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Challenges

Comments on Lone ​​​​​​ones

Parent

Lone ​​​​​​ones

+7
−0

Given an integer, output the number of lone 1s in its binary representation.

Input

  • An integer from $0$ to $4294967295$ (that is, $2^{32}-1$). In binary that's from 00000000000000000000000000000000 to 11111111111111111111111111111111.
  • You may take input in any form that does not contribute to solving the challenge. For example:
    • A signed or unsigned integer of $32$ bits or more.
    • A string of decimal digits.
    • A binary string.
    • A sequence of bits.

Output

  • The number of 1s in the binary representation that do not have a 1 to their left or right.
  • For avoidance of doubt:
    • A 1 at the left end of the binary representation does not have a 1 to its left.
    • A 1 at the right end of the binary representation does not have a 1 to its right.

Examples

Input $4$ corresponds to 100 in binary. The output is $1$ because there is $1$ 1 bit that is not adjacent to another 1 bit.

Input $5$ corresponds to 101 in binary. The output is $2$ because there are $2$ 1 bits that are not adjacent to another 1 bit.

Input $6$ corresponds to 110 in binary. The output is $0$ because each of the 1 bits is adjacent to another 1 bit.

Test cases

Test cases are in the format input : output.

Unsigned integer test cases

0 : 0
4294967295 : 0
2147483648 : 1
1 : 1
2147483649 : 2
98304 : 0
1310720 : 2
4294639615 : 1
4294967294 : 0
2147483647 : 0
3221225471 : 1
4294967293 : 1
2863311530 : 16
1431655765 : 16
1717986918 : 0

Binary test cases

00000000000000000000000000000000 : 0
11111111111111111111111111111111 : 0
10000000000000000000000000000000 : 1
00000000000000000000000000000001 : 1
10000000000000000000000000000001 : 2
00000000000000011000000000000000 : 0
00000000000101000000000000000000 : 2
11111111111110101111111111111111 : 1
11111111111111111111111111111110 : 0
01111111111111111111111111111111 : 0
10111111111111111111111111111111 : 1
11111111111111111111111111111101 : 1
10101010101010101010101010101010 : 16
01010101010101010101010101010101 : 16
01100110011001100110011001100110 : 0

Scoring

This is a code golf challenge. Your score is the number of bytes in your code. Lowest score for each language wins.

Explanations are optional, but I'm more likely to upvote answers that have one.

History

0 comment threads

Post
+1
−0

C (gcc), 44 bytes

Borrowing/modifying test harness from @Lundin:

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

Try it online!

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.

History

1 comment thread

Relevant discussion about using a global instead of `return` (4 comments)
Relevant discussion about using a global instead of `return`
trichoplax‭ wrote 7 months ago

Since this answer borrows some aspects from Lundin's C answer, the discussion Missing default rule for using globals might apply here too.

jhnc‭ wrote 7 months ago

Is using #define to define the function allowed?

trichoplax‭ wrote 7 months ago

No objection from me. I don't recall seeing a default rule about macro functions on Meta. Anyone is free to start a discussion there if they have an opinion either way.

Lundin‭ wrote 6 months ago

Technically, a macro parameter is not a global and equivalent to passing a function parameter by reference, and the result of a macro should be equivalent to return. So I think this is fine - usually typing out #define for a function-like macro isn't ideal for golfing, but sometimes it is.