Find The Sum of the First n Bouncy Numbers

Question

Terminology

An increasing number is one where each digit is greater than or equal to all the digits to the left of it (ex. 12239)

A decreasing number is one where each digit is less than or equal to all the digits to the left of it (ex. 95531)

A bouncy number is any number that is not increasing or decreasing. Since this requires at least 3 digits, the first bouncy number is 101

The task

Given an integer n greater than or equal to 1, find the sum of the first n bouncy numbers

Rules

• This is code golf, so the answer with the shortest amount of bytes wins
• If your language has limits on integer size (ex. 2^32-1) n will be small enough that the sum will fit in the integer
• Input can be any reasonable form (stdin, file, command line parameter, integer, string, etc)
• Output can be any reasonable form (stdout, file, graphical user element that displays the number, etc)

Test cases

1 > 101
10 > 1065
44701 > 1096472981

Answer 1

Jelly, 10 8 bytes

ṢeṚƬ¬µ#S

Try it online!

How it works

ṢeṚƬ¬µ#S  Main link. No arguments.

      #   Read an integer n from STDIN and call the chain to the left with argument
          k = 0, 1, 2, ... until n of them return a truthy result.
          Yield the array of successful values of k.
     µ    Monadic chain. Argument: k (integer)
Ṣ           Sort, after promoting k to its digit array.
  ṚƬ        Reverse 'til the results are no longer unique and yield unique results.
            Calling Ṛ on k promotes it to its digit array. If k = 14235, the 
            result is [14235, [5,3,2,4,1], [1,4,2,3,5]].
 e          Check if the result to the left appears in the result to the right.
    ¬       Negate the resulting Boolean.
       S  Take the sum.

Answer 2

Pyth, 10 bytes

s.f!SI#_B`

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How it works?

s.f!SI#_B` – Full program. Takes an integer Q from STDIN and outputs to STDOUT.
 .f        – Find the first Q positive integers that satisfy a certain condition.
   !SI#_B  – The condition. Returns true for bouncy numbers only.
       _B` – Cast the number to a string and bifurcate (pair) it with its reverse.
      #    – Filter-keep those...
     I     – That are invariant under...
    S      – Sorting.
           – To clarify, I (invariant) is a Pyth operator that takes two inputs, a 
             function and a value and checks whether function(value) == value, so
             this is technically not a built-in.
   !       – Logical not. The empty list gets mapped to true, other values to false.
s          – Sum.

Answer 3

K (ngn/k), 37 bytes

{+/x{{(a~&\a)|a~|\a:10\x}(1+)/x+1}\0}

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{ } is a function with argument x

x{ }\0 applies the {} on 0 x times, preserving the intermediate results

(1+) is the successor function

{ }(1+)/x+1 applies the successor function starting from x+1 until the {} returns true

10\x are the decimal digits of x

a: assign to a

|\ is the max-scan (partial maxima) of a

&\ analogously, is the min-scan

a~|\a does a match its max-scan?

| or

a~&\a its min-scan?

+/ sum

Answer 4

Python 2, 84 bytes

i=s=0
n=input()
while n:b=`i`>`sorted(`i`)`[2::5]<`i`[::-1];n-=b;s+=b*i;i+=1
print s

Try it online! Or see a test-suite.

Answer 5

JavaScript (ES6), 77 bytes

f=(i,n=0,k,p)=>i&&([...n+''].map(x=>k|=x<p|2*(p<(p=x)))|k>2&&i--&&n)+f(i,n+1)

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Commented

f = (                     // f = recursive function taking:
  i,                      //   i = number of bouncy numbers to find
  n = 0,                  //   n = current value
  k,                      //   k = bitmask to flag increasing/decreasing sequences
  p                       //   p = previous value while iterating over the digits
) =>                      //
  i && (                  // if there's still at least one number to find:
    [...n + '']           //   turn n into a string and split it
    .map(x =>             //   for each digit x in n:
      k |=                //     update k:
        x < p |           //       set bit #0 if x is less than the previous digit
        2 * (p < (p = x)) //       set bit #1 if x is greater than the previous digit
                          //       and update p
    )                     //   end of map()
    | k > 2               //   if both bits are set (n is bouncy):
    && i--                //     decrement i
    && n                  //     and add n to the total
  ) + f(i, n + 1)         //   add the result of a recursive call with n + 1

Answer 6

Python 2, 110 92 89 bytes

n=input()
x=s=0
while n:b={-1,1}<=set(map(cmp,`x`[:-1],`x`[1:]));s+=x*b;n-=b;x+=1
print s

Try it online

This function determines if a number is bouncy:

lambda x:{-1,1}<=set(map(cmp,`x`[:-1],`x`[1:]))

Answer 7

Retina, 93 bytes

K`:
"$+"{/(?=.*;(_*);\1_)(?=.*_(_*);\2;)/^{`:(#*).*
:$1#;$.($1#
)`\d
*_;
)`:(#+).*
$1:$1
\G#

Try it online! Explanation:

K`:

Initialise s=i=0. (s is the number of #s before the :, i the number of #s after.)

