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There is a strange printer with the following two special requirements:

1. The printer can only print a sequence of the same character each time.
2. At each turn, the printer can print new characters starting from and ending at any places, and will cover the original existing characters. 

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.

有台奇怪的打印机有以下两个特殊要求：

1. 打印机每次只能打印同一个字符序列。
2. 每次可以在任意起始和结束位置打印新字符，并且会覆盖掉原来已有的字符。

给定一个只包含小写英文字母的字符串，你的任务是计算这个打印机打印它需要的最少次数。

示例 1:

输入: "aaabbb"
输出: 2
解释: 首先打印 "aaa" 然后打印 "bbb"。

示例 2:

输入: "aba"
输出: 2
解释: 首先打印 "aaa" 然后在第二个位置打印 "b" 覆盖掉原来的字符 'a'。

提示: 输入字符串的长度不会超过 100。

116ms

 1 class Solution {
 2     func strangePrinter(_ s: String) -> Int {
 3         var c = [Character]()
 4         var pre : Character = " "
 5         for (_, char) in s.enumerated() {
 6             if char != pre {
 7                 c.append(char)
 8                 pre = char
 9             }
10         }
11         var l = [[Int]](repeating:[Int](repeating: 0, count:c.count) , count:c.count)
12         return turn(c, 0, c.count - 1, &l)
13     }
14     
15     func turn(_ c: [Character], _ i: Int, _ j: Int, _ l: inout [[Int]]) -> Int {
16         if i > j {
17             return 0
18         }
19         if l[i][j] > 0 {
20             return l[i][j]
21         }
22         var ans = turn(c, i, j-1, &l) + 1
23         for k in i ..< j {
24             if c[k] == c[j] {
25                 ans = min(ans, turn(c, i, k, &l) + turn(c, k+1, j-1, &l))
26             }
27         }
28         l[i][j] = ans
29         return ans
30     }
31 }

 1 class Solution {
 2     func strangePrinter(_ s: String) -> Int {
 3         if s.isEmpty {return 0}
 4         var arr:[Character] = Array(s)
 5         var n:Int = s.count
 6         var dp:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:n),count:n)
 7         for i in (0...(n - 1)).reversed()
 8         {
 9             if(n < i) {continue}
10             for j in i..<n
11             {
12                 dp[i][j] = (i == j) ? 1 : (1 + dp[i + 1][j])
13                 if(j < i + 1) {continue}
14                 for k in (i + 1)...j
15                 {
16                     if arr[k] == arr[i]
17                     {
18                         dp[i][j] = min(dp[i][j], dp[i + 1][k - 1] + dp[k][j])
19                     }
20                 }
21             }
22         }
23         return (n == 0) ? 0 : dp[0][n - 1]
24     }
25 }