The 'Winning' Gene
의견: 0
*Note: The memory limit for this problem is 512MB, twice the default.*
After years of hosting games and watching Bessie get first place over and over,
Farmer John has realized that this can't be accidental. Instead, he concludes
that Bessie must have winning coded into her DNA so he sets out to find this
"winning" gene.
He devises a process to identify possible candidates for this "winning" gene. He
takes Bessie's genome, which is a string \(S\) of length \(N\) where
\(1 \leq N \leq 3000\). He picks some pair \((K,L)\) where \(1 \leq L \leq K \leq N\)
representing that the "winning" gene candidates will have length \(L\) and will be
found within a larger \(K\) length substring. To identify the gene, he takes all
\(K\) length substrings from \(S\) which we will call a \(k\)-mer. For a given
\(k\)-mer, he takes all length \(L\) substrings, identifies the lexicographically
minimal substring as a winning gene candidate (choosing the leftmost such
substring if there is a tie), and then writes down the \(0\)-indexed position
\(p_i\) where that substring starts in \(S\) to a set \(P\).
Since he hasn't picked \(K\) and \(L\) yet, he wants to know how many candidates
there will be for every pair of \((K,L)\).
For each \(v\) in \(1\dots N\), help him determine the number of \((K,L)\) pairs with
\(|P|=v\).
Problem credits: Suhas Nagar
SCORING
- Inputs 2-4: \(N \leq 100\)
- Inputs 5-7: \(N \leq 500\)
- Inputs 8-16: No additional constraints.
Problem credits: Suhas Nagar
\(N\) representing the length of the string. \(S\) representing the given string.
All characters are guaranteed to be uppercase characters where \(s_i \in A-Z\)
since bovine genetics are far more advanced than ours.
For each \(v\) in \(1\dots N\), output the number of \((K,L)\) pairs with \(|P|=v\),
with each number on a separate line.
8
AGTCAACG11
10
5
4
2
2
1
1In this test case, the third line of the output is 5 because we see that there are exactly 5 pairs of \(K\) and \(L\) that allow for
three "winning" gene candidates. These candidates are (where \(p_i\) is \(0\)-indexed):
(4,2) -> P = [0,3,4]
(5,3) -> P = [0,3,4]
(6,4) -> P = [0,3,4]
(6,5) -> P = [0,1,3]
(6,6) -> P = [0,1,2]
To see how (4,2) leads to these results, we take all \(4\)-mers
AGTC
GTCA
TCAA
CAAC
AACG
For each \(4\)-mer, we identify the lexicographically minimal length 2 substring
AGTC -> AG
GTCA -> CA
TCAA -> AA
CAAC -> AA
AACG -> AA
We take the positions of all these substrings in the original string and add
them to a set \(P\) to get \(P = [0,3,4]\).
On the other hand, if we focus on the pair \((4,1)\), we see that this only leads
to \(2\) total "winning" gene candidates. If we take all \(4\)-mers and identify the
lexicographically minimum length \(1\) substring (using A and A' and A* to
distinguish the different As), we get
AGTC -> A
GTCA' -> A'
TCA'A* -> A'
CA'A*C -> A'
A'A*CG -> A'
While both A' and A* are lexicographically minimal in the last 3 cases, the
leftmost substring takes precedence so A' is counted as the only candidate in
all of these cases. This means that \(P = [0,4]\).
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The 'Winning' Gene
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The 'Winning' Gene