C (gcc), 67 bytes

a;r;i;f(long n){for(;i<33;)a=n&1l<<i++?a?2:1:a&1?r++,0:0;return a;}

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General algorithm:

• This is a little state machine where the variable 'alone' a can have the value 0=zero, 1=alone candidate, 2=not alone. Essentially a 'carry' but with several states.

  The non-golfed code would look something like if(!a) a=1; else a=2; which becomes a=a?2:1.

• Then the outer if-else non-golfed would look something like

  if(n&1l<<i)
    a=a?2:1;
  else
  {
    if(a&1)
      r++;
    a=0;
  }
  

  ...which turns into the one-liner ?: mess.

• When the state machine has found an alone candidate and transits from value 1 to zero, increase count r

• Whenever there's a zero, reset the state machine.

• To solve the special case at the end where there might be a lone 1, iterate one more than 32 times to get comparison with a zero in there. Which will only work if the function parameter is not sign extended! To cheat I used long which is 64 bit on Linux that tio.run uses - won't work on 32 bit long systems. Making the parameter unsigned would be portable but unsigned has more digits than long. Same thing with the 1l prefix, which needs to be 1ull to be portable.

C-specific stuff:

• Plain int is 32 bits.
• Operator precedence ++ over << over & over ?: over = over ,.
• Shifting negative numbers is of course wildly poorly-specified behavior but the code gets away with it on gcc/Linux.

Test cases:

• Taken from the challenge. An ugly macro takes a digit and expected output, then prints OK or FAIL.
• As usual, the function-only solution relying on zero init of global variables need to be reset by the caller between tests.
• Also print test cases as hex because the decimal numbers really don't tell me much :)

posted 6 days ago

CC BY-SA 4.0

3d ago

Lundin‭

1254 reputation 17 47 145 82

Bash, 53 bytes

sed 's/\(.\)/\1\n/g' /dev/stdin|uniq -c|grep -c '1 1'

1. Add a newline after each character.
2. Count the number of characters in each group with uniq
3. Count the number of groups with only one '1' with grep

posted 6 days ago

CC BY-SA 4.0

daviewales‭

31 reputation 0 2 3 5

Vyxal, 4 bytes

Ġċ†∑

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Takes input as a string of 32 bits.

Explained

Ġċ†∑­⁡​‎‎⁡⁠⁡‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁢‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁣‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁤‏‏​⁡⁠⁡‌­
Ġ     # ‎⁡Group the input by consecutive values. Basically, split the input on runs.
 ċ    # ‎⁢Push whether each run != 1. Basically, an inverted check for "is this run a '1' or multiple 1s/0s".
  †   # ‎⁣Logical not each comparison, because the above check was inverted.
   ∑  # ‎⁤Summate the list, returning the count of truthy comparisons.
💎

Created with the help of Luminespire.

posted 8 days ago

CC BY-SA 4.0

lyxal‭

626 reputation 2 26 66 21

Perl, 28 bytes

Takes input as newline-separated list of binary strings.

perl -ple's/11+//g;$_=y/1//'

• -p - assign each line of input to $_, manipulate, then print
• -l - strip trailing newline from input, add back to output
• -e - program follows
• s/11+//g - strip runs of two or more 1s
• y/1// - count remaining 1s, assign to $_

posted 4 days ago

CC BY-SA 4.0

4d ago

jhnc‭

71 reputation 0 7 7 27

awk, 35 bytes

Takes input as newline-separated list of binary strings.

awk '{gsub(/11+|0+/,_);$0=length}1'

• strip all runs of 0s and two or more 1s
• print length of remaining string

posted 4 days ago

CC BY-SA 4.0

4d ago

jhnc‭

71 reputation 0 7 7 27

Bash, 28 bytes

Takes input as a single binary string.

grep -Po '(^|0)1(?!1)'|wc -l

• -P (non-standard) - use PCRE-compatible regex syntax
• -o (non-standard) - output each matching substring as separate line
• wc -l - count lines

posted 4 days ago

CC BY-SA 4.0

4d ago

jhnc‭

71 reputation 0 7 7 27

C (gcc), 44 bytes

Borrowing/modifying test harness from @Lundin:

#define f(n)__builtin_popcount(n&~(n*2|n/2))

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n & ~(n<<1) & ~(n>>1) is non-zero for any bit precisely when the original bit of n is non-zero and the preceding and following bits aren't.

posted 4 days ago

CC BY-SA 4.0

3d ago

jhnc‭

71 reputation 0 7 7 27