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Make formation of array into multiple arrays with having minimum items - JavaScript

I have an array of teams, I am trying to split that array into multiple arrays with having minimum players in each array. I mean, form the array into multiple teams from array. I forgot what this feature is called, but let me tell you with an explanation:
minimum players: 3
team array: [4, 6, 5, 8]
This array will form like this:
[
[4, 6, 5],
[4, 6, 8],
[4, 5, 8],
[6, 5, 8],
[4, 6, 5, 8]
]
So, I have tried doing this problem, but I got stuck that what to do and how to do.
function teams(minPlayers, arr) {
for (let i = 0; i < arr.length; i++) {
const subArray = []
let startMax = minPlayers - 2
let end = minPlayers
for (let j = 0; j < arr.length; j++) {
if (j < startMax) {
subArray[i] = arr[i]
}
}
}
}
console.log(teams(3, [4, 6, 5, 10]))
It will be a pleasure if you help me.

Here is a generator that generates the combinations through recursion. Array.from will consume the iterator:
function teams(minPlayers, arr) {
// Execute a generator, and pass it to Array.from to turn its
// yielded values into an array and return it
return Array.from(function* iter(minPlayers, start) {
// When there are as many remaining elements (from start)
// as we need players, then there are no other options
// than selecting all those values, so yield that slice of
// the array and return (base case)
if (arr.length - minPlayers == start) return yield arr.slice(start);
// Produce the cases where the element at start is not selected.
// Use recursion for this, by reducing the slice that is still available
yield* iter(minPlayers, start + 1);
// Produce the cases where the element at start is selected.
// Use recursion to get the combinations after that, and
// prefix the start element to each of those
for (let combi of iter((minPlayers || 1) - 1, start + 1)) {
yield [arr[start], ...combi];
}
}(minPlayers, 0)); // Immediately execute the generator function
}
console.log(teams(3, [4, 6, 5, 10]));
So the idea of the generator is to either take the current value or not (current is arr[start]). In either case, defer the other selections to recursion. The recursive call will get an increased start value so there are fewer elements to choose from.

This is not exactly permutation quite the opposite since we don't care for order, but we do care for picking n different items.
The recursive function picks works by picking an item (length times) then running picks on the rest - concatenating the results. A care is given not to pick numbers that are "left" to the position of our current item position since we already counted them.
// this loops over minPlayers to maxPlayers (length)
function teams(minPlayers, arr) {
var result = [];
for (var i = minPlayers; i <= arr.length; i++) {
result = result.concat(picks(i, arr))
}
return result;
}
// picks all possible n items from arr (hopefully)
function picks(n, arr) {
// will hold all possible series (pick of n items from array)
var result = [];
if (n == 1) {
// base case return n arrays of 1 item
for (var i = 0; i < arr.length; i++) {
result.push([arr[i]])
}
return result;
}
// else we will loop over each of items
for (var i = 0; i < arr.length; i++) {
// make a copy since we are going to remove items from it
var copy = [...arr];
// pick an item, the first one of our series
var item = copy.splice(i, 1);
// ignore those before it
for (j = 0; j < i; j++) {
// remove an item from beginning of array
copy.shift();
}
// recursion! get picks of the rest n-1 items
var others = picks(n - 1, copy);
// combine each pf those picks with our item
others.forEach(function(other) {
// push to result the series which is [item, ...rest]
result.push(item.concat(other))
})
}
// return array of arrays
return result;
}
console.log(teams(3, [4, 6, 5, 10]))
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max-height: 100% !important;
}
Last function can be refactored a little since base case looks similar to the other steps and no need to mutate the array. Here's the shorter version:
// loop for each possible minPlayers and above
function teams(minPlayers, arr) {
var result = [];
for (var i = minPlayers; i <= arr.length; i++) {
result = result.concat(picks(i, arr))
}
return result;
}
// recursive function to pick all possible n items from arr
function picks(n, arr) {
var result = [];
for (var i = 0; i < arr.length; i++) {
// foreach item
var item = arr[i];
if (n == 1) {
// if we need pick 1, then pick item
result.push([item])
} else {
// pick n-1 possible except our item (and those before it)
var others = picks(n - 1, arr.slice(i + 1));
// combine with our item n times
others.forEach(function(other) {
result.push([item, ...other])
})
}
}
return result;
}
console.log(teams(3, [4, 6, 5, 10]))
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max-height: 100% !important;
}

Algorithm to return all combinations of 3 combinations from a array of length 6 without the number appearing with the same set of numbers

I was trying to write a algorithm in javascript that returns all the possible 3 digit numbers numbers from a given array of length 6
For Example
var arr = [1, 2, 3, 4, 5, 6];
I have already got the combinations with the same sets of numbers in different positions in the 2D array.
(The code which I took the help of)
If I have the same numbers in different combinations then I would like to remove them form the array. like I have [1, 2, 3] at index i in the array comtaining all the possible combinations then I would like to remove other combination with the same numbers like [2, 1, 3], [1, 3, 2] and so on..
Note the array also contains numbers repeated like [3, 3, 3], [2, 2, 2], [3, 2, 3] and so on
I expect an 2d array which has the values : [[1,2,3],[1,2,4],[1,2,5],[1,2,6],[1,3,4]] and so on (24 possibilities)
Is there any way to do this?

