For instance, this is 5 choose 2:
var array = [0,1,2,3,4];
var result = array.flatMap(
(v, i) => array.slice(i+1).map(w => [v, w])
);
console.log(result);How would I be able to do 5 choose 3 using this method?
-
1What's the expected output?ibrahim mahrir– ibrahim mahrir2020-10-18 15:04:25 +00:00Commented Oct 18, 2020 at 15:04
-
[[0, 1, 2], [0, 1, 3], [0, 1, 4], [0, 2, 3], [0, 2, 4], [0, 3, 4], [1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]]Yuval Levental– Yuval Levental2020-10-18 15:05:25 +00:00Commented Oct 18, 2020 at 15:05
-
@YuvalLevental Rather than adding a comment for your expected result, edit your question.Keith– Keith2020-10-18 15:13:11 +00:00Commented Oct 18, 2020 at 15:13
Add a comment
|
1 Answer
Just add another level of nesting:
var array = [0,1,2,3,4];
var result = array.flatMap((v, i) =>
array.slice(i+1).flatMap((w, j) =>
array.slice(i+1+j+1).map(u =>
[v, w, u]
)
)
);
console.log(result);
At this point, it might be easier to do with recursion though:
function choose(arr, k, prefix=[]) {
if (k == 0) return [prefix];
return arr.flatMap((v, i) =>
choose(arr.slice(i+1), k-1, [...prefix, v])
);
}
console.log(choose([0,1,2,3,4], 3));
4 Comments
mplungjan
Anyway we can do the last one with strings too? As in pass it (["ab","cd","ef"],3) and get abcdef, cdabef, etc?
Bergi
@mplungjan That example sounds like you're looking for permutations not combinations, but sure enough
choose also works with string values (or really anything). Use choose(["ab","cd","ef","gh"],3).map(c => c.join('')). You could build a string-specific version of choose as well that uses string concatenation instead of array consing.mplungjan
Hi I did find a permutation script. Yours only does
[ "AABBCC"] when given an array of AA,BB,CCBergi
@mplungjan Well that's because the OP didn't ask for permutations here… There's lots of SO threads about permutations, including some with implementions of mine.