Develop Reference

Sum all numbers in array until a certain value is reached

I'm writing a function and this is the last piece of the puzzle, which I thought would be easy to deal with, but after some unsuccessful tinkering with for and while loops, I tried looking it up online and still couldn't find an answer. I came across some obscure and complex solutions, but I think there should be a more straightforward way to solve this. For example, if I have an array
[1, 2, 3, 5, 7, 9]
And the argument is 10, the function should return 6 (1 + 2 + 3), because the sum of the values should be <= 10 (or whatever number is passed). If the argument is 4, the function should return 3 (1 + 2) and so on.

You can use a for loop:
const arg = 10
const arr = [1, 2, 3, 5, 7, 9]
let res = 0;
const calc = (arr, limit) => {
for (num of arr) {
if (num + res > limit) {
break;
}
res += num;
}
return res;
}
console.log(calc(arr, arg))

.reduce() each current value (cur) added to the accumulator (acc) and checked vs. the limit (max). If acc + cur is less than limit then return acc+ cur, otherwise return acc. Added .sort() if the array happens to be out of order, as per Spectric's comment.
const array = [1,2,3,5,6,8];
const mixed = [0,7,3,8,2,1];
const A = 10;
const B = 15;
const closest = (arr, max) => {
return arr.sort((a, b) => a - b).reduce((acc, cur) =>
(acc + cur) > max ? acc : (acc + cur));
}
console.log(closest(array, A));
console.log(closest(array, B));
console.log(closest(mixed, B));

One interesting approach is to realize that we want to do a fold (something like Array.prototype.reduce) but one which we can escape early. The answer from zer00ne does this by choosing to check the condition on every iteration, and just continually returning the accumulator each time the condition is not met. That's fine for many use-cases, but it would be nice to make it more explicit.
I know of two ways to do this. One is to have some signal value that we would return -- probably a symbol -- to say, "We're done, just return the accumulator." The downside is that this function now depends on that external signal value. It's not terrible, and often might be the right solution. It's not hard to write, and I'll leave it as an exercise.
The other technique is to require the callback to explicitly choose whether to continue the iteration or stop by supplying it with next and done functions. If we want to continue, we call next with the next accumulator value. If we're finished, we just call done. Here's a version of that:
const fold = (fn) => (a) => ([x, ...xs]) =>
x == undefined ? a : fn (a, x, (r) => fold (fn) (r) (xs), () => a)
const sumUpTo = (n) =>
fold ((a, x, next, done) => a + x > n ? done () : next (a + x)) (0)
console .log (sumUpTo (10) ([1, 2, 3, 5, 7, 9])) //=> 6
console .log (sumUpTo (4) ([1, 2, 3, 5, 7, 9])) //=> 3
console .log (sumUpTo (10) ([1, 2, 3, 4, 5, 6])) //=> 10
sumUpTo takes our total value and returns a function that takes a list of numbers. It does this by calling fold using a callback function, the initial value (0 for a sum), and eventually passing our list of numbers. That callback does the work we care about. Then fold repeatedly calls it until it runs out of values or done is called.
We can break down the one-liner version above to focus specifically on the callback, if it makes it more clear:
const callback = (n) => (a, x, next, done) =>
a + x > n
? done ()
: next (a + x)
const sumUpTo = (n) => fold (callback (n)) (0)
It's a very elegant pattern, to my mind.

Quickest way with smallest code footprint:
const dataset = [1, 2, 3, 5, 7, 9]
const getMaxSum = (dataset, max) => {
var sum = 0, num;
for( num of dataset ) {
if( sum + num > max ) break // very important to break once satisfied
sum += num
}
return sum
}
console.log( getMaxSum(dataset, 10) )
console.log( getMaxSum(dataset, 20) )

How to un-nest this recursive algorithm in javascript?

