package com.fishercoder.solutions;

/**
 * 664. Strange Printer
 *
 * There is a strange printer with the following two special requirements:
 * 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 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.
 */
public class _664 {
    public static class Solution1 {
        /**
         * reference: https://discuss.leetcode.com/topic/100137/java-solution-dp
         */
        public int strangePrinter(String s) {
            int n = s.length();
            if (n == 0) {
                return 0;
            }

            int[][] dp = new int[101][101];
            for (int i = 0; i < n; i++) {
                dp[i][i] = 1;
            }

            for (int i = 1; i < n; i++) {
                for (int j = 0; j < n - i; j++) {
                    dp[j][j + i] = i + 1;
                    for (int k = j + 1; k <= j + i; k++) {
                        int temp = dp[j][k - 1] + dp[k][j + i];
                        if (s.charAt(k - 1) == s.charAt(j + i)) {
                            temp--;
                        }
                        dp[j][j + i] = Math.min(dp[j][j + i], temp);
                    }
                }
            }
            return dp[0][n - 1];
        }
    }
}

package com.fishercoder.solutions;

public class _664 {
    public static class Solution1 {
        /**
         * reference: https://discuss.leetcode.com/topic/100137/java-solution-dp
         */
        public int strangePrinter(String s) {
            int n = s.length();
            if (n == 0) {
                return 0;
            }

            int[][] dp = new int[101][101];
            for (int i = 0; i < n; i++) {
                dp[i][i] = 1;
            }

            for (int i = 1; i < n; i++) {
                for (int j = 0; j < n - i; j++) {
                    dp[j][j + i] = i + 1;
                    for (int k = j + 1; k <= j + i; k++) {
                        int temp = dp[j][k - 1] + dp[k][j + i];
                        if (s.charAt(k - 1) == s.charAt(j + i)) {
                            temp--;
                        }
                        dp[j][j + i] = Math.min(dp[j][j + i], temp);
                    }
                }
            }
            return dp[0][n - 1];
        }
    }
}

refactor 664