Extending the answer you linked, just filter out the results with the help of a Set.
Sort an individual result, convert them into a String using join(), check if it's present in set or not, and if not, then store them in the final result.
function cartesian_product(xs, ys) {
var result = [];
for (var i = 0; i < xs.length; i++) {
for (var j = 0; j < ys.length; j++) {
// transform [ [1, 2], 3 ] => [ 1, 2, 3 ] and append it to result []
result.push([].concat.apply([], [xs[i], ys[j]]));
}
}
return result;
}
function cartesian_power(xs, n) {
var result = xs;
for (var i = 1; i < n; i++) {
result = cartesian_product(result, xs)
}
return result;
}
function unique_cartesian_power(xs, n) {
var result = cartesian_power(xs, n);
var unique_result = [];
const set = new Set();
result.forEach(function(value) {
var representation = value.sort().join(' ');
if (!set.has(representation)) {
set.add(representation);
unique_result.push(value);
}
});
return unique_result;
}
console.log(unique_cartesian_power([1, 2, 3, 4, 5, 6], 3));

const arr = [1, 2, 3, 4, 5, 6];
const result = arr.reduce((a, v) => arr.reduce((a, v2) => {
arr.reduce((a, v3) => {
const current = [v, v2, v3].sort().join(",");
!a.find(_ => _.sort().join() === current) && a.push([v, v2, v3]);
return a;
}, a);
return a;
}, a), []);
console.log(result.length);
console.log(...result.map(JSON.stringify));

You could take an iterative and recursive approach by sorting the index and a temporary array for the collected values.
Because of the nature of going upwards with the index, no duplicate set is created.
function getCombination(array, length) {
function iter(index, right) {
if (right.length === length) return result.push(right);
if (index === array.length) return;
for (let i = index, l = array.length - length + right.length + 1; i < l; i++) {
iter(i + 1, [...right, array[i]]);
}
}
var result = [];
iter(0, []);
return result;
}
var array = [1, 2, 3, 4, 5, 6],
result = getCombination(array, 3);
console.log(result.length);
result.forEach(a => console.log(...a));
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This is a good example, that it is usually worthwhile not asking for a specific answer for a generic problem shown with a specific question; however as you've requested - if you really have the above constraints which kind of don't make much sense to me, you could do it like that:
function combine(firstDigits, secondDigits, thirdDigits) {
let result = [];
firstDigits.forEach(firstDigit => {
// combine with all secondDigitPermutations
secondDigits.forEach(secondDigit => {
// combine with all thirdDigitPermutations
thirdDigits.forEach(thirdDigit => {
result.push([firstDigit, secondDigit, thirdDigit])
})
})
});
// now we have all permutations and simply need to filter them
// [1,2,3] is the same as [2,3,1]; so we need to sort them
// and check them for equality (by using a hash) and memoize them
// [1,2,3] => '123'
function hashCombination(combination) {
return combination.join('ಠ_ಠ');
}
return result
// sort individual combinations to make them equal
.map(combination => combination.sort())
.reduce((acc, currentCombination) => {
// transform the currentCombination into a "hash"
let hash = hashCombination(currentCombination);
// and look it up; if it is not there, add it to cache and result
if (!(hash in acc.cache)) {
acc.cache[hash] = true;
acc.result.push(currentCombination);
}
return acc;
}, {result: [], cache: {}})
.result;
}
console.log(combine([1,2,3,4,5,6],[1,2,3,4,5,6],[1,2,3,4,5,6]).length);
console.log(...combine([1,2,3,4,5,6],[1,2,3,4,5,6],[1,2,3,4,5,6]).map(JSON.stringify));
This does not include some super-clever assumptions about some index, but it does abuse the fact, that it's all about numbers. It is deliberately using no recursion, because this would easily explode, if the amount of combinations is going to be bigger and because recursion in itself is not very readable.
For a real world problem™ - you'd employ a somewhat similar strategy though; generating all combinations and then filter them. Doing both at the same time, is an exercise left for the astute reader. For finding combinations, that look different, but are considered to be the same you'd also use some kind of hashing and memoizing.

let arr1 = [1,2,3,4,5,6];
function getCombination(arr){
let arr2 = [];
for(let i=0; i<arr.length; i++){
for(let j=i; j<arr.length; j++){
for(let k=j; k<arr.length; k++){
arr2.push([arr[i],arr[j],arr[k]]);
}
}
}
return arr2;
}
console.log(getCombination(arr1));