In the game idle heroes, demon bells can be level 1,2,3,4. A level 4 is made of 4 level 1, a level 3 is made of 3 level 1 and so on.
I want to find all arrangements of db given a fixed number. I made this recursive algorithm in javascript:
Closer with a more simplified approach:
function findDB(numDB, arr) {
console.log("findDB"+numDB);
if (numDB == 1) {
console.log("HERE2");
return 1;
} else {
for (let i = 1; i < numDB; i++) {
console.log("FOR"+i);
console.log("COND" +(numDB + (numDB-i)));
if((numDB + (numDB-i)) > numDB+1)
continue;
arr= arr.concat([numDB,[i,findDB(numDB - i, arr)]]);
}
return arr;
}
}
var final = []
var y = findDB(3, final);
console.log(JSON.stringify(y));
Output:
findDB(2) CORRECT!
findDB2
FOR1
COND3
findDB1
HERE2
[2,[1,1]]
FindDB(3) is missing 1,1,1,
findDB3
FOR1
COND5
FOR2
COND4
findDB1
HERE2
[3,[2,1]]
here is intended output for input 1 through 6 (algo needs to scale for any number input)
/1/ (1)
/2/ (2),
(1,1)
/3/ (3),
(2,1),
(1,1,1)
/4/ (4),
(3,1),
(2,2),(2,1,1),
(1,1,1,1)
/5/ (4,1),
(3,2),(3,1,1),
(2,2,1),(2,1,1,1),
(1,1,1,1,1)
/6/ (4,2),(4,1,1),
(3,3),(3,2,1),(3,1,1,1),
(2,2,2),(2,2,1,1),(2,1,1,1,1)
(1,1,1,1,1,1)

This is called the partitions of a number, and is a well-known problem. I'm sure computer scientists have more efficient algorithms than this, but a naive recursive version might look like this:
const partitions = (n, m = n) =>
m > n
? partitions (n, n)
: m == 1
? [Array (n) .fill (1)]
: m < 1
? [[]]
: [
... partitions (n - m, m) .map (p => [m, ...p]),
... partitions (n, m - 1)
];
[1, 2, 3, 4, 5, 6] .forEach ((n) => console .log (`${n}: ${JSON .stringify (partitions (n))}`))
And if you're worried about the default parameter (there sometimes are good reasons to worry), then you can just make this a helper function and wrap it in a public function like this:
const _partitions = (n, m) =>
m > n
? _partitions (n, n)
: m == 1
? [Array (n) .fill (1)]
: m < 1
? [[]]
: [
... _partitions (n - m, m) .map (p => [m, ...p]),
... _partitions (n, m - 1)
];
const partitions = (n) => _partitions (n, n);
[1, 2, 3, 4, 5, 6] .forEach ((n) => console .log (`${n}: ${JSON .stringify (partitions (n))}`))
in either case, n is the integer we're summing to, and m is the maximum integer we can use. If m is too large, we simply call again with an appropriate m. If it equals 1, then we can only have an array of n 1's. If m reaches zero, then we have only the empty partition. Finally, we have two recursive cases to combine: When we choose to use that maximum number, we recur with the remainder and that maximum, prepending the maximum to each result. And when we don't use the maximum, we recur with the same target value and a decremented maximum.
I feel as though this has too many cases, but I don't see immediately how to combine them.
The time is exponential, and will be in any case, because the result is exponential in the size of n. If we added memoization, we could really speed this up, but I leave that as an exercise.
Update
I was bothered by those extra cases, and found an Erlang answer that showed a simpler version. Converted to JS, it might look like this:
const countdown = (n) => n > 0 ? [n , ...countdown (n - 1)] : []
const _partitions = (n, m) =>
n < 0
? []
: n == 0
? [[]]
: countdown (m) .flatMap (x => _partitions (n - x, x) .map (p => [x, ...p]))
We have a quick helper, countdown to turn, say 5 into [5, 4, 3, 2, 1]. The main function has two base cases, an empty result if n is negative and a result containing only the empty partition if n is zero. Otherwise, we countdown the possibilities for the maximum value in a single partition, and recur on the partitions for the target less than this new maximum, adding the maximum value to the front of each.
This should have similar performance characteristics as the above, but it somewhat simpler.

