Note $$$maxp$$$ as the maximum index of the maximum value in $$$a$$$.

For $$$1 \leq k \leq maxp-1$$$,they do not meet the condition.

For $$$maxp+1 \leq k \leq n$$$,they all meet the condition.

You need to be careful of $$$k==maxp$$$,it meet the condition if and only if there is only one maximum value in $$$a$$$.

Note $$$a_1$$$ as the first $$$\lceil{n/2}\rceil$$$ numbers of $$$a$$$,and $$$a_2$$$ as the last $$$n-\lceil{n/2}\rceil$$$ numbers of $$$a$$$.

Through observation,we found that the following processes will be repeated cyclically:

• erase the leftmost number of $$$a_1$$$

• erase the leftmost number of $$$a_2$$$

• erase the rightmost number of $$$a_1$$$

• erase the rightmost number of $$$a_2$$$

In conclusion:

• If $$$|a_1|>|a_2|$$$(i.e. $$$n$$$ is odd),the answer is the number in the middle of $$$a_1$$$

• If $$$|a_1|=|a_2|$$$(i.e. $$$n$$$ is even),the answer is the number in the middle of $$$a_2$$$

We notice we can convert condition $$$3$$$ to $$$c0(a,l_1,r_1)-c1(a,l_1,r_1)=c1(b,l_2,r_2)-c0(b,l_2,r_2)$$$.

This provides us with convenience: we can independently calculate the left and right value ranges.

Let's calculate the left value ranges.Make all $$$a_i=0$$$ to $$$1$$$ and $$$a_i=1$$$ to $$$-1$$$,it is the value range of all subarrays of appropriate length.

In fact,we only need to find max subsegment sum and min subsegment sum.It's a classic problem.

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