Slice into chunks and average (reduce) array in JS

I have an array:
var data = [0,1,2,3,4,5];
that I would like to splice into [0,1,2] and [3,4,5] followed by averaging it so the final result would be:
var dataOptimised = [1,4];
this is what I have found so far:
function chunk (arr, len) {
var chunks = [];
var i = 0;
var n = arr.length;
while (i < n) {
chunks.push(arr.slice(i, i += len)); // gives [[0,1,2] [3,4,5]]
}
return chunks;
};
How to reduce it?
Thanks

Sum each chunk using Array.reduce() and divide by the chunk's length.
var data = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30];
function chunkAverage(arr, len) {
var chunks = [];
var i = 0;
var n = arr.length;
var chunk;
while (i < n) {
chunk = arr.slice(i, i += len);
chunks.push(
chunk.reduce((s, n) => s + n) / chunk.length
);
}
return chunks;
};
console.log(chunkAverage(data, 3));

You can map the array and reduce it.
function chunk(arr, len) {
var chunks = [];
var i = 0;
var n = arr.length;
while (i < n) {
chunks.push(arr.slice(i, i += len)); // gives [[0,1,2] [3,4,5]]
}
return chunks;
};
var data = [0, 1, 2, 3, 4, 5];
var result = chunk(data, 3).map(o => (o.reduce((c, v) => c + v, 0)) / o.length);
console.log(result);

Split the array in half using splice. And use .reduce to take sum and average finally
var arrR = [0, 1, 2, 3, 4, 5],
arrL = arrR.splice(0, Math.ceil(arrR.length / 2)),
results = [getAverave(arrL), getAverave(arrR)];
console.log(results)
function getAverave(arr) {
return arr.reduce(function(a, b) {
return a + b;
}) / arr.length;
}

Here is the sortest answer possible for this question.
n is the index you want to slice from.
function chunk (arr, n) {
return [Math.sum.apply(null, arr.slice(0, n)), Math.sum.apply(null, arr.slice(n, arr.length))];
};

If you don't mind using underscore.js, you can use the _.chunk() function to easily chunk your array, then map() each chunk to a reduce() function which averages each chunk.
When importing the underscore.js library, you can reference the library using the _ symbol.
const arr = [0, 1, 2, 3, 4, 5];
const len = 3;
const result = _.chunk(arr, len).map(chunk => chunk.reduce((a, b) => a + b, 0) / chunk.length);
console.log(result); // Outputs [1, 4]
If you have an odd-length array; say that arr = [0, 1, 2, 3, 4, 5, 6], then result would be [1, 4, 6].
In HTML, you can include the library in a <script> tag:
<script src="http://underscorejs.org/underscore.js"></script>
Here's a working jsfiddle where you can see it in action (you'll have to open the F12 tools to see console output; the StackOverflow embedded snippets are kind of funky and don't work right).

Agreeing with both Ori Drori and Eddie, but I thought I might also provide some additional minor changes to your chunk function for consistency and maintainability's sake...
When writing JavaScript, I would recommend using function names that won't collide with common/expected variable names. For example, with a function like chunk() that returns a "chunk", it's likely you would want to create a variable called chunk to hold its return value. A line of code like var chunk = chunk() is an obvious problem, but if it gets any less direct it can easily wreak havoc down the line. Using the const var = function pattern (see snippet) helps you avoid writing over the original function by throwing an error on the correct line, but I would argue it's also still good to get in the habit of using a naming convention that doesn't have this drawback just in case you can't use something like const. My approach is to always include a verb in the function name. In your case, "chunk" can also be considered a verb, but it conflicts. So, I prefixed it with "get".
const getChunk = (arr, len) => {
const chunks = []
const n = arr.length
let i = 0
while (i < n) {
chunks.push(arr.slice(i, i += len))
}
return chunks
}
const data = [0,1,2,3,4,5]
const optimizedData =
getChunk(data, 3).map(chunk =>
chunk.reduce((total, val) => total + val) / chunk.length
)
console.log(optimizedData)

How to find all subsets of a set in JavaScript? (Powerset of array)

I need to get all possible subsets of an array.
Say I have this:
[1, 2, 3]
How do I get this?
[], [1], [2], [3], [1, 2], [2, 3], [1, 3], [1, 2, 3]
I am interested in all subsets. For subsets of specific length, refer to the following questions:
Finding subsets of size n: 1, 2
Finding subsets of size > 1: 1

Here is one more very elegant solution with no loops or recursion, only using the map and reduce array native functions.
const getAllSubsets =
theArray => theArray.reduce(
(subsets, value) => subsets.concat(
subsets.map(set => [value,...set])
),
[[]]
);
console.log(getAllSubsets([1,2,3]));

We can solve this problem for a subset of the input array, starting from offset. Then we recurse back to get a complete solution.
Using a generator function allows us to iterate through subsets with constant memory usage:
// Generate all array subsets:
function* subsets(array, offset = 0) {
while (offset < array.length) {
let first = array[offset++];
for (let subset of subsets(array, offset)) {
subset.push(first);
yield subset;
}
}
yield [];
}
// Example:
for (let subset of subsets([1, 2, 3])) {
console.log(subset);
}
Runtime complexity is proportional to the number of solutions (2ⁿ) times the average length per solution (n/2) = O(n2ⁿ).

