Consecutive 1-Bits are Incremented

Question

Given a pattern (string or array format) of Bits : [0,1,1,1,0,1,1,0,0,0,1,1,1,1,1,1]

The tasks is to replace any number of consecutive 1-Bits with an ascending number sequence starting at 1.

Input

• Pattern (can be received as an string or array) Example:
  • String: 1001011010110101001
  • Array: [1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1]

Output

• Ascending number sequence (can be returned as an string or array) Example:
  • String: 1 0 0 1 0 1 2 0 1 0 1 2 0 1 0 1 0 0 1
  • Array: [1, 0, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 0, 1]

Rules

• (only apply for strings) Input wont contain spaces between 1 and 0
• Assume Input length > 0
• (only apply for strings) Output is separated by space (use any other separator if you need as long as is not a number or a letter from the alphabet)

Example:

Given [0,1,1,1,0,1,1,0,0,0,1,1,1,1,1,1] 
Output [0,1,2,3,0,1,2,0,0,0,1,2,3,4,5,6]

--------------------------------------------------------------------------

Given 0110101111101011011111101011111111     
Output 0 1 2 0 1 0 1 2 3 4 5 0 1 0 1 2 0 1 2 3 4 5 6 0 1 0 1 2 3 4 5 6 7 8

---------------------------------------------------------------------------

Given 11111111111101    
Output 1 2 3 4 5 6 7 8 9 10 11 12 0 1

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Winning criteria: Codegolf

Answer 1

Retina, 14 bytes

rv`0|(1)+
$#1¶

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Answer 2

C (gcc), 57 52 51 bytes

f(a,l,c,i)int*a;{for(c=i=0;i<l;)a[i++]=c=a[i]*-~c;}

Port of Arnauld's JavaScript answer, modifies the array in-place. Try it online here.

Answer 3

Java, 65 bytes

void a(int[]q){for(int i=1;i<q.length;i++)q[i]+=q[i]>0?q[i-1]:0;}

Answer 4

MATL, 11 10 bytes

0w"@+@*t]x

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(-1 byte thanks to Giuseppe)

The obvious "accumulate y(x+y) over the array" approach.

16 bytes

n:G~f!-tO<YYw(X<

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Longer, more matrix-y approach: get the overall index, broadcast-subtract the indices of all zeros, and take the difference from the index of the nearest zero.

Answer 5

Zig, 65 bytes

fn f(b:[]u8)[]u8{for(b)|_,i|{if(i>0)b[i]+=b[i-1]*b[i];}return b;}

Expanded version

fn f(b:[]u8) []u8 {
    // Foreach loop where only index is used.
    for(b) |_,i| {
        // First value is always going to be 0 or 1,
        if(i > 0)
            // b[1] will either be 0 or 1,
            // so this is basically an if statement
            // that we can use to reset the count.
            b[i] += b[i-1] * b[i];
    }
    return b;
 }

I would add a Try it online!, but their compiler seems to lack some crucial language features. (Language still being in development and all that) And I am not going to attempt to make it compatible through trial and error.

But you can copy paste this into Zig playground (Confirmed to work as of writing this post)

const std = @import("std");

fn f(b:[]u8)[]u8{for(b)|_,i|{if(i>0)b[i]+=b[i-1]*b[i];}return b;}

pub fn main() anyerror!void {
    var bits = [_]u8{0,1,1,1,0,1,1,0,0,0,1,1,1,1,1,1};
    var array = f(bits[0..]);

    try std.io.getStdOut().writer().print("{any}\n", .{array});
}

Answer 6

Vyxal, 4 bytes

Ġv¦f

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-6 thanks to Lyxal.

Ġ    # Group consecutive identical elements
 v¦  # Take the cumulative sum of each
   f # Flatten the result

Answer 7

Thunno, \$ 6 \log_{256}(96) \approx \$ 4.94 bytes

zy.z(S

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Explanation

zy      # Group by consecutive
  .     # Apply to each:
   z(   #  Cumulative sums
     S  # Flatten

Answer 8

K (ngn/k), 11 9 bytes

(*/+\|,)\

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-2 bytes thanks to ovs

Answer 9

Jelly, 6 bytes

ŒgJ€F×

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Answer 10

Retina 0.8.2, 15 bytes

.(?<=(1*))
$.1 

Try it online! Link includes test cases. Alternative version, also 15 bytes:

.(?<=(1)*)
$#1 

Try it online! Link includes test cases Explanation:

.

