Timeline for Search for the longest sequence of toggled consecutive bits in an integer
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 21, 2017 at 16:54 | vote | accept | MauroAlmeida | ||
| Mar 21, 2017 at 13:00 | answer | added | Toby Speight | timeline score: 2 | |
| Mar 21, 2017 at 11:44 | comment | added | CiaPan |
Input like 0xF is not 'all 1's' because it actually is 0x0000000F for 32-bit int, hence 3 is a correct answer: the longest chain of alternating bits in the given number is a two-bit sequence 01 at positions 4 and 3.
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| Mar 21, 2017 at 11:26 | comment | added | CiaPan | The two masks given by @TobySpeight are alternating binary sequences: zero-one-zero-one... or one-zero-one-zero-... So if you XOR one of them with your number, every alternating subsequence would be converted either into a contiguous block of zeros or a block of ones. | |
| Mar 21, 2017 at 11:14 | history | edited | MauroAlmeida | CC BY-SA 3.0 |
deleted 2 characters in body
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| Mar 21, 2017 at 11:13 | comment | added | MauroAlmeida | @TobySpeight I don't understand your observation can you explain it better in an answer? | |
| Mar 21, 2017 at 9:03 | comment | added | Toby Speight | One observation: If you already have a function that will find a sequence of identical bits, you can transform the input by XOR with 0x5555... (to the length of your type) or 0xAAAA... and then the problem is equivalent. | |
| Mar 21, 2017 at 4:32 | history | edited | 200_success |
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| Mar 21, 2017 at 0:36 | review | First posts | |||
| Mar 21, 2017 at 2:28 | |||||
| Mar 21, 2017 at 0:35 | history | edited | Jamal | CC BY-SA 3.0 |
deleted 93 characters in body; edited title
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| Mar 21, 2017 at 0:30 | history | asked | MauroAlmeida | CC BY-SA 3.0 |