Is it a balanced number?

Question

A number is balanced if the sum of the digits on each half of the number is equal, so: 1423 is balanced because 1+4 = 2+3, so is: 42615 because 4+2=1+5. Note that the middle digit is not included on either side (or it's included on both sides) if there's an odd number of digits.

Challenge:

Take a positive integer as input, and output a truthy value if it's balanced and a falsy value if it's unbalanced.

Test cases (true)

1
6
11
141
1221
23281453796004414
523428121656666655655556655656502809745249552466339089702361716477983610754966885128041975406005088

Test cases (false)

10
12
110
15421
5234095123508321
6240911314399072459493765661191058613491863144152352262897351988250431140546660035648795316740212454

There will not be numbers starting with zero, for instance 00032 instead of 32. You must support numbers up to at least 100 digits (so larger than 2^64-1). As always, optional input format, so you may surround the number with apostrophes if desired.

rnso · Accepted Answer · 2016-09-23 17:07:45Z

Racket 204 bytes

(define(f n)(let*((s(number->string n))(g(λ(x y)(apply +(map(λ(x)(string->number(string x)))
(string->list(substring s x y))))))(l(string-length s))(h(/ l 2)))(if(=(g 0(floor h))(g(ceiling h)l))#t #f)))

Detailed version:

(define (f1 n)
(let* (  (s (number->string n))
         (g(λ(x y)
              (apply + (map
                        (λ(x)
                          (string->number
                           (string x)))
                        (string->list
                         (substring s x y))))))
         (l (string-length s))
         (h (/ l 2)))
    (if(= (g 0 (floor h)) (g (ceiling h) l)) 
       #t #f  ) ) )

Testing:

(f 23281453796004414)
(f 523428121656666655655556655656502809745249552466339089702361716477983610754966885128041975406005088)
(f 15421)
(f 5234095123508321)

Output:

#t
#t
#f
#f

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Pyke, 20 bytes

[QleQ`)>[R<]mmbms$X!

Try it here!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 23, 2016 at 12:45

Blue

28.4k8 gold badges49 silver badges99 bronze badges

\$\begingroup\$ @WeeingIfFirst you were right - I forgot not [0] was 0 not 1 \$\endgroup\$
– Blue
Sep 23, 2016 at 19:48

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Mathematica, 69

Tr@#[[;;⌊l=Length@#/2⌋]]==Tr@#[[⌈l⌉+1;;]]&@*IntegerDigits

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 23, 2016 at 12:55

shrx

4622 silver badges4 bronze badges

2

\$\begingroup\$ You can save a couple of bytes by changing the end to ...;;]]&@*IntegerDigits \$\endgroup\$
– Martin Ender
Sep 23, 2016 at 15:01
\$\begingroup\$ @MartinEnder thanks, but how does this work? \$\endgroup\$
– shrx
Sep 24, 2016 at 7:32
\$\begingroup\$ @* is short for Composition. f@*g is f[g[##]]&. \$\endgroup\$
– Martin Ender
Sep 24, 2016 at 7:44

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Jörg Hülsermann · Accepted Answer · 2016-09-24 11:07:54Z

1

PHP, 66 Bytes

for(;2*$i<$l=strlen($t=$argv[1]);)$s+=$t[$i++]-$t[$l-$i];echo+!$s;

-4 Bytes by @Titus Thank You

edited Sep 24, 2016 at 11:07

answered Sep 23, 2016 at 16:11

Jörg Hülsermann

13.3k3 gold badges13 silver badges36 bronze badges

1

\$\begingroup\$ That $t[$i++]-$t[$l-$i] trick is great! It saves 4 bytes in my JS answer :-) \$\endgroup\$
– ETHproductions
Sep 23, 2016 at 16:26
\$\begingroup\$ unnecessary parentheses around the $l assignment; always returns 1 for odd lengths with PHP < 7. \$\endgroup\$
– Titus
Sep 24, 2016 at 8:35
\$\begingroup\$ @Titus the result for php -r "for(;$i<($l=strlen($t=$argv[1]))/2;)$s+=$t[$i++]-$t[$l-$i];echo$s?0:1;" 110 is 0 under PHP 7. You are sure that your input is a string? \$\endgroup\$
– Jörg Hülsermann
Sep 24, 2016 at 10:15
1

\$\begingroup\$ Oops. You´re right. Use 2*$i<$l=strlen($t=$argv[1]) instead. \$\endgroup\$
– Titus
Sep 24, 2016 at 10:59
1

\$\begingroup\$ And echo+!$s; for the output. (-2) \$\endgroup\$
– Titus
Sep 24, 2016 at 11:01

| Show 2 more comments

GolfIsAGoodWalkSpoilt · Accepted Answer · 2016-09-27 07:32:58Z

1

Groovy - 74 69 bytes

Represented as a Groovy closure. Please note that the parameter passed in (i) is a String, as with the Java version.

