Find all permutations of 2 arrays in JS

Question

I'm trying to find every permutations of 2 arrays like this:

// input
lowerWords = ['one', 'two', 'three' ] 
upperWords = [ 'ONE', 'TWO', 'THREE' ]

// output
keywords = {
  'one two three': true,
  'ONE two three': true,
  'ONE TWO three': true,
  'ONE TWO THREE': true,
  'ONE two THREE': true,
  'one TWO three': true,
  'one two THREE': true,
  'one TWO THREE': true,
}

It should function with more than 3 items, both arrays will always be same length. This is my code:

const keywords = {}
const lowerWords = ['one', 'two', 'three' ] 
const upperWords = [ 'ONE', 'TWO', 'THREE' ]
const wordCount = lowerWords.length

let currentWord = 0
let currentWords = [...upperWords]
while (currentWord < wordCount) {
  currentWords[currentWord] = lowerWords[currentWord]
  let keyword = currentWords.join(' ')
  keywords[keyword] = true
  currentWord++
}

currentWord = 0
currentWords = [...lowerWords]
while (currentWord < wordCount) {
  currentWords[currentWord] = upperWords[currentWord]
  let keyword = currentWords.join(' ')
  keywords[keyword] = true
  currentWord++
}

result is missing some

ONE TWO THREE: true
ONE TWO three: true
ONE two three: true
one TWO THREE: true
one two THREE: true
one two three: true

Answer 1

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);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

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.