There is a strange printer with the following two special requirements:
Given a string consists of lower English letters only, your job is to count the minimum number of turns the printer needed in order to print it.
Example 1:
Input: "aaabbb" Output: 2 Explanation: Print "aaa" first and then print "bbb".
Example 2:
Input: "aba" Output: 2 Explanation: Print "aaa" first and then print "b" from the second place of the string, which will cover the existing character 'a'.
Hint: Length of the given string will not exceed 100.
Intuition
\nIt 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
\nWith the above deductions, the algorithm is straightforward.
\nTo 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
\nTime 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.
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