5

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!

AbsoluteBeginner
  • 2,160
  • 3
  • 11
  • 21
Cem
  • 51
  • 3
  • Can you post what you have done to achieve with 2? – Yadab May 13 '21 at 06:35
  • 1
    Why is `[1, 5, 5]` not in your expected output? – Nick May 13 '21 at 06:39
  • Yes, I have done several things but the simplest and the shortest way to achieve this was to change line 121 to `tailcombs = k_combinations(set.slice(i), k - 1);` @ [combination_k](https://gist.github.com/axelpale/3118596) – Cem May 13 '21 at 06:41
  • 1
    @Nick I see that I have missed that one, woops! I have edited my initial post and added that item in the expected list. – Cem May 13 '21 at 06:43
  • [Here is a dynamical programming solution with no recursion](https://stackoverflow.com/a/40961020/4543207). Recursion in JS might result call stack overflow problems after a certain range. – Redu May 13 '21 at 08:21

2 Answers2

5

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));
Robby Cornelissen
  • 91,784
  • 22
  • 134
  • 156
4

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('|'));
A1rPun
  • 16,287
  • 7
  • 57
  • 90
  • 1
    I voted because your answer has a nicer functional style than mine. Look at us! Making the world a better place one comment at a time :) – Robby Cornelissen May 13 '21 at 07:12