I present my solution below, it is however hidden because I donâ€™t want to spoil the solution for anyone who wantâ€™s to think about it themselves.

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Solution

It is obvious that at most n-1 replacements are needed: replace everything by A_1.

Letâ€™s now look at the following case: 1,0,0,1, This can be solved by one replacement: 1,0,0,1\implies 1,1,1,1. This was because the zeroes were surrounded by the same value, and this meant one fewer replacement was needed.

We want to take advantage of as many such â€śsurroundingsâ€ť as we can. For this we find all potential surroundings, ranges (i,j) such that A_i=A_j and A_i\neq A_k for all i<k<j. This can be done in \mathcal{O}(n) time

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Finding all surroundings

Convert the input array into one that also has the indeces of each element. For example [5,3,5] goes to [(5,0),(3,1),(5,2)]. Then sort this array: [(3,1),(5,0),(5,2)]. In the sorted array we look at adjacent elements whoâ€™s value is the same and take the indeces as a surrounding.

Now the question becomes when we can use surroundings together. I will present three examples:

- 0,0,1,1
- 1,0,0,1
- 1,0,1,0

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When surroundings can be used together

Example 1 and 2 can be done in 1 replacements, while 2 are needed for the last example. All these examples do have the same number of â€śsurroundingsâ€ť. Two surroundings can be used together in two cases:

- one surrounding comes entirely before the other: surroundings (a,b),(c,d) can work if b<c or d<a
- one surrounding is in the other surrounding: (a,b),(c,d) can work if a<c,d<b or c<a,b<d

Next we will need to find the maximum number of surroundings that can be used together. This is the hardest part. To simplify we will first only look at the first case.

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Maximum non-nested surroundings

First sort all ranges [(a_i,b_i)] in increasing a_i and when there are ties decreasing b_i. Then we go trough the ranges in order and keep track of the last range we can use. Itâ€™s hard to explain the algorithm so instead Iâ€™m going to give an illustrative example:

[(1,5),(2,3),(4,6)]

First take (1,5), itâ€™s the only one weâ€™ve considered so far so we rember it.

Then itâ€™s (2,3)'s turn. Itâ€™s end comes before the end of (1,5)\implies remember (2,3) instead

Next is (4,6). This one is entirely after (2,3) so we use (2,3) and remember (4,6)

Now weâ€™ve seen all ranges. We still have (4,6) in our memory so we also use that one.

As we only loop trough all ranges once this algorithm is \mathcal{O}(n). It can be adapted, using DP to also take into account ranges with some weight and we want the maximum weight compatible ranges.

And then the algorithm for when we include nested surroundings into the mix:

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Maximum nested surroundings

We first assign each range we found a value of 1. Then we go trough all ranges in the order small to large and compute how many compatible surroundings would fit in it. This is essentially the previous segment: the weighted compatible segments problem. Then we update the weight of this range as the calculated value + 1 (the +1 for the range itself).

We compute small to large as a large segment will never be able to fit into a smaller one.

As we apply an \mathcal{O}(n) algorithm n times the entire algorithm has efficiency \mathcal{O}(n^2)

Given the maximum number of surroundings k our answer will be n-1-k