Simple solution without recursion:
function getAllSubsets(array) {
const subsets = [[]];
for (const el of array) {
const last = subsets.length-1;
for (let i = 0; i <= last; i++) {
subsets.push( [...subsets[i], el] );
}
}
return subsets;
}
How does it work?
If we have some subsets generated from input numbers and we want to add one more number to our input array, it means that we can take all already existing subsets and generate new ones by appending the new number to each of the existing.
Here is an example for [1, 2, 3]
Start with an empty subset: []
Create new subsets by adding "1" to each existing subset. It will be:[] [1]
Create new subsets by adding "2" to each existing subset. It will be:[], [1] [2], [1, 2]
Create new subsets by adding "3" to each existing subset. It will be: [], [1], [2], [1, 2] [3], [1, 3], [2, 3], [1, 2, 3]

Another simple solution.
function getCombinations(array) {
function fork(i, t) {
if (i === array.length) {
result.push(t);
return;
}
fork(i + 1, t.concat([array[i]]));
fork(i + 1, t);
}
var result = [];
fork(0, []);
return result;
}
var data = [1, 2, 3],
result = getCombinations(data);
console.log(result);
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You can easily generate the powerset from an array, using something like the following:
var arr = [1, 2, 3];
function generatePowerSet(array) {
var result = [];
result.push([]);
for (var i = 1; i < (1 << array.length); i++) {
var subset = [];
for (var j = 0; j < array.length; j++)
if (i & (1 << j))
subset.push(array[j]);
result.push(subset);
}
return result;
}
console.log(generatePowerSet(arr));
Throughout the main loop of the function, subsets are created and then pushed into the result array.

I set out to understand what is happening with the examples in this post. While the function generator example, bit-wise operator example, and the example use of the array map and reduce functions are very elegant and impressive, I found it tough to mentally visual what precisely was happening. I have 2 examples below that I believe are easy to visualize both a non-recursive and a recursive solution. I hope this helps others attempting to wrap their heads around the process of finding all subsets.
NON-RECURSIVE:
For each value of the array clone all existing subsets (including the empty set) and add the new value to each of the clones, pushing the clones back to the results.
const PowerSet = array => {
const result = [[]] // Starting with empty set
for (let value of array) { // For each value of the array
const length = result.length // Can't use result.length in loop since
// results length is increased in loop
for (let i = 0; i < length; i++){
let temp = result[i].slice(0) // Make a clone of the value at index i
temp.push(value) // Add current value to clone
result.push(temp) // Add clone back to results array
}
}
return result;
}
console.log(PowerSet([1,2,3]))
RECURSIVELY:
Build the powerset by recursively pushing a combination of the current index value concatenated with an ever increasing prefix array of values.
const powerSetRecursive = (arr, prefix=[], set=[[]]) => {
if(arr.length === 0) return// Base case, end recursion
for (let i = 0; i < arr.length; i++) {
set.push(prefix.concat(arr[i]))// If a prefix comes through, concatenate value
powerSetRecursive(arr.slice(i + 1), prefix.concat(arr[i]), set)
// Call function recursively removing values at or before i and adding
// value at i to prefix
}
return set
}
console.log(powerSetRecursive([1,2,3]))

function subSets(num){
/*
example given number : [1,3]
[]
1: copy push 1
[] [1]
3: copy push 3
[] [1] [3] [1,3]
*/
let result = [];
result.push([]);
for(let i=0; i < num.length;i++){
let currentNum = num[i];
let len = result.length;
for(let j=0; j < len; j++){
let cloneData = result[j].slice();
cloneData.push(currentNum);
result.push(cloneData)
}
}
return result;
}
let test = [1,3];
console.log(subSets(test))//[ [], [ 1 ], [ 3 ], [ 1, 3 ] ]

let subsets = (n) => {
let result = [];
result.push([]);
n.forEach(a => {
//array length
let length = result.length;
let i =0;
while(i < length){
let temp = result[i].slice(0);
temp.push(a);
result.push(temp);
i++;
}
})
return result;
}

Using flatMap and rest/spread, this can be fairly elegant:
const subsets = ([x, ...xs]) =>
x == undefined
? [[]]
: subsets (xs) .flatMap (ss => [ss, [x, ...ss]])
console .log (subsets ([1, 2, 3]))
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This version does not return them in the requested order. Doing that seems slightly less elegant, and there's probably a better version:
const subset = (xs = []) => {
if (xs.length == 0) {return [[]]}
const ys = subset (xs .slice (0, -1))
const x = xs .slice (-1) [0]
return [... ys, ... ys .map (y => [... y, x])]
}
Or, the same algorithm in a different style,
const subsets = (
xs = [],
x = xs .slice (-1) [0],
ys = xs.length && subsets (xs .slice (0, -1))
) =>
xs .length == 0
? [[]]
: [... ys, ... ys .map (y => [... y, x])]

