saved_chats/#275

USER: https://codeforces.com/contest/1987/problem/D

BOT: Thanks! Now could you provide a link to the editorial.



USER: https://codeforces.com/blog/entry/131053

BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 



USER: editorial

BOT: **Hints**
**Hint 1**
Alice's optimal strategy is simple (and kind of fixed).
**Hint 2**
Alice always eats the least tasty cake that she can eat.
**Hint 3**
Convert the given array to it's count array $$$c_i$$$, for example, the array $$$[1, 5, 1, 2, 1, 5, 1]$$$ becomes $$$[4, 1, 2]$$$.
**Hint 4**
Consider a set of $$$k$$$ indices that Bob will choose to eat ($$$c_{i_1} + ... + c_{i_k}$$$ cakes in total). What are the necessary conditions for him to be able to do that?
**Hint 5**
For a set of $$$k$$$ indices $$$1 \le i_1 \le i_2 \le \ldots \le i_k \le |c|$$$, the condition $$$\sum_{j=1}^{p}{c_{i_j}} \le i_p - p$$$ must hold for all $$$1 \le p \le k$$$.
**Tutorial**
Let's consider Alice's strategy. Notice that if both players play optimally, Alice eating a cake with a tastiness value of $$$t$$$ is equivalent to her eating all remaining cakes with $$$a_i \le t$$$. Since it is better for her to have more cakes to choose from later on, she will choose the minimum possible $$$t$$$ on each turn.
Now let's consider Bob's strategy. Suppose Bob ate at least one cake with a tastiness value of $$$t$$$. Then he has to have eaten all of them, because eating only some of them does not affect Alice's possible moves, resulting in wasted turns (he might be forced to waste moves at the end of the game when there are some leftover cakes).
Let $$$A_1, \ldots, A_m$$$ be the sorted unique values of $$$a_1, \ldots, a_n$$$, and let $$$c_i$$$ be the number of occurrences of $$$A_i$$$ in $$$a$$$. Then the game is reduced to Alice finding the first $$$c_i > 0$$$ and assigning $$$c_i := 0$$$, and Bob choosing any $$$c_i > 0$$$ and decreasing it by $$$1$$$.
Since Bob's optimal strategy is for each $$$c_i$$$ either not to touch it or make $$$c_i = 0$$$, we can model it as him selecting some set of indices $$$1 \le i_1 < \ldots < i_k \le m$$$ and zeroing out $$$c_{i_1}, c_{i_2}, \ldots, c_{i_k}$$$ in this order. To be able to zero out $$$c_{i_p}$$$ ($$$1 \le p \le k$$$), Alice must not have gotten to the value $$$c_{i_p}$$$. In $$$c_{i_1} + \ldots + c_{i_p}$$$ turns, Alice will zero out exactly that many values. Additionally, Bob would have zeroed out $$$p - 1$$$ values before index $$$i_p$$$. So, $$$c_{i_1} + \ldots + c_{i_p} + p - 1$$$ must be less than $$$i_p$$$. Transforming this a bit, we get that the condition $$$\sum_{j=1}^{p}{c_{i_j}} \le i_p - p$$$ must hold for all $$$1 \le p \le k$$$.
Bob's objective to maximize the size of the set of incides $$$1 \le i_1 < \ldots < i_k \le m$$$. Let $$$dp[i][k] =$$$ the minimum possible $$$\sum_{j=1}^{k}{c_{i_j}}$$$ over all valid sets of indices $$$1 \le i_1 < \ldots < i_k \le i$$$. Initialize $$$dp[0][0] = 0$$$, and everything else to $$$+\infty$$$. The main idea is to grow valid sets of indices one index at a time.
We will iterate over all $$$i$$$ from $$$1$$$ to $$$m$$$, and then for each $$$k$$$ from $$$0$$$ to $$$m$$$. The transitions are: 
 $$$dp[i][k] = \min(dp[i][k], d[i - 1][k])$$$, which corresponds to not using the current index in the set. 
 Let $$$s := dp[i - 1][k - 1] + c_i$$$. If $$$s \le i_k - k$$$, update $$$dp[i][k] = \min(dp[i][k], s)$$$, which is equivalent to adding the current index to the set. We only need to check the condition on the last index because it is already satisfied for all the previous indices. 
The answer to the problem is $$$m - y$$$, where $$$y$$$ is the largest value where $$$dp[m][y] \le +\infty$$$.
Complexity: $$$\mathcal{O}(n^2)$$$
Note: it is possible to solve in $$$\mathcal{O}(n \log n)$$$ by making an additional observation.
**Solution**
**Feedback**
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Average problem 