"$+"{
...
)`

Repeat n times.

/(?=.*;(_*);\1_)(?=.*_(_*);\2;)/^{`
...
)`

Repeat while i is not bouncy.

:(#*).*
:$1#;$.($1#

Increment i and make a copy in decimal.

\d
*_;

Convert the digits of the copy to unary. The bounciness test uses the unary copy, so it only works once i has been incremented at least once.

:(#+).*
$1:$1

Add i to s and delete the copy of the unary digits, so that for the next pass of the inner loop the bounciness test fails and i gets incremented at least once.

\G#

Convert s to decimal.

121 byte version calculates in decimal, so might work for larger values of n:

K`0:0
"$+"{/(?=.*;(_*);\1_)(?=.*_(_*);\2;)/^{`:(\d+).*
:$.($1*__);$.($1*__)
)+`;\d
;$&*_;
)`\d+:(\d+).*
$.(*_$1*):$1
:.*

Try it online! Explanation:

K`0:0

Initialise s=i=0.

"$+"{
...
)`

Repeat n times.

/(?=.*;(_*);\1_)(?=.*_(_*);\2;)/^{`
...
)

Repeat while i is not bouncy.

:(\d+).*
:$.($1*__);$.($1*__)

Increment i and make a copy.

+`;\d
;$&*_;

Convert the digits of the copy to unary. The bounciness test uses the unary copy, so it only works once i has been incremented at least once.

\d+:(\d+).*
$.(*_$1*):$1

Add i to s and delete the copy of the unary digits, so that for the next pass of the inner loop the bounciness test fails and i gets incremented at least once.

:.*

Delete i.

Answer 8

05AB1E, 12 bytes

µN{‚Nå_iNO¼

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Explanation

µ              # loop over increasing N until counter equals input
     Nå_i      # if N is not in
 N{‚          # the pair of N sorted and N sorted and reversed
         NO    # sum N with the rest of the stack
           ¼   # and increment the counter

Answer 9

Java 8, 114 112 bytes

n->{int i=0,s=0;for(;n>0;++i)s+=(""+i).matches("0*1*2*3*4*5*6*7*8*9*|9*8*7*6*5*4*3*2*1*0*")?0:--n*0+i;return s;}

Uses a regular expression to check if the number is increasing or decreasing. Try it online here.

Ungolfed:

n -> { // lambda
    int i = 0, // the next number to check for bounciness
        s = 0; // the sum of all bouncy numbers so far
    for(; n > 0; ++i) // iterate until we have summed n bouncy numbers, check a new number each iteration
        s += ("" + i) // convert to a String
             .matches("0*1*2*3*4*5*6*7*8*9*" // if it's not an increasing  number ...
             + "|9*8*7*6*5*4*3*2*1*0*") ? 0 // ... and it's not a decreasing number ...
             : --n*0 // ... we have found another bouncy number ...
               + i; // ... add it to the total.
    return s; // return the sum
}

Answer 10

Python 2, 250 bytes

n=input()
a=0
b=0
s=0
while a<n:
    v=str(b)
    h=len(v)
    g=[int(v[f])-int(v[0]) for f in range(1,h) if v[f]>=v[f-1]]
    d=[int(v[f])-int(v[0]) for f in range(1,h) if v[f]<=v[f-1]]
    if len(g)!=h-1 and len(d)!=h-1:
       a+=1
       s+=b
    b+=1
print s

Answer 11

Nibbles, 9 bytes

+<$|,~~?`.`<;`p$\

(it would be 8.5 bytes if filesystems could store half bytes)

Similar to the Jelly answer.

+                # sum of
 <               # first n elements of
  $              # first input integer
   |  ~          # filter.not
    ,~           # range from 1 to infinity
       ?         # index of (begin of filter body)
        `.       # iterate while uniq
        `<       # sort
          ;`p$   # int to str of filter arg, also save value
              \  # reverse
                 # implicit $ (arg of iterate while uniq)
                 # implicit $ (saved value from `p)

Answer 12

Stax, 14 bytes

Ö²àDæ▲☼Ty≡°lû↕

Run and debug it

Answer 13

Vyxal, 7 bytes

fÞȮ¬)ȯ∑

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      ∑ # Sum of...
     ȯ  # First n integers...
----)   # Where...
f       # Digits
   ¬    # Are not
 ÞȮ     # Sorted

Answer 14

Red, 108 bytes

func[n][i: s: 0 until[t: form i
if(t > u: sort copy t)and(t < reverse u)[s: s + i n: n - 1]i: i + 1
0 = n]s]

Try it online!

