A quite fast permutation algorithm that does not use a lot of memory, O(n).

Instead of returning a whole array of permutations, the method next() returns the next permutation. Each permutation is based on the previous and is calculated quickly.

The boolean method hasNext() checks if there is a next permutation.

The array returned by next() can be used as the indices for the actual array to be permuted. For example we can use map as in the following example to get a permutation of the actual array:

const arr = ['A', 'B', 'C'];

const p = new Permutation(arr.length);

while (p.hasNext()) {
  const arrPermutation = p.next().map((i) => arr[i]);
  ...
}

If you need to generate ALL permutations with a loop like the one above, the complexity is O(n!):

• Performs well with n < 13.
• n = 13 takes several minutes to complete.
• No problems with memory, O(n).

The Steinhaus–Johnson–Trotter algorithm or Johnson–Trotter algorithm generates all of the permutations of n elements. Each permutation in the sequence that it generates differs from the previous permutation by swapping two adjacent elements of the sequence.

An improvement of the Steinhaus–Johnson–Trotter algorithm by Shimon Even provides an improvement to the running time of the algorithm by storing additional information for each element in the permutation: its position, and a direction (positive, negative, or zero) in which it is currently moving.

This algorithm takes time O(i) for every step in which the greatest number to move is n − i + 1.

For more information visit: https://en.wikipedia.org/wiki/Steinhaus%E2%80%93Johnson%E2%80%93Trotter_algorithm

npm i permutation-sjt

import { Permutation } from 'permutation-sjt';

const p = new Permutation(3);

const permutations = [];

while (p.hasNext()) {
  permutations.push(p.next());
}

console.log(permutations);

outputs

    [
      [ 0, 1, 2 ],
      [ 0, 2, 1 ],
      [ 2, 0, 1 ],
      [ 2, 1, 0 ],
      [ 1, 2, 0 ],
      [ 1, 0, 2 ]
    ]

import { Permutation } from 'permutation-sjt';

const p = new Permutation(2, 1); // starts numbers from 1

const permutations = [];

while (p.hasNext()) {
  permutations.push(p.next());
}

console.log(permutations);

outputs

[ [ 1, 2 ], [ 2, 1 ] ]