How to sum all elements in a nested array?

Question

this is an array which is given:

let arrayNumbers = [ 5, 7, 
        [ 4, [2], 8, [1,3], 2 ], 
        [ 9, [] ], 
        1, 8
      ];

and the output must be 50

This code must be WITHOUT Array.isArray;

let sum = 0;
function treeSum(array) {
    for (let i = 0; i < array.length; i++) {
        if (array[i] === false) {
            treeSum(array[i]);
        } else {
            sum += array[i];
        }
    }
    return sum;
}

treeSum(arrayNumbers);

Help, please with an idea or code, thanks.

why either "count array elements" or get sum of all? and why 5? — Nina Scholz
– Nina Scholz, Commented Jan 28, 2020 at 13:34
Is it worth explaining why you can't use Array.isArray? It's supported all the way down to IE9. — customcommander
– customcommander, Commented Jan 28, 2020 at 13:51
If Array.isArray cannot be used, then what else cannot be used either? flat, flatMap, reduce, recursion, typeof, instanceof, constructor, ...? Such requirements make no sense without more context. — trincot
– trincot, Commented Jan 28, 2020 at 15:09

Community · Accepted Answer · 2020-06-20 09:12:55Z

2

You need to use a local sum variable as well to prevent duplicate adding of nested sums.

You could check if the item

is an instance of Array with the instanceof operator, of

function treeSum(array) {
    var sum = 0;
    for (let i = 0; i < array.length; i++) {
        sum += array[i] instanceof Array ? treeSum(array[i]) : array[i];
    }
    return sum;
}

let arrayNumbers = [5, 7, [4, [2], 8, [1, 3], 2],  [9, []], 1, 8];

console.log(treeSum(arrayNumbers));

has Array as Object#constructor.

function treeSum(array) {
    var sum = 0;
    for (let i = 0; i < array.length; i++) {
        sum += array[i].constructor === Array ? treeSum(array[i]) : array[i];
    }
    return sum;
}

let arrayNumbers = [5, 7, [4, [2], 8, [1, 3], 2],  [9, []], 1, 8];

console.log(treeSum(arrayNumbers));

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Jan 28, 2020 at 13:40

Nina Scholz

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Comments

shashank · Accepted Answer · 2020-01-28 16:06:36Z

2

Flatten and reduce the array as below:-

var sum = arrayNumbers.flat(Infinity).reduce((a,b)=>a+b)

answered Jan 28, 2020 at 16:06

shashank

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Comments

Mihai Alexandru-Ionut · Accepted Answer · 2020-01-28 13:46:33Z

1

You could flat the list then use reduce method to obtain the sum.

You can use if (item.map) condition in order to check if it's array.

let arrayNumbers = [ 5, 7, 
        [ 4, [2], 8, [1,3], 2 ], 
        [ 9, [] ], 
        1, 8
      ];
const flatArray = (arr) => {
    const res = []
    for (const item of arr) {
        if (item.map) {
            const subRes = flatArray(item)
            res.push(...subRes)
        } else {
            res.push(item)
        }
    }
    return res;
}      
console.log(flatArray(arrayNumbers).reduce((a, b) => a + b));

edited Jan 28, 2020 at 13:46

answered Jan 28, 2020 at 13:39

Mihai Alexandru-Ionut

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6 Comments

10100011001 Over a year ago

I cannot do it like this also, that's the problem(

customcommander Over a year ago

Array#reduce throws an error against an empty array if you don't have an initial value. So you should probably set 0 as an initial value of your reducer. Note that Number([]) returns 0 so this would be consistent with how JavaScript would convert an empty array into a number.

Mihai Alexandru-Ionut Over a year ago

@PStarter, I used if (item.map) to check if it's array.