{i->e=i.size();k=0;s=0;for(;s<e;){k+=i.charAt(s++)-i.charAt(--e)};!k}

edited Sep 27, 2016 at 7:32

answered Sep 26, 2016 at 17:00

GolfIsAGoodWalkSpoilt

1112 bronze badges

\$\begingroup\$ Hi, and welcome to the site! This is a nice first post, but assuming the input is stored in a predefined variable is not an allowed input method. I don't know much Groovy, but could you add something like i=input() or turn it into a function? \$\endgroup\$
– DJMcMayhem
Sep 26, 2016 at 17:04
\$\begingroup\$ Rats, sorry - should be a bit better now; please do let me know. \$\endgroup\$
– GolfIsAGoodWalkSpoilt
Sep 27, 2016 at 7:21

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user470370 · Accepted Answer · 2016-09-29 16:00:23Z

1

JavaScript (ES5), 70 66 bytes

function g(n){for(a=b=0;(l=n.length)>b*2;)a+=n[b]-n[l-++b];return!a}


// tests:
var falses = ['10', '12', '110', '15421', '5234095123508321', '6240911314399072459493765661191058613491863144152352262897351988250431140546660035648795316740212454']
var i=falses.length;
while(i--) console.info(!g(falses[i]) ? 'ok' : 'not false: ' + falses[i])

var trues = ['1', '6', '11', '141', '1221', '23281453796004414', '523428121656666655655556655656502809745249552466339089702361716477983610754966885128041975406005088'];
i = trues.length;
while(i--) console.info(g(trues[i]) ? 'ok' : 'not true: ' + trues[i])

edited Sep 29, 2016 at 16:00

answered Sep 28, 2016 at 20:41

user470370

2011 silver badge3 bronze badges

1

\$\begingroup\$ Nice answer! I count 67 bytes, not including the g (it doesn't count toward your byte count). You can also save a byte or two by switching to a for loop. \$\endgroup\$
– ETHproductions
Sep 28, 2016 at 21:05
\$\begingroup\$ thx - very useful!! \$\endgroup\$
– user470370
Sep 29, 2016 at 16:00

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Magic Octopus Urn · Accepted Answer · 2016-09-30 16:21:10Z

1

Groovy (85 73 Bytes)

f={n->x=n.collect{[it,it]}.flatten().collate(n.size())*.sum();x[0]==x[1]}

.collect{[it,it]} - Dupe all elements into a 2D array.
.flatten() - Turn 2D array into 1D array.
.collate(n.size()) - Split new array in half.
*.sum() - Sum both sub-arrays.
x[0]==x[1] - Check for equality / return.

edited Sep 30, 2016 at 16:21

answered Sep 30, 2016 at 16:09

Magic Octopus Urn

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DELETE_ME · Accepted Answer · 2018-08-01 03:49:04Z

1

Jelly, 7 bytes

Dx2ŒH§E

Try it online!

Same algorithm as 05AB1E answer.

answered Aug 1, 2018 at 3:49

DELETE_ME

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Python 3, 69 67 bytes

I'm quite proud of this one, since it's smaller than any other python answers in this thread :). I owe the idea of printing the equivalence of the two sums to tuukkaX's answer though, I don't think I would have thought of that myself.

*a,=map(int,input())
h=(len(a)+1)//2
print(sum(a[:h])==sum(a[-h:]))

Try it online!

The +1 on the second line is necessary to make my slices work properly for single digit inputs. I realised that even though it overslices, it does so on both sides so it still works.

On a list of len 1, slicing a[:0] gives an empty list but a[-0;] gives the whole list. Every 1 digit input became itself==0 so it was always false.

thanks to david for shaving off 2 bytes

\$\begingroup\$ h=(len(a)+1)//2 saves a couple of bytes \$\endgroup\$
– david
Dec 14, 2018 at 9:45 — david, Dec 14, 2018 at 9:45

dana · Accepted Answer · 2018-12-14 23:35:27Z

1

C# (Visual C# Interactive Compiler), 58 bytes

s=>s.Zip(s.Reverse(),(a,b)=>a-b).Take(s.Length/2).Sum()==0

Try it online!

edited Dec 14, 2018 at 23:35

answered Dec 14, 2018 at 23:06

dana

2,6811 gold badge13 silver badges11 bronze badges

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evanstar3 · Accepted Answer · 2022-06-09 20:09:19Z

1

C (gcc), 98 bytes

V=2;main(A,v)char**v;{char*a=*++v;while(a[1])a++;while(*v<a)V+=*(*v)++,A+=*a--;puts(A^V?"0":"1");}

There's a shorter submission for a C function, but mine compiles on its own and handles command line arguments.

Try it online!

edited Jun 9, 2022 at 20:09

answered Jun 9, 2022 at 19:58

evanstar3

3387 bronze badges

\$\begingroup\$ 92 bytes \$\endgroup\$
– ceilingcat
Jun 12, 2022 at 4:43

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zoomlogo · Accepted Answer · 2023-03-01 08:19:35Z

1

flax, 7 bytes

EṠXFz˘D

Port of the 05AB1E answer. Ties with jelly. No online link as the latest commit is not on ATO.

