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

Intuition

It is natural to consider letting dp(i, j) be the answer for printing S[i], S[i+1], ..., S[j], but proceeding from here is difficult. We need the following sequence of deductions:

• Whatever turn creates the final print of S[i] might as well be the first turn, and also there might as well only be one print, since any later prints on interval [i, k] could just be on [i+1, k].

• Say the first print is on [i, k]. We can assume S[i] == S[k], because if it wasn't, we could print up to the last occurrence of S[i] in [i, k] for the same result.

• When correctly printing everything in [i, k] (with S[i] == S[k]), it will take the same amount of steps as correctly printing everything in [i, k-1]. This is because if S[i] and S[k] get completed in separate steps, we might as well print them first in one step instead.

Algorithm

With the above deductions, the algorithm is straightforward.

To compute a recursion for dp(i, j), for every i <= k <= j with S[i] == S[k], we have some candidate answer dp(i, k-1) + dp(k+1, j), and we take the minimum of these candidates. Of course, when k = i, the candidate is just 1 + dp(i+1, j).

To avoid repeating work, we memoize our intermediate answers dp(i, j).

Complexity Analysis

• Time Complexity: where is the length of s. For each of possible states representing a subarray of s, we perform work iterating through k.

• Space Complexity: , the size of our memo.

Analysis written by: @awice.