Mihai Alexandru-Ionut Over a year ago

@customcommander, the OP says This code must be WITHOUT Array.isArray;

Mihai Alexandru-Ionut Over a year ago

@customcommander, one possible alternative could be typeof arr.map === 'function'

|

StepUp · Accepted Answer · 2020-01-28 13:56:51Z

0

You can use recursion. So you need to check whether iterated element is array. If it is array, then run sumArray method again, otherwise you can sum up value:

const sumArray = (arr) => {
  let sum = 0;
  arr.forEach(element => {
    if (typeof element != 'number') {
        sum += sumArray(element);
    } else {
      sum += element
      return sum;
    }
  });
  return sum;
}


console.log(sumArray(arrayNumbers));

An example:

let arrayNumbers = [ 5, 7,
  [ 4, [2], 8, [1,3], 2 ],
  [ 9, [] ],
  1, 8
];

const sumArray = (arr) => {
  let sum = 0;
  arr.forEach(element => {
    if (typeof element != 'number') {
        sum += sumArray(element);
    } else {
      sum += element
      return sum;
    }
  });
  return sum;
}

console.log(sumArray(arrayNumbers));

edited Jan 28, 2020 at 13:56

answered Jan 28, 2020 at 13:51

StepUp

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Comments

Zulqarnain Jalil · Accepted Answer · 2020-01-28 14:02:46Z

0

you can use the combination of 'join', 'split' and reduce

var t =[ 5, 7, 
        [ 4, [2], 8, [1,3], 2 ], 
        [ 9, [] ], 
        1, 8
      ].join().split(',').reduce(function(a,b) { return parseInt(a?a:"0") + parseInt(b?b:"0")})
      
 console.log(t)

answered Jan 28, 2020 at 14:02

Zulqarnain Jalil

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Comments

LukStorms · Accepted Answer · 2020-01-28 15:47:30Z

0

A simple trick that will work on any array depth is to join to a string.
Then use match to get the numbers.
Then reduce those numbers to a sum.

Example snippet:

let arrayNumbers = [ 5, 7, 
    [ 4, [2], 8, [1,3], 2 ], 
    [ 9, [] ], 
    1, 8
  ];
  
let sum = arrayNumbers
        .join()
        .match(/\d+/g)
        .reduce((acc, x) => acc + parseInt(x), 0);

console.log(sum)

edited Jan 28, 2020 at 15:47

answered Jan 28, 2020 at 15:38

LukStorms

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Comments

Equal · Accepted Answer · 2020-01-28 16:29:06Z

Recursion version using `Array.prototype.reduce`

let arrayNumbers = [5, 7, [4, [2], 8, [1, 3], 2], [9, []], 1, 8];

function treeSum(array) {
  return array.reduce(
    (sum, n) =>
      sum +
      (typeof n === "number"
        ? // n is a number
          n
        : // n is an array
          treeSum(n)),
    0
  );
}

console.log(treeSum(arrayNumbers) === 50);

Recursion version using `for...of`

let arrayNumbers = [5, 7, [4, [2], 8, [1, 3], 2], [9, []], 1, 8];

function treeSum(array, sum = 0) {
  for (const n of array) {
    sum += typeof n === "number" ? n : treeSum(n);
  }
  return sum;
}

console.log(treeSum(arrayNumbers) === 50);

Iteration version

let arrayNumbers = [5, 7, [4, [2], 8, [1, 3], 2], [9, []], 1, 8];

function treeSum(array) {
  let sum = 0;

  do {
    // not summed values
    let values = [];
    for (const n of array) {
      if (typeof n === "number") sum += n;
      else values = values.concat(n);
    }
    array = values;
  } while (array.length > 0);

  return sum;
}

console.log(treeSum(arrayNumbers) === 50);

Oneline solution using `Array.prototype.flat`

let arrayNumbers = [5, 7, [4, [2], 8, [1, 3], 2], [9, []], 1, 8];

function treeSum(array) {
  return array.flat(Infinity).reduce((sum, n) => sum + n, 0);
}

console.log(treeSum(arrayNumbers) === 50);

Equal · Accepted Answer · 2020-01-30 15:00:05Z

-1

Use flat function:

var flat = arrayNumbers.flat(Infinity);
var sum = flat.reduce((x, y) => x + y);
console.log(sum);

edited Jan 30, 2020 at 15:00

Equal

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answered Jan 28, 2020 at 14:37

Ali Adravi

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2 Comments

customcommander Over a year ago

Why .flat().flat() when you can specify a depth? A reducer without an initial value will throw an error if the array is empty. You should specify 0 as an initial value.

Ali Adravi Over a year ago

Thanks for the depth suggestion, and will not give error because it is already have value, otherwise you can use the default value

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Recursion version using `Array.prototype.reduce`

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