 6, 8 ]
[ 1, 3, 5, 7 ]
[ 1, 3, 5, 8 ]
[ 1, 3, 6, 7 ]
[ 1, 3, 6, 8 ]
[ 1, 4, 5, 7 ]
[ 1, 4, 5, 8 ]
[ 1, 4, 6, 7 ]
[ 1, 4, 6, 8 ]
It should call the permutation callback with the number of elements as there are non-empty inner arrays.
I see questions regarding how to find all permutations of elements in a fixed number of arrays. But in this case, I don't know how many inner arrays are present.

Answer 1

As I hinted in the question, here is a solution that uses recursion and generics:
function recursePermutations<T>(arrOfArrays: T[][], callback: (permutation: T[]) => void, i: number = 0, previousElements: T[] = []) {
if (i < arrOfArrays.length) {
const currentElements = arrOfArrays[i];
for (var element of currentElements) {
recursePermutations(arrOfArrays, callback, i + 1, previousElements.concat(element));
}
if (currentElements.length < 1) {
recursePermutations(arrOfArrays, callback, i + 1, Array.from(previousElements));
}
}
else if (previousElements.length > 0) {
callback(previousElements);
}
}
It handles the case of the outer array not having any inner arrays by not calling the callback.
It also handles the case of one or more of the inner arrays having zero elements. In that case, it calls the permutation callback with only the number of elements as there are non-empty inner arrays.