_MASTER__'s blog

Why my solution got accepted in this DIV1 B:

By _MASTER__, history, 2 years ago, In English

my submission: https://codeforces.me/contest/126/submission/190247628 //

as far as I can see, My solution is O(n²) when there is no solution and it got AC. and the string length is ~~ 1e6

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2 years ago, # |
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This problem is very old (Beta round) so test cases are likely to be weak.

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2 years ago, # |
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It's actually not $$$n^2$$$. The hashes are done in $$$O(1)$$$ so comparing each suffix with the prefix is $$$n \cdot O(1)$$$. Suppose $$$A$$$ is the shortest substring of $$$S$$$ such that $$$A$$$ is both a suffix and a prefix. Then $$$S = ATA$$$. Suppose $$$AB$$$ is the second shortest substring such that $$$AB$$$ is both a suffix and a prefix. Then $$$S = ABT^\prime AB$$$. This implies that $$$A$$$ was also an infix of $$$S$$$. So when we actually ran the algorithm, we should've stopped at just $$$A$$$.

What all this implies is that there can be at most one candidate prefix/suffix. If there is more than one, we know that the first candidate also appears somewhere inside of $$$S$$$ and we'd just print it. This makes the algorithm $$$O(n)$$$.

(Edit: simplified argument slightly)

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    2 years ago, # ^ |
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    But in the code I m not stopping at the shortest, the opposite, I m searching for the longest ones , starting from the prefix at string.length()-2

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      2 years ago, # ^ |
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      My bad for not noticing but the idea is the same. Here we could call the longest suffix/prefix $$$AB$$$ and the second longest $$$A$$$ and the rest of the argument is identical.

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