Heap's algorithm to return array instead of printing

Question

I am trying to get this heaps algorithm to return an array of permutations instead of printing as below. I know it could be done by declaring an array outside of the function and pushing to it but, I want to avoid that approach. How can I make this return the array of permutations without using an outside array?

function heaps(arr, n) {
  if (n === undefined) n = arr.length;
  if (n <= 1) console.log(arr);
  else {
    for (let i = 0; i <= n - 1; i++)
     {
      heaps(arr, n-1);
      if 
        (n % 2 === 0) [arr[n-1], arr[i]] = [arr[i], arr[n-1]];
      else             
        [arr[n-1], arr[0]] = [arr[0], arr[n-1]];
    }
  }
}

Answer 1

Just make it return that result as an array, and collect return values from your recursive calls in the loop:

function heaps(arr, n = arr.length) {
  if (n <= 1) return [arr.slice()];
  let result = [];
  for (let i = 0; i <= n - 1; i++) {
    result.push(...heaps(arr, n-1));
    if (n % 2 === 0)
      [arr[n-1], arr[i]] = [arr[i], arr[n-1]];
    else             
      [arr[n-1], arr[0]] = [arr[0], arr[n-1]];
  }
  return result;
}

Alternatively, make an inner helper function that does push to an outer array

function heaps(arr) {
  let result = [];
  function helper(n) {
    if (n <= 1)
      result.push(arr.slice());
    else
      for (let i = 0; i <= n - 1; i++) {
        heaps(n-1);
        if (n % 2 === 0)
          [arr[n-1], arr[i]] = [arr[i], arr[n-1]];
        else             
          [arr[n-1], arr[0]] = [arr[0], arr[n-1]];
      }
  }
  helper(arr.length);
  return result;
}

If you insist on not using an "outer" array, make the result array and the input arr explicit parameters of the helper function.