Letter most:

For each lowercase letter count number of its occurrences in $$$s$$$, maximum of this value for all letters is the answer. All can be done in $$$O(n)$$$.

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Array modification:

Assume  will be the maximum. it can be proven that all operation should be done in direction of $$$x$$$ ($$$j=i+1$$$ for all operation on its left and $$$j=i-1$$$ for all operations on its right).

Now consider $$$pref_x$$$ is maximum value of $$$a_x$$$ if all operations be done in direction of right for all $$$i=1, 2, ..., x$$$.

And consider $$$suf_x$$$ is maximum value of $$$a_x$$$ if all operations be done in direction of left for all $$$i=x, x+1, ..., n$$$.

It can bee seen that:

• $$$pref_x = \frac{perf_{x-1}}{2} + a_x$$$.

• $$$suf_x = \frac{suf_{x+1}}{2} + a_x$$$.

And so the answer is $$$max_{i=1}^{n}(pref_i + suf_i - a_i)$$$.

All can be done in $$$O(n)$$$.

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Plan for nothing:

Consider an array $$$c_1, c_2, .., c_n$$$, for all intervals like $$$[l, r]$$$ do the following operation:

• $$$c_l = c_l + 1$$$.

• if $$$r < n$$$, $$$c_{r+1} = c_{r+1} - 1$$$.

Then consider $$$pref_i = c_1 + c_2 + ... c_i$$$. a day is good for date if and only if $$$pref_i = 0$$$.

So the answer will be minimum $$$i$$$ that $$$pref_i = 0$$$ and if it doesn't exist answer is $$$-1$$$.

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Lonely M's array:

First reverse $$$a_1, a_2, .., a_n$$$.

Consider two following arrays:

• $$$less_i, 1 \leq i \leq 3\times 10^5$$$: maximum length of all subsequence which ends with $$$... \leq i$$$.

• $$$greater_i, 1 \leq i \leq 3\times 10^5$$$: maximum length of all subsequence which ends with $$$... \geq i$$$.

• initially both are filled with zero.

Start sweep line form $$$1$$$ to $$$n$$$:

• for each $$$a_i$$$ maximum length subsequence that ends with $$$... \leq a_i$$$ is $$$max_{j=1}^{a_i}(greater_j) + 1$$$ lets call it $$$X$$$.

• for each $$$a_i$$$ maximum length subsequence that ends with $$$... \geq a_i$$$ is $$$max_{j=1}^{a_i}(less_j) + 1$$$ lets call it $$$Y$$$.

• after calculating $$$X$$$ and $$$Y$$$, $$$less_{a_i}$$$ should be updated with $$$X$$$ and also greater_{a_i} should be updated with $$$Y$$$. The answer is $$$max_{i=1}^{3 \times 10^5}(less_i)$$$ because the subsequence should finish with $$$\leq$$$.

There are some RMQ(range maximum queries) and some single elements updates that all can be done using segment trees or fenwick trees (because queries are all either a prefix or suffix) in $$$log(3 \times 10^5)$$$.

So all can be done in $$$n \times log(3 \times 10^5)$$$.