Here is a recursive function that produces the results you want. It attempts to break down the input (numDB) into parts up to the maximum number (maxDB, which defaults to 4). It does this by taking the numbers from numDB down to 1 and adding all the possible results from a recursive call to that number, noting that the value of maxDB has to change to be no more than the first number.
const findDB = function(numDB, maxDB = 4) {
if (numDB == 0) return [ [] ];
let result = [];
let thisDB = Math.min(numDB, maxDB);
for (let i = thisDB; i > 0; i--) {
findDB(numDB - i, Math.min(i, thisDB)).forEach(n => {
result.push([i].concat(n));
});
}
return result;
}
;
[6, 5, 4, 3, 2, 1].forEach((i) => console.log(JSON.stringify(findDB(i))))
.as-console-wrapper {
min-height: 100% !important;
}
I've written the above function in the style in your question, with the use of various ES6 Array methods it can be simplified:
const DBlist = (n) => [...Array(n).keys()].map(k => n - k)
const findDB = (numDB, maxDB = 4) => {
if (numDB == 0) return [ [] ];
const thisDB = Math.min(numDB, maxDB);
return DBlist(thisDB).flatMap((i) => findDB(numDB - i, Math.min(i, thisDB)).map(a => [i, ...a]))
}
DBlist(6).forEach((n) => console.log(JSON.stringify(findDB(n))))
.as-console-wrapper {
min-height: 100% !important;
}

Best way to get combination of array of elements in Ramda?

What is the best way to implement a function that takes three arguments
smallest length of combinations
highest length of combinations
array of values
and returns all combinations of length l (arg1 <= l <= arg2). E.g.
getComb (2, 2, [1, 2, 3]) === [[1,2], [2,3], [3,1]]
getComb (0, 3, [1, 2, 3]) === [[],[1],[2],[3],[1,2],[2,3],[3,1],[1,2,3]]
(=== is defined here as deep equals without respect to order (almost set equality for both depths of the array) Also duplicate values should be ignored (e.g. getComb(a, b, [x,x,y]) === getComb(a, b, [x,y]) for all a,
b, x, y)
Then a fn to get all combinations can be implemented:
getAllComb = arr => getComb (0, arr.length, arr)
Thanks!

Here's another recursive solution, structured slightly differently from the answer by Nina Scholz. It has a function to choose exactly n elements from the list, and then uses that in the main function, which calls it for each value from min to max:
const choose = (n, xs) =>
n < 1 || n > xs .length
? []
: n == 1
? [...xs .map (x => [x])]
: [
...choose (n - 1, xs .slice (1)) .map (ys => [xs [0], ...ys]),
...choose (n , xs .slice (1))
]
const getCombs = (min, max, xs) =>
xs .length == 0 || min > max
? []
: [...choose (min, xs), ...getCombs (min + 1, max, xs)]
console .log (
getCombs (0, 3, [1, 2, 3]),
getCombs (2, 2, [1, 2, 3])
)
Here getCombs is the main function, and should be fairly clear, just concatenating the result of choose (min, xs) with the result of the recursive call to getCombs (min + 1, max, xs). choose is a nicely reusable function which operates on a double recursion, the first one selecting all those combinations which use the initial element and the second all those that don't.
This doesn't quite match Nina's solution, as it ignores the empty list when min is zero. If you want one that includes the empty list, you could change choose to the (slightly uglier, IMHO) version:
const choose = (n, xs) =>
n < 1 || n > xs .length
? [[]]
: [
...choose (n - 1, xs .slice (1)) .map (ys => [xs [0], ...ys]),
...(n + 1 > xs .length ? [] : choose (n , xs .slice (1)))
]

one way to implement getComb is :
[1,2,3].reduce( (acc, v, i, original) =>
acc.concat(original.slice(i+1).map( w => [w, v] )),
[]);

