Javascript Challenges - Array Sum

Javascript Challenges - Array Sum
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[@bulatbaltin5975]

Thanks a lot! One more solution function arraySum(arr) { let sum = 0; let max = 0; arr.forEach((num) => { if (max < num) { max = num; } sum += num; }) return (sum - max) === max; }

[@DEV_XO]

Amazing man, im enjoying this challenges. Hope you keep uploading stuff like this, some of us "knows the javascript surface", but didn't really knows how to deal with this kind of things, and well it really helps. Thanks a lot.

[@thelostvorg7805]

Answers can differ but here is mine function arraySum(arr){ const arrayMax = Math.max(...arr); return arr.reduce((t,p)=>t+p) - arrayMax === arrayMax }

[@victorvlas1241]

function ArraySum(arr) { const max = Math.max.apply(null, arr) return arr .filter(el => el !== max) .reduce((curr, acc) => curr + acc, 0) === max }

[@noirsociety]

function arraySum(arr) { const high = Math.max(...arr); const sum = arr.reduce((acc,num) => acc + num, 0) - high; return sum === high; } console.log(arraySum([1,2,4,6,13])); // true console.log(arraySum([1,2,4,34,22])); // false

[@mounirelardi1201]

const arraySum = arr => { let result = 0; for (const num of arr) { result += num; } return result - Math.max(...arr) === Math.max(...arr) } console.log(arraySum([1,2,3,7]))

[@SaliAalaNabiyak]

my little solution : function arraysum(arr) { let maxvalue = Math.max(...arr) let newarr = arr.filter((e) => { return e !== maxvalue }).reduce((pr, cu) => { return pr + cu }) if (newarr !== maxvalue){ return "the sum of the array except the max value doesn't equal the max value" } return 'yaaay it worked' } console.log(arraysum([2, 5, 7, 6, 1, 20])) > the output is: the sum of the array except the max value doesn't equal the max value console.log(arraysum([2, 5, 7, 6, 20])) > the output is: yaay it worked

[@TheJenasandeep]

I learn how to sort , difference between map and forEach and solutions to this type of coddding problem ❤️

[@EMartinezOliver]

function sum(arr){ let others = [] let restSum let max = Math.max(...arr) arr.filter(item => { if (item != max){ others.push(item) } }) restSum = others.reduce((a,b)=>{ return a + b }) if (max === restSum){ return true } else { return false} }

[@eskandarabedini6389]

Lots of thanks for your amazing course.

[@moulaoui1279]

const sum = (...values) => { const max = Math.max(...values); return max === values.reduce((acc,current) => acc+current ,0) / 2 }

[@_wayne7928]

function arraySum(arr){ toSumArray = arr.slice(0,(arr.length)-1) sumResult = toSumArray.reduce((x,y)=>x+y) if(sumResult == arr[arr.length-1]) return true else return false }

[@oguzakankan518]

Thank you for Vanilla js. I'll buy your js course.

[@CodingAddict]

Thank you for your support!

[@entertainment11378]

How can we do sum of uniques items in object array [ {id :1,chat: 2},{id :1, chat:3}] result should be [{ id:1, chat : 5}] ?

[@mickaelrichard7255]

function ArraySum(arr){ let largest = 0 const sum = arr.reduce((acc,item)=>{ if(item>largest) largest = item return acc += item },0) return largest === sum-largest } console.log(ArraySum([1,2,4,6,13]))

[@bikramchettri9405]

Thanks

[@ogzhn]

function arraySum(arr) { const maxNumber = Math.max(...arr) let total = 0; for (let i = 0; i < arr.length; i++) total+= arr[i] return total % maxNumber === 0 }

[@tigrangrigoryan6157]

function ArraySum(arr) { let tempArr = arr.sort((a, b) => a - b), sum = 0; for (let i = 0; i < tempArr.length - 1; i++) { sum += tempArr[i]; } return sum === tempArr[tempArr.length - 1]; }

[@hareeshsapparad2845]

function ArraySum(arr){ var result = 0; var maxVal = Math.max(...arr); for(let i = 0; i < arr.length; i++){ result += arr[i]; } if(result == (maxVal*2)){ return true } return false; } ArraySum([1,2,3,6]);//true ArraySum([1,2,3,6,7]);//false

[@roman4889]

const getSumEqualsMaxNum = arr => { const sum = arr.reduce((sum, curr) => sum + curr); const max = Math.max(...arr); return max === sum - max; }

[@bishalshah1465]

Using filter and reduce const arraySum = (arr) => { const max = Math.max(...arr); return arr .filter((el) => el !== max) .reduce((a, b) => a + b)=== max; };

[@ovuobachidera3987]

this is really good thnks for the share

[@ganeshpbharadwaj8773]

function arraySum(arr){ let total = 0; for(let num of arr){ if (num < Math.max(...arr)){ total += num; } } return total == Math.max(...arr); } console.log(arraySum([1,2,4,6,13])); console.log(arraySum([1,2,4,34,22]));

[@davidpocsai3419]

Can someone come up with a shorter solution than this? const Arraysum = (arr) => Math.max(...arr) * 2 === arr.reduce((a, b) => a + b,0);

[@mahmoudelkhadragy8154]

function arraySum(arr) { let max = Math.max(...arr); let newArr = arr.filter((num) => num !== max); let acc = newArr.reduce((acc, num) => (acc += num)); if (acc === max) { return true; } return false; } console.log(arraySum([1, 2, 4, 6, 13])); //true console.log(arraySum([1, 2, 4, 34, 22])); //false

[@sagartechnosoft2807]

let arr = [1,2,3,4,5]; let arrSum = (arr) => { let total = 0; arr.forEach(element => { total = total + element; }); return total } console.log(arrSum(arr));

[@emre-mz4sc]

my solution: function sumAll(arr) { let sum = 0; let biggest = Math.max(...arr); let array = arr.filter((item) => item !== biggest); for (number of array) { sum = sum + number; } if (sum === biggest) return true; else { return false; } } console.log(sumAll([1, 2, 4, 6, 13]));

[@KhanhPham-jq7cf]

Just one line of code to solve the problem, take a look: const ArraySum = (arr) => arr.reduce((sum, value) => sum+value, 0) === 2*arr.sort((a,b) => b-a)[0] console.log(ArraySum([1, 6, 4, 2, 13])) // true console.log(ArraySum([1, 2, 4, 34, 22])) // false

[@rajatgupta9582]

function ArraySum(arr) { let sum = 0 var max = Math.max(...arr) for (let i = 0; i < arr.length; i++) { if (arr[i] === max) { continue } sum += arr[i] } return sum === max }