Match the 0s and 1s.

(?<=(1*))
(?<=(1)*)

Count the number of 1s from the current digit backwards. The first one counts by capturing the run of 1s as a subtring, while the second one counts the number of times the 1 was captured.

$.1 
$#1 

Replace the digit with the number of 1s, either via the length of the capture or by the number of captures as appropriate.

Answer 11

Pyth, 9 bytes

t.u*YhNQZ

Verify all the test cases

t.u*YhNQZ   Implicit: Q=eval(input())

 .u    QZ   Cumulative reduce over Q, starting value 0
     hN       Increment the current value
   *Y         Multiply by the next value
t           Remove 1st element of list, implicit print

Answer 12

racket, 107 bytes

(λ(i)(let c((x i)(z 0))(if(empty? x)'()(if(= (car x)1)(cons(+ 1 z)(c(cdr x)(+ 1 z)))(cons 0(c(cdr x)0))))))

Answer 13

Brain-Flak, 78 bytes

(<>)<>([]){{}{{}<>(({})())<>}([]){{}(<>)<>}{}([])}{}<>([]){{}({}<>)<>([])}<>{}

Try it online!

Readable version:

(<>)<>
([])

{

    {}

    {
        {}<>(({})())<>
    }

    ([])
    {

        {}(<>)<>
    }

    {}

    ([])

}{}<>

([])

{
    {}({}<>)<>([])
}<>{}

Answer 14

Julia 0.6, 28 bytes

t->(x=0;[(x=(x+y)y)for y=t])

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Answer 15

C++, 46 45 41 bytes

Generic lambda, any int container.

[](auto&a){int p=0;for(int&i:a)p=i*=p+i;}

Saved a byte with the *= operator, saved 4 bytes by removing unneeded braces and parenthesis.

Note: The p+i part can be replaced with p+1 or ++p, I'm not sure why I did p+i but it wouldn't save any bytes to change it.

Old:

[](auto&a){int p=0;for(int&i:a){p=i=i*(p+i);}}

Answer 16

C (gcc), 52 51 bytes

Thanks to ceilingcat for the suggestion.

c;f(char*i){for(;*i;printf("%d ",c=~c*(48-*i++)));}

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Answer 17

Arturo, 22 bytes

$=>[0map&=>[<=&*<=1+]]

Try it

$=>[        ; a function, assign input to &
    0       ; push 0
    map&=>[ ; map over input, assign current elt to &
        1+  ; increment tos
        <=  ; dup
        &*  ; multiply by current elt
        <=  ; dup
    ]       ; end map
]           ; end function

Answer 18

TI-Basic, 25 bytes

Prompt B
For(I,1,dim(⁻B
⁻B(I)(Ans+1
Disp Ans
End

Input is taken as a list. Output is as numbers separated by newlines.

Answer 19

Nibbles, 2.5 bytes (5 nibbles)

=\$+*

=\$+*       # full function
=\$+*$@$    # (with implicit arguments shown):
=\          # scan from left over
  $         # input
            # with function:
   +        #   add
       $    #   each element 
            #   to
    *       #   product of 
     $      #   each element
            #   and
      @     #   result so far

Answer 20

BQN, 6 bytes

(×+⊢)`

Try it at BQN online!

Explanation

Port of Dominic van Essen's Nibbles answer.

(×+⊢)`
(    )`  Left scan on this function:
  +       Add
   ⊢      each element to
 ×        the product of the element and the previous intermediate value

Answer 21

KamilaLisp v0.2, 14 (APL SBCS)

$(⍀.←[+ * #1])

Equivalent to $(scanl1[+ * #1]).

