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

Question

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!

Answer 1

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;
};

Answer 2

Brute force O(N*2^N) 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]]

Answer 3

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.

Answer 4

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));

Answer 5

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]]

Answer 6

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.

Answer 7

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));

Answer 8

@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

Answer 9

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]