Find all permutations of 2 arrays in JS

Thought I'd give it a try. I used binary to get the possible combinations, as this problem needed a base 2 solution:

const low = ["one", "two", "three"];
const up = ["ONE", "TWO", "THREE"];

const words = [low, up]
const len = words[0].length


function getCombinations(noOfArrays, len) {
  var temp, newCombo, combos = [];
  for (var i = 0; i < (noOfArrays ** len); i++) {
    temp = new Array(len).fill(0)
    newCombo = i.toString(noOfArrays).split('');
    newCombo.forEach((el, i) => temp[temp.length - newCombo.length + i] = +el);
    combos.push(temp);
  }
  return combos;
}

function setCombinations(combos) {
  return combos.map(combo => combo.map((el, i) => words[el][i]))
}


var combos = getCombinations(words.length, len)
combos = setCombinations(combos)


console.log(combos)

Explanation of loop:

1. temp = new Array(len).fill(0)
2. newCombo = i.toString(2).split("");
3. newCombo.forEach((el, i) => temp[temp.length - newCombo.length + i] = +el);

1. Creates temp array [0,0,0]
2. Grabs loop number (i) and converts it to binary e.g:

1 -> 1
2 -> 10
3 -> 11
4 -> 100
etc...

Then split the binary into an array 100 -> [1,0,0].

3. Then for each element push it in that new array. This gave a problem with pushing the 1 and 2 element arrays (10 -> [1,0]) into the back of the array. I used temp.length - newCombo.length + i to fix that.

That function then returns:

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

Then, I can map over each combination, and grab each array depending on the value, and get the words ('one' or 'ONE') via loop index.

Note this code works with more than one array, as long as the arrays are all the same length.

You could transpose the arrays for getting an array of pairs and then get all combinations of the pairs.

const
    transpose = array => array.reduce((r, a) => a.map((v, i) => [...(r[i] || []), v]), []),
    combinations = array => array.reduce((a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []));

var lowerWords = ['one', 'two', 'three'],
    upperWords = ['ONE', 'TWO', 'THREE'],
    pairs = transpose([lowerWords, upperWords]),
    result = combinations(pairs);
    
console.log(result);

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You need to get a total of 2 ^ 3 combinations. If you create a 2D matrix from the 2 arrays, the below table represents the row number from which of the item should be taken.

0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1

If you analyze indexes of the combinations, each of this a binary number from 0 to 2 ^ 3 with leading zeros.

So, you could

• loop from 0 to 8
• create binary number using toString(2)
• Add leading zero using padStart
• split each digit to get an array
• Get each item from matrix[digit-from-binary][position-of-each-split]
• join the array of items with a ' ' separator to get the key
• Add the key to the output object

Working snippet:

function getAllCombinations(matrix) {
  const combinations = 2 ** 3,
        output = {};
  
  for(let i = 0; i < combinations; i++) {
      const key = i.toString(2)
                    .padStart(3, 0)
                    .split('')
                    .map((n, j) => matrix[n][j])
                    .join(" ")
                    
      output[key] = true;
  }
  
  return output
}

console.log(getAllCombinations([['one', 'two', 'three' ],[ 'ONE', 'TWO', 'THREE' ]]))

You can generalize this for m x n matrix. Instead of converting each to a binary number, you need to convert it to base-m and padStart to length n

function getAllCombinations(matrix) {
  const rows = matrix.length,
        columns = matrix[0].length,
        combinations = rows ** columns;

  return Array.from({ length: combinations },
    (_, i) => i.toString(rows)
                .padStart(columns, 0)
                .split('')
                .map((n, j) => matrix[n][j])
  )
}

console.log(JSON.stringify(
    getAllCombinations( [[1, 2, 3], [4, 5, 6], [7, 8, 9]] ) // 3 x 3 matrix
));

console.log(JSON.stringify(
    getAllCombinations( [[1, 2], [3, 4], [5, 6], [7, 8]] ) // 4 x 2 matrix
));

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