Answer 22

Thunno 2 S, 4 bytes

ġıż×

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Explanation

ġıż×  # Implicit input
ġ     # Group consecutive items
 ı    # Map over this list of lists:
  ż   #  Push [1..length] without popping
   ×  #  Multiply the two lists elementwise
      # Implicit output, summed (flattened)

Answer 23

Python 2, 52 46 bytes

f=lambda x,y=1:x and[y*x[0]]+f(x[1:],y*x[0]+1)

Saved 6 bytes thanks to Jo King and Rod!

Answer 24

C# (Visual C# Interactive Compiler), 75 bytes

---

Golfed Try it online!

a=>{for(int i=1;i<a.Length;i++)if(a[i-1]>0&a[i]>0)a[i]=a[i-1]+1;return a;};

---

Ungolfed

a => {
    for( int i = 1; i < a.Length; i++ )
        if( a[ i - 1 ] > 0 & a[ i ] > 0 )
            a[ i ] = a[ i - 1 ] + 1;

    return a;
};

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Full code

using System;

namespace Namespace {
    class Program {
        static void Main( String[] args ) {
            Func<Int32[], Int32[]> f = a => {
                for( int i = 1; i < a.Length; i++ )
                    if( a[ i - 1 ] > 0 & a[ i ] > 0 )
                        a[ i ] = a[ i - 1 ] + 1;

                return a;
            };

            List<Int32[]>
                testCases = new List<Int32[]>() {
                    new Int32[] { 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1 },
                    new Int32[] { 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1 },
                    new Int32[] { 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1 },
                };

            foreach( Int32[] testCase in testCases ) {
                Console.WriteLine( $"{{ {String.Join(", ", testCase)} }}\n{f( testCase )}" );
            }

            Console.ReadLine();
        }
    }
}

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Releases

• v1.0 - 75 bytes - Initial solution.

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Notes

• None

Answer 25

Dodos, 152 bytes

	_ b _ F >
b
	dab
t
	dot
(
	t b
	t
X
	X dip
<
	b X (
/
	<
	< b
0
	t 0 b
>
	
	0
.
	dip <
	< b
	t b
x
	x .
H
	b b x >
	b
G
	t H /
	b b
F
	<
	F G
_
	_ b
	<

Try it online!

Answer 26

Python 2.7, 217 Bytes

def f(d):
    c = 0
    r = []
    for e in d:
        if e == 1:
            r.append(1 + c)
            c += 1
        else:
            c = 0
            r.append(0)
    return r

I'm new so.

Answer 27

Kotlin, 73 bytes

var i=0
readLine()?.forEach{(it-48).toInt().let{c->i=i*c+c;print("$i ")}}

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Answer 28

Perl 5 -p, 22 20 bytes

-2 bytes thanks to @DomHastings

s/./$i=$&*++$i.$"/ge

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Method:

Replace each character with itself multiplied by the incremented value of $i, then store that value back in $i. Thus, $i starts at 0 (undef) and gets reset to 0 every time the character in the bit string is a 0.

Finally, append a space to the result of the above step to make the output format correct.

Answer 29

Lua, 63 54 bytes

function f(j)for i=2,#j do j[i]=j[i]*(j[i-1]+1)end end

Written by Jo King. Try it online!

---

Explanation

function f(j) -- declares the function 'f' that will receive a variable 'j' (we know it's a table)
  for i=2,#j do -- from 2 to the number of elements in j
    j[i]=j[i]*(j[i-1]+1) -- change the current element of j to itself
              (j[i-1]+1) -- multiplied by the one before it plus one
  end
end

Original (invalid) in expanded form:

for i=2,#input do -- a loop that goes from 2 to the number of elements in the input table
  if input[i] == 0 then -- if the element at 'i' is 0
    input[i] = 0 -- then keep it as 0
  else
    input[i] = input[i-1]+1 -- change it to the element that comes before + 1
  end
end

Feel free to ask or suggest anything!

Answer 30

C++ (gcc), 37 bytes

[](auto&a){for(int&i:a)i*=*(&i-1)+1;}

Try it online!

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A lambda function which works with any int container supporting range-based for loops.