A shorter version of #koorchik's answer.
var getAllSubsets = (nums) => {
const subsets = [[]];
for (n of nums) {
subsets.map((el) => {
subsets.push([...el, n]);
});
}
return subsets;
};
console.log(getAllSubsets([1, 2, 3]));
// [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]

For loop:
function powerSet(numbers) {
const subsets = [[]]
for (const number of numbers) {
subsets.forEach(subset => subsets.push([...subset, number]))
}
return subsets
}
Recursion:
function powerSet(numbers) {
const subsets = [[]]
if (numbers.length === 0) return subsets
for (let i = 0; i < numbers.length; i++) {
subsets.push(...powerSet(numbers.slice(i + 1)).map(subset => [numbers[i], ...subset]))
// Or step by step:
// const number = numbers[i]
// const otherNumbers = numbers.slice(i + 1)
// const otherNumbersSubsets = powerSet(otherNumbers)
// const otherNumbersSubsetsWithNumber = otherNumbersSubsets.map(subset => [number, ...subset])
// subsets.push(...otherNumbersSubsetsWithNumber)
}
return subsets
}

Using reduceRight:
const subsets = array =>
array.reduceRight(
(accumulator, a) => [...accumulator, ...accumulator.map(b => [a, ...b])],
[[]]
);
console.log(subsets([1, 2, 3])); // [[], [3], [2], [2, 3], [1], [1, 3], [1, 2], [1, 2, 3]]

This one is with recursion
var subsets = function(s){
if(s.length === 0) {
return [[]]
}
var h,t,ss_excl_h;
var ss_incl_h = [];
[h,...t] = s;
ss_excl_h = subsets(t)
for(ss of ss_excl_h) {
let hArr = [];
hArr.push(h);
let temp = hArr.concat(ss)
ss_incl_h.push(temp);
}
return ss_incl_h.concat(ss_excl_h)
}
console.log(subsets([1,2,3])) // returns distinct subsets

Update ES2020
With ES2020 BigInts have become available.
Bigints don’t have a fixed storage size in bits; their sizes adapt to the integers they represent.
- Dr. Axel Rauschmayer; JavaScript for impatient programmers - Chapter 18.2 BigInts
See source.
Using BitInts we can use a binary counter to calculate the power set and are no longer limited by the maximum integer size.
Using a generator we can additionally loop over a power set with constant memory requirement which is important if you want to generate a huge power set.
Here an example using you original array [1, 2, 3].
/**
* Generate power set from a given array
* #param {Array<any>} array array to create power set from
*/
function* powerSet(array){
// use BigInt to be no longer limited by maximum length of 53-bits
const size = 2n ** BigInt(array.length);
for (let i = 0; i < size; i++) {
const cur = [];
for(let j = 0; j < array.length; j++){
// check if i-th bit is set to 1
if((i & (1 << j)) > 0){
// push array value (represented by that 1-bit) to result
cur.push(array[j]);
}
}
// generate next result
yield cur;
}
}
// generate power set for [1, 2, 3] and print results
console.log([...powerSet([1, 2, 3])]);
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Here how you could loop over a very large power set with constant memory and no upper bound (theoretically, there will be an upper bound in terms of compute time) for the array length.
/**
* Generate power set from a given array
* #param {Array<any>} array array to create power set from
*/
function* powerSet(array){
// use BigInt to no longer limited by maximum length of 53-bits
const size = 2n ** BigInt(array.length);
for (let i = 0; i < size; i++) {
const cur = [];
for(let j = 0; j < array.length; j++){
// check if i-th bit is set to 1
if((i & (1 << j)) > 0){
// push array value (represented by that 1-bit) to result
cur.push(array[j]);
}
}
// generate next result
yield cur;
}
}
/**
* Helper function to generate an array containing more than 53 elements
* #param {number} start
* #param {number} end
*/
function* range(start, end){
for (let i = start; i < end; i++) {
yield i;
}
}
// create an array containing elments 1 through 60 ([1, 2, 3, ..., 60])
const oneToSixty = [...range(1, 61)];
let i = 0;
const max = 1000;
// loop over whole powerSet with constant memory requirement
// abort after 1000 subsets, otherwise this will take a very long time to complete
for(const subset of powerSet(oneToSixty)){
console.log(subset);
if(i++ === max) break;
}
.as-console-wrapper { max-height: 100% !important; top: 0; }

Return all possible combinations of numbers in an array whose sum is less than or equal to n

var a = [1,3,6,10,-1];
function combinations(array, n) {
}
combinations(a, 9) // should return...
[[1], [3], [6], [-1], [1,3], [1,6], [1,-1], [3,6], [3,-1], [6, -1], [10, -1], [1,3,-1], [3,6,-1], [1,6,-1], [1,3,6,-1]]
maybe i'm missing some correct answers but you get the idea. Really dying to know how to solve this!