    



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Let me know if you have any questions.



USER: interactive editorial

BOT: The problem revolves around Alice and Bob taking turns to eat cakes with the goal of Alice maximizing the number of cakes she eats and Bob minimizing that number. Alice can only eat a cake if its tastiness is greater than any cake she has previously eaten, while Bob can eat any remaining cake. The key strategy for Alice is to always eat the least tasty cake that she can eat, ensuring she has more options later. Bob's optimal strategy is to either not touch a cake or eat all cakes of a certain tastiness value to prevent Alice from eating them.

To solve the problem, transform the array of tastiness values into a count array, where each element represents the number of cakes with a specific tastiness. The game can then be modeled as Alice and Bob zeroing out elements of this count array. Using dynamic programming, define `dp[i][k]` as the minimum sum of cakes Alice can eat over all valid sets of indices up to `i` with `k` elements. Iterate through the count array, updating the `dp` table based on whether Bob chooses to zero out a specific index or not. The final answer is derived from the largest value of `k` for which the condition holds, giving the number of cakes Alice eats as `m - y`, where `y` is the largest valid `k`.



USER: ok let's go

BOT: Alice and Bob are playing a game where they take turns eating cakes with the goal of Alice maximizing the number of cakes she eats and Bob minimizing that number. Alice can only eat a cake if its tastiness is greater than any cake she has previously eaten, while Bob can eat any remaining cake. Alice's optimal strategy is to always eat the least tasty cake she can, ensuring she has more options later. Bob's optimal strategy is to either not touch a cake or eat all cakes of a certain tastiness value to prevent Alice from eating them.

To solve the problem, transform the array of tastiness values into a count array, where each element represents the number of cakes with a specific tastiness. Using dynamic programming, define `dp[i][k]` as the minimum sum of cakes Alice can eat over all valid sets of indices up to `i` with `k` elements. Iterate through the count array, updating the `dp` table based on whether Bob chooses to zero out a specific index or not. The final answer is derived from the largest value of `k` for which the condition holds, giving the number of cakes Alice eats as `m - y`, where `y` is the largest valid `k`. This approach ensures that both players play optimally, and the solution is found efficiently.



USER: what we should after that&

BOT: After transforming the array of tastiness values into a count array, the next step is to set up a dynamic programming (DP) table to model the game. Define `dp[i][k]` as the minimum sum of cakes Alice can eat over all valid sets of indices up to `i` with `k` elements. Initialize `dp[0][0] = 0` and all other entries to infinity. Iterate over each index `i` from 1 to `m` (where `m` is the number of unique tastiness values) and for each `k` from 0 to `m`. Update the DP table with two transitions: one where the current index is not used (`dp[i][k] = min(dp[i][k], dp[i-1][k])`) and one where the current index is used (`dp[i][k] = min(dp[i][k], dp[i-1][k-1] + c_i)`), ensuring the condition `dp[i-1][k-1] + c_i <= i - k` holds.

Finally, determine the largest `k` for which `dp[m][k]` is finite. This `k` represents the maximum number of indices Bob can zero out while playing optimally. The number of cakes Alice eats is then calculated as `m - y`, where `y` is this largest valid `k`. This approach ensures that both players are playing optimally, and the solution is derived efficiently by leveraging the DP table to keep track of the minimum sum of cakes Alice can eat for each possible set of indices.