Get all (number of) combinations of an array

Question

I have been trying to accomplish this since yesterday, though no luck yet. I have found solutions where there always is a slight difference in what I want to accomplish.

I am trying to get all possible combinations, slightly like this: combination_k , but I also want the same items to pair up with itself, so given the following:

input [1, 4, 5] and 2 (number of combinations) should return:

[1, 1], [1, 4], [1, 5], [4, 4], [4, 5], [5, 5]

input [1, 4, 5] and 3 should return:

[1, 1, 1], [1, 1, 4], [1, 1, 5], [1, 4, 4], [1, 4, 5], [4, 4, 4], [4, 4, 5], [5, 5, 5], [5, 5, 4], [5, 5, 1] (The order is not important).

I have been adjusting combination_k, it got me far enough that it worked with 2 but it didn't work when I provided 3 as a parameter.

const combinations = getAllCombinations([1, 4, 5], 2);
// combinations = [1, 1], [1, 4], [1, 5], [4, 4], [4, 5], [5, 5]

Any tips are welcome!

Answer 1

The problem is commonly referred to as k-combinations with repetitions .

Here's a solution that relies on recursion to get the desired result:

 const combinations = (array, r) => { const result = []; const fn = (array, selected, c, r, start, end) => { if (c == r) { result.push([...selected]); return; } for (let i = start; i <= end; i++) { selected[c] = array[i]; fn(array, selected, c + 1, r, i, end); } } fn(array, [], 0, r, 0, array.length - 1); return result; } console.log(combinations([1, 4, 5], 3));

Answer 2

A modified version of the code you provided:

 function getAllCombinations(arr, n) { if (n <= 0) return []; if (n === 1) return [...arr]; return arr.reduce((acc, cur, i) => { const head = arr.slice(i, i + 1); const combinations = getAllCombinations(arr.slice(i), n - 1).map(x => head.concat(x)); return [...acc, ...combinations]; }, []); } console.log(getAllCombinations([1, 4, 5], 2).join('|')); console.log(getAllCombinations([1, 4, 5], 3).join('|'));