I would say the problem here is to take the power set of an array, and filter it down to only the elements whose sum is greater than a certain number.
The power set of a set is the set of all subsets of that set. (Say that five times fast and you'll be a mathematician)
For example, the power set of [1] is [[], [1]] and the power set of [1, 2] is [[], [1], [2], [1, 2]].
First I would define a powerSet function like this:
var powerSet = function (arr) {
// the power set of [] is [[]]
if(arr.length === 0) {
return [[]];
}
// remove and remember the last element of the array
var lastElement = arr.pop();
// take the powerset of the rest of the array
var restPowerset = powerSet(arr);
// for each set in the power set of arr minus its last element,
// include that set in the powerset of arr both with and without
// the last element of arr
var powerset = [];
for(var i = 0; i < restPowerset.length; i++) {
var set = restPowerset[i];
// without last element
powerset.push(set);
// with last element
set = set.slice(); // create a new array that's a copy of set
set.push(lastElement);
powerset.push(set);
}
return powerset;
};
Then I would define a function that takes the power set of the array and only includes elements whose sum is less than or equal to some amount:
var subsetsLessThan = function (arr, number) {
// all subsets of arr
var powerset = powerSet(arr);
// subsets summing less than or equal to number
var subsets = [];
for(var i = 0; i < powerset.length; i++) {
var subset = powerset[i];
var sum = 0;
for(var j = 0; j < subset.length; j++) {
sum += subset[j];
}
if(sum <= number) {
subsets.push(subset);
}
}
return subsets;
};
This might not be fast on large arrays, but it works well for small ones.
It looks like it gives the right answer for console.log(subsetsLessThan([1,3,6,10,-1], 9));
edit: a little more about the power set function as implemented here
The only subset of [] is [], so the power set of [] is a set containing only []. That would be [[]].
The initial if statement in the powerSet function immediately returns [[]] if you pass in [].
var powerSet = function (arr) {
if(arr.length === 0) {
return [[]];
}
If you pass in a set with at least one element, the powerSet function begins by removing the last element. For example, if you call powerSet on [1, 2], the variable lastElement will be set to 2 and arr will be set to [1].
var lastElement = arr.pop();
Then the powerSet function recursively calls itself to get the power set of the "rest" of the list. If you had passed in [1, 2], then restPowerset is assigned to powerSet([1]) which is [[], [1]].
var restPowerset = powerSet(arr);
We define a variable that's going to hold the power set of what was passed in, here [1, 2]
var powerset = [];
We loop through every set in restPowerset.
for(var i = 0; i < restPowerset.length; i++) {
var set = restPowerset[i];
Any subset of [1] is also a subset of [1, 2] so we add it to the list. That is, [] and [1] are both subsets of [1, 2].
powerset.push(set);
If you add the element 2 to any subset of [1], that is also a subset of [1, 2], so we add it to the list. Both [2] and [1, 2] are subsets of [1, 2].
set = set.slice(); // copy the array
set.push(lastElement); // add the element
powerset.push(set);
That's all. At this point, the variable powerset is [[], [2], [1], [1, 2]]. Return it!
}
return powerset;
};

Brute force O(N*2N) solution, where N = a.length < 31.
This uses the index i as a bit field to filter the elements of a in each iteration into a sublist.
var a = [1,3,6,10,-1];
function combinations(array, n) {
var lists = [], M = 1<<array.length;
for( var i = 1 ; i < M ; ++i ) {
var sublist = array.filter(function(c,k){return i>>k & 1});
if( sublist.reduce(function(p,c){return p+c},0) <= n )
lists.push(sublist);
}
return lists;
}
console.log(JSON.stringify(combinations(a,9)));
[[1],[3],[1,3],[6],[1,6],[3,6],[-1],[1,-1],[3,-1],[1,3,-1],[6,-1],[1,6,-1],[3,6,-1],[1,3,6,-1],[10,-1]]

Similar to Matt's answer, but uses Array.filter() and Array.reduce() to pack a punch. The variable, mask is incremented from 1 to 32-1 in this example (because array length is 5 and count = 1 << 5, which is 32). The array is filtered for each mask increment, producing a new array or permutation where only certain values are included.
A value is included in the permutation if the mask shifted right by the value's index is odd. Think binary here, because either a value is supposed to be in the permutation or it isn't (0 or 1) and since the mask will go through all possible numbers, all of the possible permutations are covered directly in the number when expressed as binary:
index: 4,3,2,1,0
mask: 0 0 0 0 1 (grab index 0, [1])
mask: 0 0 0 1 0 (grab index 1, [3])
mask: 0 0 0 1 1 (grab index 0 and 1, [1,3])
mask: 1 1 0 0 0 (grab index 3 and 4, [10,-1])
var a = [1,3,6,10,-1];
function combinations(array, n) {
var mask, len = array.length, count = 1 << len, permutations = [];
var indexVisible = function(v, i) { return ((mask >> i) & 1) == 1 }
var sum = function(a, b) { return a + b }
for (mask = 1; mask < count; ++mask) {
permutations.push(array.filter(indexVisible))
}
return permutations.filter(function(p) { return p.reduce(sum) <= n })
}
console.log(JSON.stringify(combinations(a, 9)));
The function, indexVisible() is used to filter the original array and return a permutation that matches the mask.
The function, sum() is used to reduce each permutation to the sum of its values, and if that sum is less than or equal to n then it is included in the final result and returned from combinations()
Here are the permutations:
[[1],[3],[1,3],[6],[1,6],[3,6],[1,3,6],[10],[1,10],[3,10],[1,3,10],[6,10],[1,6,10],[3,6,10],[1,3,6,10],[-1],[1,-1],[3,-1],[1,3,-1],[6,-1],[1,6,-1],[3,6,-1],[1,3,6,-1],[10,-1],[1,10,-1],[3,10,-1],[1,3,10,-1],[6,10,-1],[1,6,10,-1],[3,6,10,-1],[1,3,6,10,-1]]
Here are the results:
[[1],[3],[1,3],[6],[1,6],[3,6],[-1],[1,-1],[3,-1],[1,3,-1],[6,-1],[1,6,-1],[3,6,-1],[1,3,6,-1],[10,-1]]
You can see how all of this works and play with different combinations in this JSFiddle.