You could take a recursive approach.
function getComb(min, max, array) {
function iter(left, right = [], push = true) {
if (push && min <= right.length && right.length <= max) result.push(right);
if (!left.length) return;
iter(left.slice(1), [...right, left[0]]);
iter(left.slice(1), right, false);
}
var result = [];
iter(array);
return result;
}
console.log(getComb(2, 2, [1, 2, 3]));
console.log(getComb(0, 3, [1, 2, 3]));
.as-console-wrapper { max-height: 100% !important; top: 0; }

Ok I have a partial solution (for a = 1, b = arr.length):
const list = R.unapply (R.identity)
const xproduct = arr => R.apply (R.liftN (arr.length) (list)) (arr)
const getPerm = arr => xproduct (R.repeat (arr) (arr.length))
const getComb = arr => R.uniq (R.map (R.uniq) (getPerm (arr)))
getComb([1,2,3]) === [[1],[2],[3],[1,2],[2,3],[3,1],[1,2,3]]
There's got to be something better ;)

Here's a solution (atleast to getAllComb) that I'm kinda proud of :) There's a lot of stuff, but most of it is boilerplate
Inspired by bitstrings
// Generic helper functions
const appendIfNotFalse = fn => (acc, val) => R.ifElse (R.equals (false)) (R.always (acc)) (R.flip (R.append) (acc)) (fn (acc, val))
const mapAndFilter = fn => arr => R.reduce (appendIfNotFalse (fn)) ([]) (arr)
// My helper fn
const has1InBitstring = n => idx => (n & 2 ** idx) > 0
// Soltuion
const indices = arr => key => mapAndFilter ((_, j) => has1InBitstring (key) (j) ? j : false) (R.range (0) (arr.length))
const getAllComb = arr => R.times (i => R.props (indices (arr) (i)) (arr)) (2 ** arr.length)
// Example
getAllComb ([1,2,3]) === [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]

Return not working and overcomplex code

As part of a bigger function, I'm trying to simply run a check to see whether the values in an array are 'increasing' - e.g.:
var a = [1,2,3,4,5] // is increasing
var a = [1,4,6,7,36] // is increasing
var a = [1,6,3,6,5] // not increasing
It's increasing if the previous value a[previous] is less than a[next]. For some reason, the problem is it doesn't return -1 when it's not increasing. And, as I'm learning JavaScript, my code seems way more complex than it should be.
My Questions:
1. why does it not return -1 when a cannot increase?
2. why is my code seemingly so complex for such a simple test? am I missing something? (that is, if you think it's overcomplex, which often my code is)
What would be a better way of writing such a function? Should I put the 'test if already increased' part inside a separate function? If someone could also give hints about writing better, simpler, readable, drier code, that would be much appreciated :)
var a = [1,2,3,4,5]
var canIncrease = 0; // boolean
// test if already increased
for(i=0;i<a.length;i++) {
if((a[i] < a[i+1] && i !== a.length-1)||(a[i] > a[i-1] && i==a.length-1)) {
console.log('index ' + i + ' cannot increase');
} else {
console.log('index ' + i + ' can increase');
canIncrease = 1;
}
}
if (!canIncrease) {
console.log('array a cannot increase');
return -1;
} else {
console.log('would continue');
// continue with main function...
}

You can use every() and return true if the element is the last element OR smaller than the next element.
function func(arr) {
return arr.every((o, i, a) => (i + 1) === a.length || o < a[i + 1] );
}
console.log(func([1, 2, 3, 4, 5]));
console.log(func([1, 4, 6, 7, 36]));
console.log(func([1, 6, 3, 6, 5]));
Doc: every()

To answer your last question, this can be simplified to the following:
var a = [1,2,3,4,5]
var b = [1,4,6,7,36]
var c = [1,6,3,6,5]
function isIncreasing(arr) {
return arr.every((n, i) => i === 0 || n[i - 1] < n);
}
isIncreasing(a); //true
isIncreasing(b); //true
isIncreasing(c); //false