More readable:

f: func [ n ] [
    i: s: 0
    until [
       t: form i  
       if ( t > u: sort copy t ) and ( t < reverse u ) [
            s: s + i
            n: n - 1
       ]
       i: i + 1
       0 = n
    ]
    s
]

A good opportunity to use form - form i is 5 bytes shorter than to-string i

Answer 15

MATL, 31 30 bytes

l9`tVdZS&*0<a?wQtG>?.]y]QT]xvs

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l       % Initialize counter to 1
9       % First value to check for being bouncy
`       % Start do-while loop
  tV    % duplicate the current number and convert to string
        % (let's use the iteration where the number is 1411)
        % stack: [1, 1411, '1411']
  d     % compute differences between the characters
        %  For non-bouncy numbers, this would be either all 
        %  positive values and 0, or all negative values and 0
        % stack: [1, 1411, [3 -3 0]]
  ZS    % keep only the signs of those differences
        % stack: [1, 1411, [1 -1 0]]
  &*    % multiply that array by its own transpose, so that
        %  each value is multiplied by every other value
        %  for non-bouncy numbers, all products will be >=0
        %  since it would have had only all -1s or all 1 (other than 0s)
        % stack: [1, 1411, [1 -1  0
                            -1  1 0
                            0  0  0]]
  0<a?  % if any value in the matrix is less than 0
    wQ    % switch the counter to top, increment it
          % stack: [1411, 2]
    tG>?  % duplicate it, check if it's gotten greater than the input limit
      .]    % if so, break out of loop
  y]    % else, duplicate bouncy number from inside stack,
        %  keeping a copy to be used for summing later
        % stack: [1411, 2, 1411]
  QT    % increment number, push True to continue loop
]     % loop end marker
x     % if we've broken out of loop, remove counter from stack
v     % concatenate the bouncy numbers we've collected in stack and 
s     % sum them

Answer 16

R, 96 bytes

function(n){while(n){y=diff(T%/%10^(0:log10(T))%%10);g=any(y<0)&any(y>0);F=F+T*g;n=n-g;T=T+1};F}

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Explanation :

while(n){                         # while n > 0

        T%/%10^(0:log10(T))%%10   # split T into digits(T==TRUE==1 at the 1st loop)
y=diff(                         ) # compute the diff of digits i.e. digits[i] - digits[i+1]

g=any(y<0)&any(y>0)               # if at least one of the diff is < 0 and 
                                  # at least one is > 0 then T is "bouncy"

F=F+T*g                           # if bouncy increment F (F==FALSE==0 at the 1st loop)

n=n-g                             # decrement n by 1 if bouncy

T=T+1}                            # increment T by 1 and loop

F}                                # return F

Answer 17

Ruby, 76 bytes

->n,s=i=0{[d=(i+=1).digits,d.reverse].index(d.sort)||(s+=i;n-=1);n<1?s:redo}

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Answer 18

C (gcc), 104 bytes

f(b,o,u,n,c,y){for(o=u=0;b;u+=y?0:o+0*--b,++o)for(n=o,y=3;n/10;)c=n%10,n/=10,y&=(c-=n%10)<0?:c?2:y;b=u;}

Try it online here.

Ungolfed:

f(n, // function: return type and type of arguments defaults to int;
     // abusing extra arguments to declare variables
  i,   // number currently being checked for bounciness
  s,   // sum of the bouncy numbers
  j,   // copy of i to be used for checking bounciness in a loop
  p,   // used for investigating the last digit of j
  b) { // whether i is not bouncy; uses the two least significant bits to indicate increasing/decreasing
    for(i = s = 0; // check numbers from zero up; initial sum is zero
        n; // continue until we have n bouncy numbers
        s += b ? 0 // not bouncy, no change to the sum
        : i + 0* --n, // bouncy, add it to the sum and one less bouncy number to go
        ++i) // either way, move to the next number
        for(j = i, b = 3; j/10; ) // make a copy of the current number, and truncate it from the right until there is just one digit left
        // bounciness starts as 0b11, meaning both increasing and decreasing; a value of 0 means bouncy
            p = j % 10, // get the last digit
            j /= 10, // truncate one digit from the right
            b &= // adjust bounciness:
                 (p -= j % 10) // compare current digit to the next
                 < 0 ? // not an increasing number, clear second to least significant bit
                 : p ? 2 // not a decreasing number, clear least significant bit
                 : b; // keep it the same
    n = s; // gcc shortcut for return s
}

Answer 19

Ruby (123 106 bytes)

o=0
x=[?0]
(z=(x=(x.join.to_i+1).to_s.chars).sort)!=x&&z!=x.reverse ? (n-=1;o+=x.join.to_i):1while n>0
p o

Looks pretty ugly to me. Bounciness is defined in this block x.sort!=x&&x.sort!=x.reverse