The following code will give you all sub-arrays summing up to 9 or less..
function getSubArrays(arr,n){
var len = arr.length,
subs = Array(Math.pow(2,len)).fill();
return subs.map((_,i) => { var j = -1,
k = i,
res = [];
while (++j < len ) {
k & 1 && res.push(arr[j]);
k = k >> 1;
}
return res;
}).slice(1)
.filter(a => a.reduce((p,c) => p+c) <= n);
}
var arr = [1,3,6,10,-1],
result = getSubArrays(arr,9);
console.log(JSON.stringify(result));

edit: giving credit where due.. borrowed the bulk of this logic from this answer
var combinations = function(a,m) {
var gc = function(a) {
var fn = function(n, src, got, all) {
if (n == 0) {
if (got.length > 0) {
all[all.length] = got;
}
return;
}
for (var j = 0; j < src.length; j++) {
fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
}
return;
}
var all = [];
for (var i = 0; i < a.length; i++) {
fn(i, a, [], all);
}
all.push(a);
return all;
}
var c = gc(a);
return c.filter(function(e) {
var n = e.length;
var sum = 0;
while(n--)
sum += parseFloat(e[n]) || 0;
return sum<=m;
},m);
}
var a = [1,3,6,10,-1];
combinations(a,9);
output
[[1], [3], [6], [-1], [1, 3], [1, 6], [1, -1], [3, 6], [3, -1], [6, -1], [10, -1], [1, 3, -1], [1, 6, -1], [3, 6, -1], [1, 3, 6, -1]]

It looked like to much fun not to play, here's what I have.
Javascript
function kCombs(set, k) {
var setLength = set.length,
combs = [],
i = 0,
tailLength,
head,
tail,
j,
t,
u;
if (k > 0 && k <= setLength) {
if (k === setLength) {
combs.push(set);
} else if (k === 1) {
while (i < setLength) {
combs.push([set[i]]);
i += 1;
}
} else {
u = k - 1;
setLength = setLength - k + 1;
while (i < setLength) {
t = i + 1;
head = set.slice(i, t);
tail = kCombs(set.slice(t), u);
j = 0;
tailLength = tail.length;
while (j < tailLength) {
combs.push(head.concat(tail[j]));
j += 1;
}
i = t;
}
}
}
return combs;
}
function combinations(array, n) {
var arrayLength = array.length,
combs = [],
combsLength,
results = [],
temp = 0,
current,
currentLength,
i,
j,
k = 1;
while (k <= arrayLength) {
i = 0;
current = kCombs(array, k);
currentLength = current.length;
while (i < currentLength) {
combs.push(current[i]);
i += 1;
}
k += 1;
}
i = 0;
combsLength = combs.length;
while (i < combsLength) {
j = 0;
current = combs[i];
currentLength = current.length;
while (j < currentLength) {
temp += current[j];
j += 1;
}
if (temp <= n) {
results.push(current);
}
temp = 0;
i += 1;
}
return results;
}
var a = [1, 3, 6, 10, -1];
console.log(JSON.stringify(combinations(a, 9)));
Output
[[1],[3],[6],[-1],[1,3],[1,6],[1,-1],[3,6],[3,-1],[6,-1],[10,-1],[1,3,-1],[1,6,-1],[3,6,-1],[1,3,6,-1]]
On jsFiddle
And a jsPerf of all these, although #jcarpenter solutions gives an ambiguity.
On a modern browser you could squeeze more out of this solution using for intead of while as they are highly optimised for for. And assign by index rather than push would also give you a performance boost.
It would be nice to extend the performance tests to include some more test sets, maybe if I get bored.