You can use the .reduce method to achieve the desired result:
The .reduce method applies a function against an accumulator and each element in the array (from left to right) to reduce it to a single value.
on every element compare the current element curr with the saved one acc. If the current is greater and not false save it in acc for the next element. Else set acc to false.
At the end, if the array is in ascending acc will be an integer (the last element in the array, i.e. the maximum). Else acc will be false.
adding !! before the statement ensures the integers are converted to bools.
So the statement will look like:
!!a.reduce((acc, curr) => (acc && (curr >= acc)) ? curr : false)
ES6 syntax:
() => {} or () => are shorthand syntax to define functions (arrow functions)
condition ? expr1 : expr2 is a ternary operator, it's equivalent to if(condition) { expr1 } else { expr2 }
Here are some tests:
const isAscOrder = (a) => !!a.reduce((acc, curr) => (acc && (curr >= acc)) ? curr : false);
console.log(
isAscOrder([2, 2, 3, 4, 5]) // is increasing
);
console.log(
isAscOrder([1, 6, 3, 6, 5]) // is not increasing
);
console.log(
isAscOrder([2, 1, 2, 3, 4, 5]) // is not increasing
);
I've just noticed the previous function didn't work if the array contained a 0: for example: [0, 2, 2, 3, 4, 5]. The reason being that 0 is false. In order to solve this, we can use NaN instead of false:
!isNaN(a.reduce((acc, curr) => (!isNaN(acc) && (curr >= acc)) ? curr : NaN))
const isAscOrder = (a) => !isNaN(a.reduce((acc, curr) => (!isNaN(acc) && (curr >= acc)) ? curr : NaN));
console.log(
isAscOrder([0, 2, 2, 3, 4, 5]) // is increasing
);
console.log(
isAscOrder([1, 6, 3, 6, 5]) // is not increasing
);
console.log(
isAscOrder([2, 1, 2, 3, 4, 5]) // is not increasing
);

Passing a math operator as a parameter

I'd like to write a function in Javascript that allows me to pass in a mathematical operator and a list of ints and for each item in that list, apply the operator to it.
Thinking of it in terms of a sum, this is what I've come up with:
function accumulate(list, operator){
var sum = 0;
for each(var item in list){
sum = accumulator(sum, item);
}
print(sum);
}
Testing this code produces the following error:
var list = new Array();
list[0] = 1;
list[1] = 2;
list[2] = 3;
js> accumulate(list, +);
js: "<stdin>", line 9: syntax error
js: accumulate(list, +);
js: ..................^
js: "<stdin>", line 9: Compilation produced 1 syntax errors.

You can't pass an operator as a parameter, but you can pass a function:
function accumulate(list, accumulator){ // renamed parameter
var sum = 0;
for(var i = 0; i < list.length; i++){ // removed deprecated for…each loop
sum = accumulator(sum, list[i]);
}
print(sum);
}
accumulate(list, function(a, b) { return a + b; });
This is pretty close to what the Array.prototype.reduce function does, though not exactly. To mimic the behavior of reduce, you'd have to get the first element from list and use that as the seed for your accumulator, rather than always using 0:
function accumulate(list, accumulator, seed){
var i = 0, len = list.length;
var acc = arguments.length > 2 ? seed : list[i++];
for(; i < len; i++){
acc = accumulator(acc, list[i]);
}
print(acc);
}
This way, you could compute the product of list (your method would always return 0):
accumulate(list, function(a, b) { return a * b; });
Update: If you're developing for newer browsers that support ECMAScript 2015 / ES6 (or using a transpiler like Babel), you can also use 'arrow function' syntax to make your code a bit more compact:
accumulate(list, (a, b) => a * b);

If all the operations you are planning to do are binary operations, then you can do this
var operations = {
"+" : function (operand1, operand2) {
return operand1 + operand2;
},
"-" : function (operand1, operand2) {
return operand1 - operand2;
},
"*" : function (operand1, operand2) {
return operand1 * operand2;
}
};
function accumulate(list, operator) {
return list.reduce(operations[operator]);
}
console.log(accumulate([1, 2, 3, 4], "+")); // 10
console.log(accumulate([1, 2, 3, 4], "-")); // -8
console.log(accumulate([1, 2, 3, 4], "*")); // 24

