488. Zuma Game

Think about Zuma Game. You have a row of balls on the table, colored red(R), yellow(Y), blue(B), green(G), and white(W). You also have several balls in your hand.

Each time, you may choose a ball in your hand, and insert it into the row (including the leftmost place and rightmost place). Then, if there is a group of 3 or more balls in the same color touching, remove these balls. Keep doing this until no more balls can be removed.

Find the minimal balls you have to insert to remove all the balls on the table. If you cannot remove all the balls, output -1.

Examples:

Input: "WRRBBW", "RB"
Output: -1
Explanation: WRRBBW -> WRR[R]BBW -> WBBW -> WBB[B]W -> WW

Input: "WWRRBBWW", "WRBRW"
Output: 2
Explanation: WWRRBBWW -> WWRR[R]BBWW -> WWBBWW -> WWBB[B]WW -> WWWW -> empty

Input:"G", "GGGGG"
Output: 2
Explanation: G -> G[G] -> GG[G] -> empty 

Input: "RBYYBBRRB", "YRBGB"
Output: 3
Explanation: RBYYBBRRB -> RBYY[Y]BBRRB -> RBBBRRB -> RRRB -> B -> B[B] -> BB[B] -> empty 

Note:
You may assume that the initial row of balls on the table won’t have any 3 or more consecutive balls with the same color.
The number of balls on the table won’t exceed 20, and the string represents these balls is called “board” in the input.
The number of balls in your hand won’t exceed 5, and the string represents these balls is called “hand” in the input.
Both input strings will be non-empty and only contain characters ‘R’,’Y’,’B’,’G’,’W’.

思路：
题目希望我们用最少的球来消掉桌上所有的球，如果不能完全消掉，返回-1。
我们使用哈希表来统计手中每种球的个数。
然后我们遍历桌上的球，我们找连续相同球的个数，在没有可以消除的情况下，连续的个数只能是1个或2个，然后我们用3减去连续个数，就是我们需要补充的球数以使其可以被消除，那么我们在哈希表表中看我们手中的该类型的球够不够，如果够就表示可以消除，我们在哈希表中减去需要使用掉的球数，然后将消掉的球移除。
然后对新的字符串调用递归，如果可以成功消除，会返回一个结果，该结果加上之前需要的球数用来更新结果res，注意调用完递归要恢复哈希表的状态(backtrack)。

Time Complexity: O(N) Space Complexity: O(N)

class Solution {
    int MAXCOUNT = 6;   // the max balls you need will not exceed 6 since "The number of balls in your hand won't exceed 5"

    public int findMinStep(String board, String hand) {
        int[] handCount = new int[26];
        for (int i = 0; i < hand.length(); i++) {
            handCount[hand.charAt(i) - 'A']++;
        }
        int result = backtrack(board + "#", handCount);  // append a "#" to avoid special process while j==board.length, make the code shorter.
        return result == MAXCOUNT ? -1 : result;
    }

    private int backtrack(String s, int[] h) {
        s = removeConsecutive(s);     
        if (s.equals("#")) return 0;
        int rs = MAXCOUNT, need = 0;
        int i = 0;
        for (int j = 0 ; j < s.length(); ++j) {
            if (s.charAt(j) == s.charAt(i)) continue;
            need = 3 - (j - i);     //balls need to remove current consecutive balls.
            if (h[s.charAt(i) - 'A'] >= need) {
                h[s.charAt(i) - 'A'] -= need;
                rs = Math.min(rs, need + backtrack(s.substring(0, i) + s.substring(j), h));
                h[s.charAt(i) - 'A'] += need;
            }
            i = j;
        }
        return rs;
    }
    //remove consecutive balls longer than 3
    private String removeConsecutive(String board) {
        int i = 0;
        for (int j = 0; j < board.length(); ++j) {
            if (board.charAt(j) == board.charAt(i)) continue;
            if (j - i >= 3)
                return removeConsecutive(board.substring(0, i) + board.substring(j));
            else i = j;
        }
        return board;
    }
}