#### Approach #1: Dynamic Programming [Accepted]

Intuition

The cost of making both sequences increasing up to the first i columns can be expressed in terms of the cost of making both sequences increasing up to the first i-1 columns. This is because the only thing that matters to the ith column is whether the previous column was swapped or not. This makes dynamic programming an ideal choice.

Let's remember n1 (natural1), the cost of making the first i-1 columns increasing and not swapping the i-1th column; and s1 (swapped1), the cost of making the first i-1 columns increasing and swapping the i-1th column.

Now we want candidates n2 (and s2), the costs of making the first i columns increasing if we do not swap (or swap, respectively) the ith column.

Algorithm

For convenience, say a1 = A[i-1], b1 = B[i-1] and a2 = A[i], b2 = B[i].

Now, if a1 < a2 and b1 < b2, then it is allowed to have both of these columns natural (unswapped), or both of these columns swapped. This possibility leads to n2 = min(n2, n1) and s2 = min(s2, s1 + 1).

Another, (not exclusive) possibility is that a1 < b2 and b1 < a2. This means that it is allowed to have exactly one of these columns swapped. This possibility leads to n2 = min(n2, s1) or s2 = min(s2, n1 + 1).

Note that it is important to use two if statements separately, because both of the above possibilities might be possible.

At the end, the optimal solution must leave the last column either natural or swapped, so we take the minimum number of swaps between the two possibilities.

Complexity Analysis

• Time Complexity: .

• Space Complexity: .

Analysis written by: @awice.