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Sum All Numbers in a Range

Mohammed Awad on May 10, 2023

DESCRIPTION: We'll pass you an array of two numbers. Return the sum of those two numbers plus the sum of all the numbers between them. T...
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Jon Randy 🎖️ • • Edited

My approach, based on the fact that the sum from 0→n is equal to (n/2)(n+1):

const sumAll =
  (arr, [a,b]=[...arr].sort((c,d)=>c-d))=>(b*b-a*a+b+a)/2
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(Could be made shorter by not copying the original array, but I wanted to avoid mutation)

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Jon Randy 🎖️ • • Edited

With mutation (and more code golf):

const sumAll=(a,[b,c]=a.sort((d,e)=>d-e))=>(c*c-b*b+c+b)/2
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Mohammed Awad •

thank you, Relly for your interest!

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donnywi •

This works fast too when you pass in:
sumAll(7, 80000)

function sumAll(arr) {
    return ( (arr[0]+arr[1]) * (Math.abs(arr[0]-arr[1])+1) ) /2 ;
}
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Mohammed Awad •

I'm suck at Math; I hope to be become good like that someday.

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donnywi profile image
donnywi •

I think you're awesome at Javascript, and I need to learn how to use fill(), map(), and reduce() properly. I'm new at Javascript, so your posts are very inspiring.

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Mohammed Awad •

Relly, glad to hear that. you can check freecodecamp.com to learn more

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Red Ochsenbein (he/him) •

Try and be amazed:

function sumAll(arr) {
  const minNumber = Math.min(...arr);
  const maxNumber = Math.max(...arr);
  return (maxNumber*(maxNumber+1)/2) - ((minNumber-1)*minNumber/2);
}
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Mohammed Awad •

this is a pattern ?

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donnywi •

I belive that's just a basic math equation.

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Mohammed Awad •

Yeah, but I mean is it design pattern.

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syeo66 profile image
Red Ochsenbein (he/him) •

Nah. It's math. The basic idea is that you can calculate the sum of all integers from 1 to n using n*(n+1)/2

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Mohammed Awad •

I need to focus on math and learn how use it because better with BigO

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Andrew Bone • 
function sumAll(arr) {
  const [a, l] = arr.sort((a, b) => a - b);

  return (l - a + 1) * (a + l) / 2;
}
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Mohammed Awad •

Any tips or edit are most welcome. share it with me on the comments. Thanks for being here!

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Jon Randy 🎖️ • • Edited

Another one (no mutation):

const sumAll = ([m,n],[a,b]=m>n?[n,m]:[m,n])=>(b*b-a*a+b+a)/2
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Mohammed Awad •

great but it's seemed little complex, right? I more into simple solutions

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Jon Randy 🎖️ • • Edited

It's interesting you would say that, as your solution is actually the most complex (timewise) here at O(N). The simplest I can now think of is a small reworking of @donnywi solution (which in itself is a slight mathematical simplification of others here):

const sumAll = ([a, b]) => (a+b) * (Math.abs(a-b)+1) / 2
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Jon Randy 🎖️ •

I suspect the way I wrote my code in posts here may have made it look more complex than it is, as it doesn't really explain itself. Sorry, my brain often tends to work in terse, code-golf like code.

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Mohammed Awad •

yes, I know my code was the worst, but it easier to read than yours, I should study the math section again, because I didn't Relly use Math methods expect min, max and random.

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donnywi •

This is a good use of the arrow function. 👍