Editorial for A Harder Contest '19 - An Alternating Problem

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Submitting an official solution before solving the problem yourself is a bannable offence.

Author: ChrisT

Subtask 1 was intended to reward bitmask solutions.

We can iterate through all 2^N subsequences of the original array, and calculate their alternating sum. The answer is the maximum of all possible sums which have a difference of at most M between adjacent elements.

Time Complexity: \mathcal{O}(N\cdot 2^N)

Subtask 2 was intended to reward unoptimized dynamic programming solutions.

Let dp[i][j] represent the maximum possible alternating sum of any subsequence using only the first i values in the array, and with length \(\equiv\:j\pmod{2}\)

Our base case is dp[i][0] = 0 for all (0 \leq i < N), as you can always choose an empty subsequence up to that point

This leads to the recurrence:

\(dp[i][0] = \max\limits_{j < i} (dp[j][1] - a_i)\:iff\:|a_i-a_j|\leq M\)

\(dp[i][1] = \max\limits_{j < i} (dp[j][0] + a_i)\:iff\:|a_i-a_j|\leq M\)

On top of this, dp[i][j] can always be at least dp[i-1][j], as you are not required to take element i

Our answer lies in \max(dp[n][0],dp[n][1])

Note that you must keep a parent array to find the indices used.

Time Complexity: \mathcal{O}(N) states and \mathcal{O}(N) transitions, thus \mathcal{O}(N^2)

For the final subtask, we must optimize this dp.

Notice that the only impact of i in the transition is the requirement |a_i-a_j|\leq M

Thus, we can hold the best dp values for different values of a_i in a data structure, and query the best dp value in the range [a_i-M,a_i+M]

For this we will use two segment trees, one for each of dp[x][0], and dp[x][1]

With this the transition is \mathcal{O}(\log N)

Time Complexity: \mathcal{O}(N \log N)


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