The important thing to note in this question is it's always optimal to go until the last day and then time travel back to the first day.

In order to prove this, let's assume there's another optimal option, $$$e_i$$$ and $$$s_i$$$, time-travelling from day $$$e_i$$$ to $$$s_i$$$. Therefore we can think of the array $$$p$$$=[$$$p_0$$$, ...., $$$p_{s_i}$$$, ...., $$$p_{e_i}$$$, $$$p_{s_i}$$$, ...., $$$p_{e_i}$$$, ...., $$$p_{n-1}$$$], and solve leave out the time travel aspect of the question. Note how we added [$$$p_{s_i}$$$, ...., $$$p_{e_i}$$$] after $$$p_{e_i}$$$ in the original array $$$p$$$.

Once we know this, we can append the contents of $$$p$$$ to $$$p$$$ and treat it like the buy and sell stock II question on Leetcode.

Time Complexity: O(n)

Space Complexity: O(n)

For this problem, if an divider has $$$k$$$ outlets, then it provides $$$k-1$$$ outlets since it has to plug into an already existing outlet. So in order to minimize the number of outlets we are using we can keep taking the divider with the largest number of outlets until the sum of their outlets ($$$-1$$$ for each divider) is above or equal to the number of students.

An edge case to be aware of is if the number of students is less or equal to 2, whereby we don't need any additional dividers.

Time Complexity: $$$O(nlgn)$$$ with normal sorting, $$$O(n + p)$$$ with counting sort ($$$p$$$ being $$$max(outlets)$$$).