We'll begin by sorting the observations in non-decreasing order of position (ordering equal-position observations arbitrarily), and re-numbering them in this order. In addition to each observation \(i\)'s \(P_i\) and \(R_i\) values, we'll store the original trip index \(I_i\) which it corresponds to.

We'll proceed with dynamic programming. Let \(D_{i,j}\) be the minimum number of renovations which could have occurred if we consider only the first \(i\) observations in this ordered list, such that \(j\) of their trips have been rearranged. We'll only concern ourselves with indices \(i\) on boundaries between different positions (specifically, such that \(i = 0\), \(i = N\), or \(P_i < P_{i+1}\)). We know that \(D_{0,0} = 0\), and our answer will be \(\min(D_{N,0..K})\).

In order to compute \(D_{i,j}\), we'll consider each possible earlier index \(k\) such that \(k < i\) and either \(k = 0\) or \(P_k < P_{k+1}\), determine what can be done with observations \((k+1)..i\), compute a candidate value for \(D_{i,j}\) accordingly, and take the minimum of these candidate values.

If \(R_o = 0\) for each \(o \in [k+1, i]\), then no renovations nor trip rearrangements are required, and our candidate value will simply be \(D_{k,j}\).

Otherwise, there must be at least one renovation amongst these observations' positions, and we'll assume that there was exactly one renovation spanning that full interval of positions (noting that other possibilities would be captured by different choices of \(k\)). We must then determine the minimum number of trip rearrangements necessary for that to be possible.

Considering each possible original trip index \(t\) just after which the renovation was made (such that \(0 \le t \le N\), with \(t = 0\) representing the renovation having been before the first original trip), it's optimal to:

- Rearrange all 1-observations before that time (observations \(o\) such that \(k+1 \le o \le i\), \(R_o = 1\), and \(I_i \le t\)) to have occurred after that time
- Rearrange all 0-observations after that time (observations \(o\) such that \(k+1 \le o \le i\), \(R_o = 0\), and \(I_i > t\)) to have occurred before that time

Given the minimum number \(x\) of observations which need to be rearranged (across all possible choices of \(t\)), our candidate value for \(D_{i,j}\) will then be \(D_{k,j-x} + 1\) (or \(\infty\) if \(j-x < 0\)).

The most direct, naive implementation of the above approach would be too slow, taking \(O(N^5)\) time (considering \(O(N^4)\) quadruples of \((i, j, k, t)\), and iterating over \(O(N)\) observations each time to count the number of trips needing rearrangement). However, the time complexity may be reduced to \(O(N^3)\) through a combination of two optimizations each removing a factor of \(N\):

1. We can observe that the value \(x\) described above is independent of \(j\), meaning that we only need to compute it once per pair of \((i, k)\).
2. For each pair of \((i, k)\), if we sweep over all observations in non-decreasing order of \(I\) value (considering only ones in that interval of positions), while keeping track of the number of 0-observations and 1-observations already encountered. This is sufficient to compute \(x\) with no additional time.