I think you can do that in several different ways, but I would suggest you something like this:
var operatorFunction = {
'+' : function(x, y) {
return x + y;
},
'-' : function(x, y) {
return x - y;
},
'*' : function(x, y) {
return x * y;
}
};
function accumul(list, neutral, operator) {
var sum = neutral;
list.forEach(function(item) {
sum = operatorFunction[operator](sum, item);
});
return sum;
}
console.log(accumul([2, 3, 4], 0, '+'));
console.log(accumul([2, 3, 4], 0, '-'));
console.log(accumul([2, 3, 4], 1, '*'));
console.log(accumul([], 0, '+'));
console.log(accumul([], 1, '*'));
In the example above, you just need something like accumul([2, 3, 4], 0, '+'); to call you function. operatorFunction[operator] calls the correspondent operator function.
Running the example in the command line, with node.js, gives:
$ node accumulate.js
9
-9
24
0
1
This version also work if the array is empty. You can not use list.reduce if the list is empty.

I know this is an old question. Just adding some more information.
If you often use operators and need to reduce the results (accumulate), it is highly recommended to develop different helpers, so you can quickly use any input form to obtain the results.
Although, this will not be always the case when you use reduce, the following helper will allow to pass the first element of your array as default value:
reducer = (list, func) => list.slice(1).reduce(func, list.slice(0, 1).pop())
The above, still has a function dependency, so you still need to declare the specific function that wraps your target operator:
sum = list => reducer(list, (a, b) => a + b)
sum([1, 2, 3, 4, 5])
You could then redefine sum, for example, as per new input formats you see will be backwards compatible. In this example by using a new helper, flat (still experimental as per now; added the code):
flat = (e) => Array.isArray(e) ? [].concat.apply([], e.map(flat)) : e
sum = (...list) => reducer(flat(list), (a, b) => a + b)
mult = (...list) => reducer(flat(list), (a, b) => a * b)
sum([1, 2, 3, 4, 5])
sum(1, 2, 3, 4, 5)
mult([1, 2, 3, 4, 5])
mult(1, 2, 3, 4, 5)
Then you can use reducer (or any variant you may find more useful) to simplify the definition of other helpers as well. Just one last example with matrix custom operators (in this case, they are functions):
zip = (...lists) => lists[0].map((_l, i) => lists.map(list => list[i]))
dot_product = (a, b) => sum(zip(a, b).map(x => mult(x)))
mx_transpose = (mx) => zip.apply([], mx)
// the operator
mx_product = (m1, m2) =>
m1.map(row => mx_transpose(m2).map(
col => dot_product(row, col) ))
// the reducer
mx_multiply = (...mxs) => reducer(mxs, (done, mx) => mx_product(done, mx))
A = [[2, 3, 4],
[1, 0, 0]]
B = [[0, 1000],
[1, 100],
[0, 10]]
C = [[2, 0],
[0, 0.1]]
JSON.stringify(AB = mx_product (A, B))
JSON.stringify(ABC = mx_product (AB, C))
JSON.stringify(ABC2 = mx_multiply(A, B, C))

just pass 1 or -1 as input then multiply all items with this after wh.

var list = { a: 3, b: 7, c: 5 }
var expression = (a) => a * 2;
function computeObject(list, bc){
for(var x in list){
list[x] = bc(list[x]);
console.log(list[x]);
}
}
computeObject(list, expression);

It can be done pretty decent using currying:
const calculate = a => str => b => {switch(str) {
case '+': return a + b
case '-': return a - b
case '/': return a / b
case '*': return a * b
default: return 'Invalid operation'
}}
const res = calculate(15)('*')(28)
console.log('15 * 28 =', res)

Unfortunately, its not really possible to do this like you are trying to do. What I would do is pass in a number, and have a if/then or a switch/case to decide what to do based on that number