Brevity is very cryptic here. How about some descriptive functions?
The approach uses binary to create maps of all the possible combinations. Then the map is used to pluck items from the array. The plucked items are summed, and that's about it.
The result of combinations([1, 3, 6, 10, -1], 9) produced is: [[-1],[10,-1],[6],[6,-1],[3],[3,-1],[3,6],[3,6,-1],[1],[1,-1],[1,6],[1,6,-1],[1,3],[1,3,-1],[1,3,6,-1]].
Here is a Fiddle.
/**
* Get an array of all the possible combinations
* of x items. Combinations are represented as binary.
* #param {Number} x - example 2
* #return {String[]} - example ['00', '01', '10', '11']
*/
function getCombinationsOfXItems(x) {
var allOn = '',
numCombos = 0,
i = 0,
combos = [];
// find upper limit
while (allOn.length < x) {
allOn += 1;
}
// number of possible combinations
numCombos = parseInt(allOn, 2) + 1;
// generate the combos
while(i < numCombos) {
combos.push(pad(toBase2(i++), allOn.length));
}
return combos;
}
/**
* Pad a string with leading zeros.
* #param {String} x - example '100'
* #param {Number} length - example 6
* #return {String} - example '000100'
*/
function pad(x, length) {
while (x.length < length) {
x = 0 + x;
}
return x;
}
/**
* Get a number as a binary string.
* #param {Number} x - example 3
* #return {String} - example '11'
*/
function toBase2(x) {
return x.toString(2);
}
/**
* Given an array and a map of its items as a binary string,
* return the items identified by 1.
* #param {Array} arr - example [1,2,3]
* #param {String} binary - example '101'
* #return {Array} - example [1,3]
*/
function pluckFromArrayByBinary(arr, binary) {
var plucked = [],
i = 0,
max = binary.length;
for (; i < max; i++) {
if (binary[i] === '1') {
plucked.push(arr[i]);
}
}
return plucked;
}
/**
* Given an array, return a multi-dimensional
* array of all the combinations of its items.
* #param {Array} - example [1, 2];
* #return {Array[]} - [ [1], [1, 2], [2] ]
*/
function getCombosOfArrayItems(arr) {
var comboMaps = getCombinationsOfXItems(arr.length),
combos = [];
// remove the "all off" combo (ex: '00000')
comboMaps.shift();
for (var i = 0; i < comboMaps.length; i++) {
combos.push(pluckFromArrayByBinary(arr, comboMaps[i]));
}
return combos;
}
/**
* Return all possible combinations of numbers in an
* array whose sum is less than or equal to n
* #param {Number[]} arr
* #param {Number} x
* return {Number[]} - stringified for readability
*/
function combinations(arr, x) {
var combos = getCombosOfArrayItems(arr),
i = 0,
max = combos.length,
combo;
for (; i < max; i++) {
if (sumArray(combos[i]) > x) {
combos.splice(i, 1);
i--;
max--;
}
}
return JSON.stringify(combos);
}
/**
* Return the sum of an array of numbers.
* #param {Number[]} arr
* #return {Number}
*/
function sumArray(arr) {
var sum = 0,
i = 0,
max = arr.length;
for (; i < max; i++) {
sum += arr[i];
}
return sum;
}
console.log(combinations([1, 3, 6, 10, -1], 9));

#jcarpenter solution was so nice I just had to rework it for those that love ECMA5. This will not be as fast as the raw power of for, the modern methods have not had the length of time to be so highly optimised (and they do quite a bit more work). But the performance results do show just how good the powerSet algorithm is (and it is a reusable function). I've also filtered out the ambiguity, which slows things slightly.
Javascript
function powerSet(arr) {
var lastElement,
val;
if (!arr.length) {
val = [[]];
} else {
lastElement = arr.pop();
val = powerSet(arr).reduce(function (previous, element) {
previous.push(element);
element = element.slice();
element.push(lastElement);
previous.push(element);
return previous;
}, []);
}
return val;
}
function combinations(array, n) {
return powerSet(array).filter(function (set) {
return set.length && set.reduce(function (previous, element) {
return previous + element;
}, 0) <= n;
});
}
var a = [1, 3, 6, 10, -1];
console.log(JSON.stringify(combinations(a, 9)));
Output
[[-1],[10,-1],[6],[6,-1],[3],[3,-1],[3,6],[3,6,-1],[1],[1,-1],[1,6],[1,6,-1],[1,3],[1,3,-1],[1,3,6,-1]]
On jsFiddle
And added to the jsPerf

Try this:
var a = [1,3,6,10,-1];
function combinations(array, n) {
var arrayCopy = [],
results = [];
// duplicate the array
for (var i in array)
arrayCopy[i] = array[i];
for (var i in array)
for (var j in arrayCopy)
if ((array[i] + arrayCopy[j]) <= n)
results.push([array[i], arrayCopy[j]]);
return results;
}
console.log(combinations(a, 9));
This logged:
[1, 1], [1, 3], [1, 6], [1, -1],
[3, 1], [3, 3], [3, 6], [3, -1],
[6, 1], [6, 3], [6, -1],
[10, -1],
[-1, 1], [-1, 3], [-1, 6], [-1, 10], [-1, -1]