【python3】leetcode 748. Shortest Completing Word（easy）

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748. Shortest Completing Word（easy）

Find the minimum length word from a given dictionary words, which has all the letters from the string licensePlate. Such a word is said to complete the given string licensePlate

Here, for letters we ignore case. For example, "P" on the licensePlate still matches "p" on the word.

It is guaranteed an answer exists. If there are multiple answers, return the one that occurs first in the array.

The license plate might have the same letter occurring multiple times. For example, given a licensePlate of "PP", the word "pair" does not complete the licensePlate, but the word "supper" does.

Example 1:

Input: licensePlate = "1s3 PSt", words = ["step", "steps", "stripe", "stepple"]
Output: "steps"
Explanation: The smallest length word that contains the letters "S", "P", "S", and "T".
Note that the answer is not "step", because the letter "s" must occur in the word twice.
Also note that we ignored case for the purposes of comparing whether a letter exists in the word.

Example 2:

Input: licensePlate = "1s3 456", words = ["looks", "pest", "stew", "show"]
Output: "pest"
Explanation: There are 3 smallest length words that contains the letters "s".
We return the one that occurred first.

注意几点：

1 提取licensePlate里的字母且 大小写不敏感 ，先转为str.lower()

2 返回满足条件（licensePlate里的字母都在word里出现且数量《word里的数量）长度最小的word，所以要先按长度排序

依旧使用collections.Counter计算字母出现的次数

class Solution:
    def shortestCompletingWord(self, licensePlate, words):
        """
        :type licensePlate: str
        :type words: List[str]
        :rtype: str
        """
        letter = [i.lower() for i in licensePlate if  'a'<= i <='z' or 'A'<= i <= 'Z']
        countle = collections.Counter(letter)
        sortwords = sorted(words,key = len)
        for x in sortwords:
            flag = 0
            countx = collections.Counter(x)
            for key,val in countle.items():
                if key not in countx.keys() or  val > countx[key]:
                    flag = 1
                    break
            if flag == 0:return x