664. Strange Printer

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.

 

Approach #1: DP. [C++]

class Solution {
public:
    int strangePrinter(string s) {
        int len = s.length();
        dp = vector<vector<int>>(len+1, vector<int>(len+1, 0));
        return turn(s, 0, len-1);
    }
    
private:
    vector<vector<int>> dp;
    
    int turn(const string& s, int i, int j) {
        if (i > j) return 0;
        if (dp[i][j] != 0) return dp[i][j];
        int ans = turn(s, i, j-1) + 1;
        
        for (int k = i; k < j; ++k)
            if (s[k] == s[j])
                ans = min(ans, turn(s, i, k)+turn(s, k+1, j-1));
        dp[i][j] = ans;
        return dp[i][j];
    }
};

　　

 

Approach #2: DP. [Java]

class Solution {
    public int strangePrinter(String s) {
        int len = s.length();
        t_ = new int[len][len];
        return turn(s.toCharArray(), 0, len-1);
    }
    
    public int turn(char[] s, int i, int j) {
        if (i > j) return 0;
        if (t_[i][j] > 0) return t_[i][j];
        
        int ans = turn(s, i, j-1) + 1;
        
        for (int k = i; k < j; ++k) {
            if (s[k] == s[j]) 
                ans = Math.min(ans, turn(s, i, k) + turn(s, k+1, j-1));
        }
        
        t_[i][j] = ans;
        
        return t_[i][j];
    }
    
    private int[][] t_;
}

　　

 

Approach #3: DP. [Python]

class Solution(object):
    def strangePrinter(self, s):
        """
        :type s: str
        :rtype: int
        """
        l = len(s)
        self._t = [[0 for _ in xrange(l)] for _ in xrange(l)]
        
        return self.turn(s, 0, l-1)
    
    def turn(self, s, i, j):
        if i > j:
            return 0
        if self._t[i][j] > 0:
            return self._t[i][j]
        
        ans = self.turn(s, i, j-1) + 1
        
        for k in xrange(i, j):
            if s[k] == s[j]:
                ans = min(ans, self.turn(s, i, k) + self.turn(s, k+1, j-1))
                
        self._t[i][j] = ans
        
        return self._t[i][j]

　　

 

Reference：

http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-664-strange-printer/