Return sum of a number (positive or negative)

Question

I need to make a function that takes a number and returns sum of it's digits, if the number is negative the first digit should be considered negative when adding numbers, this is what I have:

var arrx = [];
var oper;
var others = 0;

function sumDigits(num) {
    // your code here
    var y = num.toString();
    var c = y.split("");
    c.forEach((h) => arrx.push(Number(h)) );
    if (num < 0){
        oper = -arrx[0];
        for (var z = 1; z < arrx.length; z++){
            others += arrx[z];
        }

        return others + oper;
    }

    return arrx.reduce((a,b) => a+b);
}

sumDigits(1234);

When given negative number, function returns NaN, what's the problem ?

Why do you push the digits to an array when `c` is already one? — Code-Apprentice, Jan 29 '17 at 23:03
I suggest developing an algorithm that works directly with numbers rather than converting the number to a String. Can you think of a way to get the ones digit from a number? How do you get the rest of the number other than the ones digit? — Code-Apprentice, Jan 29 '17 at 23:04

Abhinav Galodha · Answer 1 · 2017-01-29T22:52:17.830

In case of negative number, the first char is '-' the minus symbol. When you are trying to convert it to number, you are getting NaN. After that if you try to add NaN to any number the result would be NaN.

As a resolution, you need to ignore the first digit if the number is negative.

So, you can make a code change like

if(z === 1){
    others = others - arrx[z];
    }else{
     others += arrx[z];
      }

and also changing the return in case of negative numbers to return others;

Following code should work.

var arrx = [];
var oper;
var others = 0;

function sumDigits(num) {
 // your code here

var y = num.toString();

var c = y.split("");

c.forEach((h) => arrx.push(Number(h)) );

if (num < 0){


for (var z = 1; z < arrx.length; z++){

  if(z === 1){
    others = others - arrx[z];
    }else{
     others += arrx[z];
      }

}

return others;
}

return arrx.reduce((a,b) => a+b);

}
console.log(sumDigits(-1234));

If I ignore it the result won't be good in case of a negative number. — , Jan 29 '17 at 22:49
Check the code in post, it should work, it is pretty much the same, except the `return` and `if` condition. — Abhinav Galodha, Jan 29 '17 at 22:55

RomanPerekhrest · Accepted Answer · 2017-01-29T23:01:01.503

1

Use optimized and short version of sumDigits() function:

function sumDigits(num) {
  var isNeg = num < 0,   // check whether the number is negative
      numbers = (isNeg? String(num).slice(1) : String(num)).split('').map(Number);
  if (isNeg) numbers[0] *= -1;   // 'recovering' the number's sign

  return numbers.reduce(function(a,b){ return a + b; });
}

console.log(sumDigits(1234));
console.log(sumDigits(-951));

edited Jan 29 '17 at 23:01

answered Jan 29 '17 at 22:53

RomanPerekhrest

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i am having trouble understanding the value in numbers. If I was to write it out, would it look like this? `if(isNeg) { String(num).slice(1); } else { String(num); } numbers.split('').map(Number);` ? – Francis Apr 12 '17 at 18:42
@FrancisNgo, you forgot to assign the result of condition to `numbers` – RomanPerekhrest Apr 12 '17 at 19:51
`var numbers = if(isNeg) { String(num).slice(1); } else { String(num); } numbers.split('').map(Number);` ? – Francis Apr 12 '17 at 20:27
no, `if(isNeg) { numbers = String(num).slice(1); } else { numbers = String(num); } `. But it doesn't look much better than original – RomanPerekhrest Apr 12 '17 at 20:29

Return sum of a number (positive or negative)

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