8.22 - hard - 81

411. Minimum Unique Word Abbreviation

利用比较基本的方法，backtracking找出全部的abbreviation，TLE

class Solution(object):
    def minAbbreviation(self, target, dictionary):
        """
        :type target: str
        :type dictionary: List[str]
        :rtype: str
        """
        # 基本想法是先把dict里的所有可能的变形都找出来
        # 然后再找target
        res = set()
        if dictionary:
            for word in dictionary:
                self.search(word, "", 0, 0, res)
        if not res:
            return str(len(target))
        target_set = set()
        self.search(target, "", 0, 0, target_set)
        new = target_set.difference(res)
        
        return sorted(new, key=self.cal)[0]
        
        
    def cal(self, s):
        l = 0
        first_digit = True
        for c in s:
            if first_digit and c.isdigit():
                l += 1
                first_digit = False
            if not c.isdigit():
                l += 1
                first_digit = True
        
        return l
        
    
    def search(self, s, cur, pos, count, res):
        if pos == len(s):
            if count > 0:
                cur += str(count)
            res.add(cur)
            return
        
        self.search(s, cur, pos+1, count+1, res) # 缩略当前的字母增加count
        if count > 0:
            cur = cur + str(count) + s[pos]
        else:
            cur = cur + s[pos]
        self.search(s, cur, pos+1, 0, res)