README_EN.md

664. Strange Printer

中文文档

Description

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

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

Given a string s, return the minimum number of turns the printer needed to print it.

 

Example 1:

Input: s = "aaabbb"
Output: 2
Explanation: Print "aaa" first and then print "bbb".

Example 2:

Input: s = "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'.

 

Constraints:

• 1 <= s.length <= 100
• s consists of lowercase English letters.

Solutions

Python3

class Solution:
    def strangePrinter(self, s: str) -> int:
        n = len(s)
        dp = [[inf] * n for _ in range(n)]
        for i in range(n - 1, -1, -1):
            dp[i][i] = 1
            for j in range(i + 1, n):
                if s[i] == s[j]:
                    dp[i][j] = dp[i][j - 1]
                else:
                    for k in range(i, j):
                        dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j])
        return dp[0][-1]

Java

class Solution {
    public int strangePrinter(String s) {
        int n = s.length();
        int[][] f = new int[n + 1][n + 1];
        for (int i = 0; i < n; ++i) {
            f[i][i] = 1;
        }
        for (int i = n - 2; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                f[i][j] = 1 + f[i + 1][j];
                for (int k = i + 1; k <= j; ++k) {
                    if (s.charAt(i) == s.charAt(k)) {
                        f[i][j] = Math.min(f[i][j], f[i + 1][k] + f[k + 1][j]);
                    }
                }
            }
        }
        return f[0][n - 1];
    }
}

class Solution {
    public int strangePrinter(String s) {
        int n = s.length();
        int[][] dp = new int[n][n];
        for (int i = n - 1; i >= 0; --i) {
            dp[i][i] = 1;
            for (int j = i + 1; j < n; ++j) {
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = dp[i][j - 1];
                } else {
                    dp[i][j] = 10000;
                    for (int k = i; k < j; ++k) {
                        dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k + 1][j]);
                    }
                }
            }
        }
        return dp[0][n - 1];
    }
}

C++

class Solution {
public:
    int strangePrinter(string s) {
        int n = s.size();
        vector<vector<int>> dp(n, vector<int>(n, INT_MAX));
        for (int i = n - 1; i >= 0; --i) {
            dp[i][i] = 1;
            for (int j = i + 1; j < n; ++j) {
                if (s[i] == s[j]) {
                    dp[i][j] = dp[i][j - 1];
                } else {
                    for (int k = i; k < j; ++k) {
                        dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);
                    }
                }
            }
        }
        return dp[0][n - 1];
    }
};

Go

func strangePrinter(s string) int {
	n := len(s)
	dp := make([][]int, n)
	for i := range dp {
		dp[i] = make([]int, n)
	}
	for i := n - 1; i >= 0; i-- {
		dp[i][i] = 1
		for j := i + 1; j < n; j++ {
			if s[i] == s[j] {
				dp[i][j] = dp[i][j-1]
			} else {
				dp[i][j] = 10000
				for k := i; k < j; k++ {
					dp[i][j] = min(dp[i][j], dp[i][k]+dp[k+1][j])
				}
			}
		}
	}
	return dp[0][n-1]
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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