Explanation

EṠXFz˘D

      D  ⊳ Digits
    z˘   ⊳ Zip with itself
   F     ⊳ Flatten
  X      ⊳ Split into two
EṠ       ⊳ Are the sum of the two halves equal?

answered Mar 1 at 8:19

zoomlogo

1,8851 gold badge9 silver badges29 bronze badges

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Grax32 · Accepted Answer · 2016-09-23 18:06:56Z

0

JavaScript 95 88 bytes

i=>(s=(i,a,b)=>[...i.slice(a,b)].reduce((p,e)=>p-e,0))(i=i+"",0,m=i.length/2)==s(i,m+.5)

Explanation:

Convert input to a string, create a function that takes string segments and sums up all of the digits, and call that function for the left and right halves of the input.

edited Sep 23, 2016 at 18:06

answered Sep 23, 2016 at 17:49

Grax32

1,3027 silver badges10 bronze badges

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Himself12794 · Accepted Answer · 2016-09-23 21:12:38Z

0

Scala, 77 bytes

Like Java, must take Strings.
I'm sure someone could find an improvement.

def f(s:String)=(for(i<-0 until s.length/2)yield s(i)-s(s.length-i-1)).sum==0

Can test online

answered Sep 23, 2016 at 21:12

Himself12794

614 bronze badges

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david · Accepted Answer · 2018-12-14 10:44:31Z

0

Tcl, 156 bytes

proc R n {set l [string le $n];puts [expr [join [split [string ra $n 0 [expr $l/2-($l%2==0)]] {}] +]==[join [split [string ra $n [expr $l/2] $l+$l] {}] +]]}

Try it online!

answered Dec 14, 2018 at 10:44

david

4993 silver badges13 bronze badges

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DeathIncarnate · Accepted Answer · 2022-06-09 19:48:55Z

0

Burlesque, 39 bytes

{iTg1!!}s9XXsa2./s1{q%9!{<-%9!}}M-)++sm

Try it online!

{
 iT   # Generate tails
 g1   # Get var 1
 !!   # Select from array
}s9   # Save as var 9
XX    # Split int to digits
sa    # Number of digits
2./   # /2
s1    # Save as 1
{
 q%9! # Eval 9 as function 
 {
  <-  # Reverse
  %9! # Eval 9 as function
 }
}M-   # Eval each contained function on the digits
)++   # Map sum
sm    # Same

answered Jun 9, 2022 at 19:48

DeathIncarnate

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The Thonnu · Accepted Answer · 2023-03-01 18:56:13Z

0

Thunno `E`, \$14\log_{256}(96)\approx\$ 11.52 bytes

eDZPiES2AP.Sze

Attempt This Online!

Port of Emigna's 05AB1E answer.

    # E flag converts to string 
e   # Map:
DZP #  Pair with itself
i   #  Convert to integer
E   # End map
S   # Flatten
2AP # Split in half
.S  # Sum each
ze  # All equal?

answered Mar 1 at 18:56

The Thonnu

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Stack Exchange Network

Is it a balanced number?

47 Answers 47

Racket 204 bytes

Pyke, 20 bytes

Mathematica, 69

PHP, 66 Bytes

Groovy - 74 69 bytes

JavaScript (ES5), 70 66 bytes

Groovy (85 73 Bytes)

Jelly, 7 bytes

Python 3, 69 67 bytes

C# (Visual C# Interactive Compiler), 58 bytes

C (gcc), 98 bytes

flax, 7 bytes

Explanation

JavaScript 95 88 bytes

Scala, 77 bytes

Tcl, 156 bytes

Burlesque, 39 bytes

Thunno `E`, \$14\log_{256}(96)\approx\$ 11.52 bytes

Your Answer

Not the answer you're looking for? Browse other questions tagged
code-golf
number
or ask your own question.

Linked

Hot Network Questions

Is it a balanced number?

47 Answers 47

Racket 204 bytes

Pyke, 20 bytes

Mathematica, 69

PHP, 66 Bytes

Groovy - 74 69 bytes

JavaScript (ES5), 70 66 bytes

Groovy (85 73 Bytes)

Jelly, 7 bytes

Python 3, 69 67 bytes

C# (Visual C# Interactive Compiler), 58 bytes

C (gcc), 98 bytes

flax, 7 bytes

Explanation

JavaScript 95 88 bytes

Scala, 77 bytes

Tcl, 156 bytes

Burlesque, 39 bytes

Thunno E, \$14\log_{256}(96)\approx\$ 11.52 bytes

Your Answer

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Not the answer you're looking for? Browse other questions tagged code-golfnumber or ask your own question.

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Hot Network Questions

Thunno `E`, \$14\log_{256}(96)\approx\$ 11.52 bytes

Not the answer you're looking for? Browse other questions tagged
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