Minimum Distance to Type a Word Using Two Fingers

You have a keyboard layout as shown above in the XY plane, where each English uppercase letter is located at some coordinate, for example, the letter A is located at coordinate (0,0), the letter B is located at coordinate (0,1), the letter P is located at coordinate (2,3) and the letter Z is located at coordinate (4,1).

Given the string word, return the minimum total distance to type such string using only two fingers. The distance between coordinates (x1,y1) and (x2,y2) is |x1 - x2| + |y1 - y2|. 

Note that the initial positions of your two fingers are considered free so don't count towards your total distance, also your two fingers do not have to start at the first letter or the first two letters.

Example 1:

Input: word = "CAKE"
Output: 3
Explanation: 
Using two fingers, one optimal way to type "CAKE" is: 
Finger 1 on letter 'C' -> cost = 0 
Finger 1 on letter 'A' -> cost = Distance from letter 'C' to letter 'A' = 2 
Finger 2 on letter 'K' -> cost = 0 
Finger 2 on letter 'E' -> cost = Distance from letter 'K' to letter 'E' = 1 
Total distance = 3

Example 2:

Input: word = "HAPPY"
Output: 6
Explanation: 
Using two fingers, one optimal way to type "HAPPY" is:
Finger 1 on letter 'H' -> cost = 0
Finger 1 on letter 'A' -> cost = Distance from letter 'H' to letter 'A' = 2
Finger 2 on letter 'P' -> cost = 0
Finger 2 on letter 'P' -> cost = Distance from letter 'P' to letter 'P' = 0
Finger 1 on letter 'Y' -> cost = Distance from letter 'A' to letter 'Y' = 4
Total distance = 6

Example 3:

Input: word = "NEW"
Output: 3

Example 4:

Input: word = "YEAR"
Output: 7

Constraints:

• 2 <= word.length <= 300
• Each word[i] is an English uppercase letter.

思路：记忆化搜索，top down dp，两个手指move，可以写一个cost function计算两个点的cost，这个没有问题，那么问题转换成下一个move由谁来完成，cost分别是多少。top down就是最后一步，如果由左边完成，cost + dfs(左边)， 由右边完成，cost + dfs(右边）建立一个三维的dp，[leftindex][rightindex][pos in word]来标记计算过的结果；Time ( n * 27 ^ 2); Space ( n * 27 ^ 2);

class Solution {
    
    public int minimumDistance(String word) {
        int n = word.length();
        int[][][] dp = new int[27][27][n]; // 27 主要是为了 null的时候，为0，从而开一个空的char在0index；
        return dfs(word, dp, 0, null, null);
    }
    
    private int dfs(String word, int[][][] dp, int pos, Character c1, Character c2) {
        if(pos >= word.length()) {
            return 0;
        }
        int index1 = c1 == null ? 0 : c1 - 'A' + 1;
        int index2 = c2 == null ? 0 : c2 - 'A' + 1;
        if(dp[index1][index2][pos] > 0) {
            return dp[index1][index2][pos];
        }
        char next = word.charAt(pos);
        dp[index1][index2][pos] = Math.min(cost(c1, next) + dfs(word, dp, pos + 1, next, c2),
                                          cost(c2, next) + dfs(word, dp, pos + 1, c1, next));
        return dp[index1][index2][pos];
    }
    
    private int cost(Character c1, Character c2) {
        if(c1 == null || c2 == null) {
            return 0;
        }
        int d1 = c1 - 'A', d2 = c2 - 'A';
        int x1 = d1 / 6, y1 = d1 % 6;
        int x2 = d2 / 6, y2 = d2 % 6;
        return Math.abs(x1 - x2) + Math.abs(y1 - y2);
    }
}