# Sum all numbers in a range

I have implemented the "Sum All Numbers in a Range" challenge from Free Code Camp quoted below.

The code works like a charm, but I don't think it's idiomatic. The challenge hints imply I should be using Math.max(), Math.min() and Array.reduce(). In addition I used the spread operator from ES6 to simplify both Math calls.

I'm looking for a more idiomatic way to solve this. I think it can be done in a one-liner, something like:

return arr.sort((function(a, b) { return a - b; })).reduce(function(a, b) { /* */ });


But I'm pretty sure I'm heading the wrong way with the above.

The challenge:

We'll pass you an array of two numbers. Return the sum of those two numbers and all numbers between them.

The lowest number will not always come first.

The code:

function sumAll(arr) {
var out = 0;
for (var i = Math.min(...arr); i <= Math.max(...arr); i++) {
out += i;
}
return out;
}


Test cases:

sumAll([1, 4])
sumAll([4, 1])
sumAll([5, 10])
sumAll([10, 5])


Expected output:

10
10
45
45

• How do you get freecodecamp to use es6? Jan 18, 2016 at 16:56
• @Barry It's not allowed for every challenge for some reason, but on this one it just gave a notification ES6 was required. Nothing else.
– Mast
Jan 18, 2016 at 17:11
• ever heard of Gauss? (en.wikipedia.org/wiki/Carl_Friedrich_Gauss#Anecdotes) Jan 18, 2016 at 21:53
• the oneliner you give is very close. in the final function you have a and b being the min and max values. just apply gauss to it: return (a+b)*(b-a+1)/2; Jan 18, 2016 at 21:58

I don't know what you would use reduce() for to be honest. But you don't need to loop. We just have a simple arithmetic sequence with a straightforward formula:

$$\sum_{i=min}^{max} i = \frac{(max-min+1)(min+max)}2$$

We just need to pick out which one is larger:

function sumFrom(min, max) {
return (max-min+1) * (min+max) / 2;
}

function sumAll(arr) {
return sumFrom(Math.min(...arr), Math.max(...arr));
}


Or we could generalize to any arithmetic sequence:

function sumArithmetic(a1, n, d) {
return n*(2*a1 + (n-1)*d) / 2;
}

function sumAll(arr) {
var min = Math.min(...arr);
return sumArithmetic(min, Math.max(...arr) - min + 1, 1);
}


I guess if you really want reduce() you'd do it this way (not using ES6 because I don't know how to get freecodecamp to use it):

function sumAll(arr) {
var min = Math.min.apply(null, arr);
var max = Math.max.apply(null, arr);

return Array.apply(null, Array(max-min+1))   // get array of correct size
.map(function(_, b) { return b+min; })   // change it to have correct values
.reduce(function(a, b) { return a+b; }); // and sum it
}

• you would use reduce because it is a way to not access the array by indexes. Jan 18, 2016 at 21:54
• $max(a,b)-min(a,b)=|a-b|$, $min+max=a+b$ saves you some steps.
– A.S.
Jan 19, 2016 at 1:09

Like Barry mentioned, you are probably expected to use Gauss and the property of triangular numbers.

reduce is a funny way of accessing your array without touching the indexes, hence without needing to verify that they exist. (consider what would happen if your function was called with [] or [0]).

The one-liner you give is very close. In the final function you have a and b being the min and max values in that order. just apply the sum formula to it: return (a+b)*(b-a+1)/2;

So:

return arr
.sort((a, b) => a - b)
.reduce((a, b) => (a + b) * (b - a + 1) / 2);


You'll notice that since the array is size 2 and known to be that way, the complexity is O(1), while if you actually sum the numbers, it becomes O(n) where n is the amount of numbers to add.

Personally, I'd change out to result. And also the structure in something like the following:

function sumAll(arr){
var smallest = 0;
var greatest = 0;

if(arr[0] < arr[1]){
smallest = arr[0];
greatest = arr[1];
} else{
smallest = arr[1];
greatest = arr[0];
}

return sumRange(smallest, greatest);
}

function sumRange(from, to){
var result = 0;

for(var i = from; i <= to; i++){
result += i;
}

return result;
}


In this way it becomes more readable IMO.

• Javascript does not support ad hoc polymorphism. only the second function exists. Jan 18, 2016 at 22:09
• @njzk2 thanks for pointing that out. Silly mistake. Now it should be ok. Jan 18, 2016 at 22:48