Almost the same code but very different answers

Question

I was writing a program in JavaScript that calculates the number of 6-digit numbers the sum of whose first 3 digits is equal to the sum of the last 3 digits. So I wrote two different solutions to the problem although only one returns the correct answer.
Also I wrote a function sumOfDigits() which simply calculates the sum of digits of the number passed to it. This function is not written here but it works correct.

function count1() {
    total=0;
    for (i = 100000; i <= 999999; i+=1) {
         part = i % 1000;
         if ((sumOfDigits(i-part)) == ((sumOfDigits(part)))) {
             total+=1;
         }
     } return total;
}

function count2() {
    array = [];
    for (i = 100000; i <= 999999; i+=1) {
        part = i % 1000;
        if ((sumOfDigits(i-part)) == ((sumOfDigits(part)))) {
            array.push(i);
        }
    } return array;
}

The count1() function does not work correctly and returns 28 as the answer while the count2() function which returns an array returns an array of length 50412 which is the correct answer. Can somebody please tell me why the first function does not work correctly.
A screenshot of the count1 function in action. A screenshot of the count1() function in action.

i'm try your `count1` function and it return `50412`, possibly in your code you have another function with name `count1` — Grundy, Dec 26 '13 at 09:29
I also tested it, and it works fine. Whatever you're doing wrong isn't in the function. Btw, go easy on the parentheses. You can just write `if (sumOfDigits(i-part) == sumOfDigits(part))`. — Marcelo Cantos, Dec 26 '13 at 09:42
@MarceloCantos It aint working for me. Check the screenshot i have uploaded. — Shaurya Gupta, Dec 26 '13 at 09:46
try to merge both the codes. I mean have a counter as well as an array at the same time. Then see what are the corresponding values. — Tejasva Dhyani, Dec 26 '13 at 09:48
You use global variables inside your functions, it is not very well :-) — Grundy, Dec 26 '13 at 09:50
I stand (self-)corrected. The problem was inside the function. — Marcelo Cantos, Dec 26 '13 at 09:53

Marcelo Cantos · Accepted Answer · 2013-12-27T00:56:42.810

5

Are you using total in the implementation of sumOfDigits()? I'm guessing that both functions are using the same global total.

You should prefix all local variables with var on their first use so that they aren't treated as global variables, e.g.:

var total = 0;

EDIT: I just confirmed this. If sumOfDigits() uses the global total, count1() returns 28.

edited Dec 27 '13 at 00:56

answered Dec 26 '13 at 09:46

Marcelo Cantos

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Yes, now it works! But why did the var matter? Can you please edit your answer to explain that too. – Shaurya Gupta Dec 26 '13 at 09:54
1

@shauryagupta when you ouse global var - on every call `sumOfDigits` it change, so as result you have 9+9+9 as total and increment it, so 28 – Grundy Dec 26 '13 at 09:58
2

http://stackoverflow.com/questions/500431/javascript-variable-scope more abuot variables – nzn Dec 26 '13 at 10:01

Ringo · Answer 2 · 2013-12-26T09:50:19.973

1

May be total is a global variable with some value on your solution. Try create another variable for return:

 function count1() {
        var count = 0;
        for (i = 100000; i <= 999999; i+=1) {
             part = i % 1000;
             if ((sumOfDigits(i-part)) == ((sumOfDigits(part)))) {
                 count+=1;
             }
         } return count;
    }

edited Dec 26 '13 at 09:50

answered Dec 26 '13 at 09:30

Ringo

3,795
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No, my code is correct because the `if` block must be true 50412 times. – Shaurya Gupta Dec 26 '13 at 09:39
3

`total += 1` and `array.push(i)` are called the same number of times, so `total` should equal `array.length`. – Marcelo Cantos Dec 26 '13 at 09:40
@shauryagupta, possibly in your code you have another function with name `count1` that return another value? – Grundy Dec 26 '13 at 09:41

KooiInc · Answer 3 · 2013-12-27T10:06:04.943

I think you could reconsider your code. Here is a simpler and faster approach (<500ms in nodejs, faster because it's all numbers, i.e. no conversions):

//helper: sum of digits in number [n]
function sumOfDigits(n){
   if (!(n^0) || n<10) {return n;}
   var sum = 0;
   while (n>=10) {
     sum += n%10;
     n = (n/10)^0;
   }
   return sum+n;
}

function sumDigitsComparer(num){
  return  sumOfDigits((num/1000)^0) === sumOfDigits(num%1000);
}

var x = 1000000, n = 0;
while ((x-=1,x>=100000)) { sumDigitsComparer(x) && (n+=1); }
//=> n = 50412

Almost the same code but very different answers

3 Answers3