Combinations Of Size N From An Array

I am trying to build a javascript function to get all combinations of size N from an array. lets say I got: const Xarray = ['19', '21','42','23', '25', '28']; const n = 4; combin

Solution 1:

Doing this by yourself might be rather tough, because I've tried that. There's already a js tool that does this for you, combinations.js

/**
 * Copyright 2012 Akseli Palén.
 * Created 2012-07-15.
 * Licensed under the MIT license.
 */functionk_combinations(set, k) {
    var i, j, combs, head, tailcombs;
    if (k > set.length || k <= 0) {
        return [];
    }
    if (k == set.length) {
        return [set];
    }
    if (k == 1) {
        combs = [];
        for (i = 0; i < set.length; i++) {
            combs.push([set[i]]);
        }
        return combs;
    }
    combs = [];
    for (i = 0; i < set.length - k + 1; i++) {
        head = set.slice(i, i+1);
        tailcombs = k_combinations(set.slice(i + 1), k - 1);
        for (j = 0; j < tailcombs.length; j++) {
            combs.push(head.concat(tailcombs[j]));
        }
    }
    return combs;
}
functioncombinations(set) {
    var k, i, combs, k_combs;
    combs = [];
    for (k = 1; k <= set.length; k++) {
        k_combs = k_combinations(set, k);
        for (i = 0; i < k_combs.length; i++) {
            combs.push(k_combs[i]);
        }
    }
    return combs;
}
	
var array = ["19", "21","42","23", "25", "28"];

document.body.innerHTML += "<pre>" + JSON.stringify(k_combinations(array, 4), false, "\t") + "</pre>";

Solution 2:

Well i have to say i hate JS sometimes. I definitely hate push() that for sure. When you do functional you need more and more a reference to the mutated objects. A method mutating an object it's called upon shall return a reference to that object.

Anyways the code could have been much simpler looking but alas... this is as far as it gets. It's nothing more than a simple recursive run. The complicated looking parts are actually the stupid parts such as;

a.slice(0,i).concat(a.slice(i+1))

in fact means delete the element at index position i and return the resulting array. When you need to use this functionality as a single instruction or a chainable instruction as an argument to a function this seems to be the only way. Just like the

(t.push(c),t)

instruction. Which means push c to t array and return t array. Silly push(c) would return the length... I hate it. You give me the reference man i can get the length from that if needed. So the rest is easy to understand.

So i have two solutions one for the permutations and one for the combinations.

var xarray = ["19", "21", "42", "23", "25", "28"],
         n = 4;

functiongetPermutations(a,n,s=[],t=[]){
  return a.reduce((p,c,i,a) => { n > 1 ? getPermutations(a.slice(0,i).concat(a.slice(i+1)), n-1, p, (t.push(c),t))
                                       : p.push((t.push(c),t).slice(0));
                                 t.pop();
                                 return p},s)
}
document.write("<pre>" + JSON.stringify(getPermutations(xarray,n),null,2) + "</pre>");

And now the combinations...

var xarray = ["19", "21", "42", "23", "25", "28"],
         n = 4;

functiongetCombinations(a,n,s=[],t=[]){
  return a.reduce((p,c,i,a) => { n > 1 ? getCombinations(a.slice(i+1), n-1, p, (t.push(c),t))
                                       : p.push((t.push(c),t).slice(0));
                                 t.pop();
                                 return p},s)
}

document.write("<pre>" + JSON.stringify(getCombinations(xarray,n),null,2) + "</pre>");

Solution 3:

I offer two solutions, the first with a result which includes the items at any place.

functioncombine(array, length) {
    functionc(l, r) {
        var i, ll;
        if (r.length === length) {
            result.push(r);
            return;
        }
        for (i = 0; i < l.length; i++) {
            ll = l.slice();
            c(ll, r.concat(ll.splice(i, 1)));
        }
    }
    var result = [];
    c(array, []);
    return result;
}

document.write('<pre>' + JSON.stringify(combine(["19", "21", "42", "23", "25", "28"], 4), 0, 4) + '</pre>');

The second which returns only one item at in original order. This result set is shorter then the above and like the answer of akinuri.

functioncombine(array, length) {
    functionc(l, r) {
        var ll = l.slice();
        if (r.length === length) {
            result.push(r);
            return;
        }
        while (ll.length) {
            c(ll, r.concat(ll.shift()));
        }
    }
    var result = [];
    c(array, []);
    return result;
}

document.write('<pre>' + JSON.stringify(combine(["19", "21", "42", "23", "25", "28"], 4), 0, 4) + '</pre>');