Timeline for Palindrome Permutation string
Current License: CC BY-SA 4.0
10 events
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| Aug 26, 2022 at 11:30 | comment | added | Toby Speight |
I was imagining counting mod 2, which is a bit shorter (hist[c] ^= 1 in the loop, then std::ranges::accumulate() to count the ones); even then, the bitset is more convenient.
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| Aug 24, 2022 at 0:21 | comment | added | Deduplicator |
Yes, std::ranges::count_if(hist, [](auto x){return x % 2; }); would be a bit more verbose.
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| Aug 23, 2022 at 7:02 | history | edited | Toby Speight | CC BY-SA 4.0 |
Accumulate valuable tests as we go. And demonstrate that.
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| Aug 22, 2022 at 20:19 | history | edited | Toby Speight | CC BY-SA 4.0 |
Another observation
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| Aug 22, 2022 at 20:12 | history | edited | Toby Speight | CC BY-SA 4.0 |
Simplify flipping one bit
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| Aug 22, 2022 at 20:11 | comment | added | Toby Speight |
@Deduplicator, yes that may well be true, and I did start with a std::array<bool>. But the convenience of bitset::count() swung me!
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| Aug 22, 2022 at 20:02 | comment | added | Deduplicator | Of course an array of some unsigned type (byte up to register size) might be more efficient (code and time, not stack). Only measuring will show. | |
| Aug 22, 2022 at 19:57 | history | edited | Toby Speight | CC BY-SA 4.0 |
Use a string-view, and give it a more conventional name
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| Aug 22, 2022 at 18:30 | comment | added | chux |
Good use of unsigned char vs. char. isPalinStr() here is poetic.
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| Aug 22, 2022 at 17:13 | history | answered | Toby Speight | CC BY-SA 4.0 |