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# Editorial - PEAKS

Author: Kanstantsin Sokal
Tester: Pavel Sheftelevich
Editorialist: Pawel Kacprzak

Medium

# PREREQUISITES:

Dynamic programming, Segment tree, Fenwick tree

# PROBLEM:

In this problem, for a given integer sequence $S$ of length $N$, your task is to count the number of non-empty subsequences of $S$ with exactly $A$ local minimums and exactly $B$ local maximums.

# QUICK EXPLANATION:

Use dynamic programming in order to compute the number of non-empty sequences S[1..j] with exactly $a$ local minimums and exactly $b$ local minimums for all $1 \leq j \leq N$, $0 \leq a \leq A$, $0 \leq b \leq B$. Next, speed up the computations using Segment tree or Fenwick tree. Finally, notice that $A$ and $B$ differ by at most 1, because all elements in $S$ are distinct. Use this fact to optimize the solution further in order to pass the last subtask.

# EXPLANATION:

In the first subtask $N$ is at most $20$, so in order to solve the problem, we can simply generate all possible $2^N - 1$ non-empty subsequences of $S$, and for each one, count the number of local minimums and local maximums in it using a linear scan over its element. This approach leads to $O(2^N \cdot N)$ running time for a single test case, which will pass the first subtask if only it's implemented efficiently.

In the second subtask $N$ can be at most $200$. This is of course way too big for any exponential method used in the first subtask to pass. One thing you should notice is that $A$ and $B$ are not so big here - both are not greater than $10$. We can use this fact to come up with the following dynamic programming approach:

Let $dp[i][a][b][0]$ be the number of subsequences of $S[1..i]$ ending in $S[i]$ with exactly $a$ local minimums, exactly $b$ local maximums and the last element smaller than the one before last element. Similarly, let $dp[i][a][b][1]$ be the number of subsequences of $S[1..i]$ ending in $S[i]$ with exactly $a$ local minimums, exactly $b$ local maximums and the last element greater than the one before last element. With these values defined, we can easily show how to compute any $dp$ entry with first index $i$ having all the $dp$ entries for first indexes $j < i$ computed.

Let's fix index $i$. We are going to iterate over all indices $1 \leq j < i$. For a fixed $j$, there are two possibilities: either $S[j] < S[i]$ or $S[j] > S[i]$. Since all elements of $S$ are distinct, equality is not possible here.

If $S[j] < S[i]$, we can create a subsequence of length $2$ using just $S[j]$ and $S[i]$, so we add $1$ to $dp[i][0][0][1]$. Similarly, if $S[j] > S[i]$, we add $1$ to $dp[i][0][0][0]$.

Next, we are going to extend any previously counted subsequences of length at least $2$ to subsequences ending with $S[i]$. In order to do this, we iterate over all possible values $0 \leq a \leq A$ and over all possible values $0 \leq b \leq B$. Again, there are two cases to consider:

If $S[j] < S[i]$, we add value of $dp[j][a][b][1]$ to $dp[i][a][b][1]$. Moreover, if $a < A$, we add $dp[j][a][b][0]$ to $dp[i][a + 1][b][1]$ creating new local minimum at index $j$.

Analogously, if $S[j] > S[i]$, we add value of $dp[j][a][b][0]$ to $dp[i][a][b][0]$. Moreover, if $b < B$, we add $dp[j][a][b][1]$ to $dp[i][a][b + 1][0]$ creating new local maximum at index $j$.

In order to compute the final result, we sum up all $dp[i][A][B][0]$ and $dp[i][A][B][0]$ values over all $1 \leq i \leq N$. At the end, if $A = 0$ and $B = 0$, we add $N$ to the result in order to count all subsequences of length $1$.

Last but not least, remember to do all necessary computations modulo $10^9 + 9$.

The total time complexity of this method is $O(N^2 \cdot A \cdot B)$, because we fill $O(N \cdot A \cdot B)$ $dp$ entries, and we fill a single one in $O(N)$ time.

In the third subtask, $A$ and $B$ are still not greater than $10$, but $N$ can be now as big as $5000$. It would be nice to get rid of one $N$ factor from the complexity of solution given for the second subtask. It turns out that we can do this using a very natural an common idea of improving dynamic programming solutions using segment trees (or any similar data structure like Fenwick tree).

In the solution given above, we update a single entry in the $dp$ table in $O(N)$ time. Now, we are going to do this in $O(\log N)$ time. Notice, that in the original $dp$ solution for a fixed $i$, we iterate over all possible $1 \leq j < i$ in order to compute the $dp$ entries for index $i$. There were two cases to consider depending on whether $S[j] < S[i]$ or $S[j] > S[i]$.

Here comes a crucial observation: if we accumulate all the values of $dp[j][a][b][0]$ for all $j$ such that $S[j] < S[i]$ and do the same with all values $dp[j][a][b][1]$ for such $j$, we can easily compute the value of $dp[i][a][b][1]$ and $dp[i][a + 1][b][1]$ using these accumulated sums. Similarly, we can speed up the computation of $dp[i][a][b][0]$ and $dp[i][a][b + 1][0]$ if we accumulate all $dp[j][a][b][0]$ for all $j$ such that $S[j] > S[i]$ and if we do the same with all $dp[j][a][b][1]$ for such $j$.

Notice that we can maintain these sums using a segment tree with the sum over a contiguous query (or even just prefix and suffix sum queries) and update operation adding a value at the given index. One thing left to do is to compress the input values in order to easily store the segment tree in memory - their absolute values are initially less or equal to $10^9$. We can compress them easily to a range $1..N$ which helps a lot here.

The total time complexity of this approach is $O(N \cdot \log N \cdot A \cdot B)$.

In the last subtask, we still have $N \leq 5000$, but $A$ and $B$ can now be as big as $200$, so the solution used for the second subtask is too slow now and we need a further improvement.

Let's consider any subsequence $P$ of $S$ and take a closer look at the difference between local minimums in $P$ and local maximums in it. Let's denote this difference as $D$. The crucial observation here is that since all elements of $S$ are distinct, $D$ can be only $-1$, $0$ or $1$.

In order to prove that, let's consider any subsequence $P$ with $1$ local minimum and $0$ local maximums. Let $i$ be the index which is the local minimum. Notice that $P$ is an increasing subsequence from index $i$, so any other subsequence $W$ which have $P$ as a prefix will be either increasing from index $i$ till the end, or it have two adjacent indexes $i_{j}$ and $i_{j+1}$ such that $S[i_j] > S[i_{j+1}]$, so the value $D$ for $W$ never grows over $1$. You can use the same argument to show why the value of $D$ is never less than $-1$.

We can use this fact in order to speed up the solution given for the third subtask. Notice that now we only need to take care of $dp$ states for which we have $0 \leq a \leq A$ local minimums and for each such $a$ there are at most $3$ possible values of $b$ - the number of local maximums: the only possible ones are $a-1$, $a$ and $a+1$. This observation leads to a solution with total time complexity $O(N \cdot \log N \cdot A)$.

# AUTHOR'S AND TESTER'S SOLUTIONS:

Author's solution can be found here.
Tester's solution can be found here.

This question is marked "community wiki".

asked 31 Jan '16, 04:11

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accept rate: 12%

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question asked: 31 Jan '16, 04:11

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last updated: 31 Jan '16, 22:01