“combination with repetition algorithm js” Code Answer

combination with repetition algorithm js

javascript by Worried Wildebeest on Nov 26 2020 Donate

(() => {    'use strict';     // COMBINATIONS WITH REPETITIONS -------------------------------------------     // combsWithRep :: Int -> [a] -> [[a]]    const combsWithRep = (k, xs) => {        const comb = (n, ys) => {            if (0 === n) return ys;            if (isNull(ys)) return comb(n - 1, map(pure, xs));             return comb(n - 1, concatMap(zs => {                const h = head(zs);                return map(x => [x].concat(zs), dropWhile(x => x !== h, xs));            }, ys));        };        return comb(k, []);    };     // GENERIC FUNCTIONS ------------------------------------------------------     // concatMap :: (a -> [b]) -> [a] -> [b]    const concatMap = (f, xs) => [].concat.apply([], xs.map(f));     // dropWhile :: (a -> Bool) -> [a] -> [a]    const dropWhile = (p, xs) => {        let i = 0;        for (let lng = xs.length;            (i < lng) && p(xs[i]); i++) {}        return xs.slice(i);    };     // enumFromTo :: Int -> Int -> [Int]    const enumFromTo = (m, n) =>        Array.from({            length: Math.floor(n - m) + 1        }, (_, i) => m + i);     // head :: [a] -> Maybe a    const head = xs => xs.length ? xs[0] : undefined;     // isNull :: [a] -> Bool    const isNull = xs => (xs instanceof Array) ? xs.length < 1 : undefined;     // length :: [a] -> Int    const length = xs => xs.length;     // map :: (a -> b) -> [a] -> [b]    const map = (f, xs) => xs.map(f);     // pure :: a -> [a]    const pure = x => [x];     // show :: a -> String    const show = x => JSON.stringify(x, null, 2);     // TEST -------------------------------------------------------------------    return show({        twoFromThree: combsWithRep(2, ['iced', 'jam', 'plain']),        threeFromTen: length(combsWithRep(3, enumFromTo(0, 9)))    });})();

Source: rosettacode.org