LWC 63:748. Shortest Completing Word

LWC 63:748. Shortest Completing Word

传送门:748. Shortest Completing Word

Problem:

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.

Note:

licensePlate will be a string with length in range [1, 7].

licensePlate will contain digits, spaces, or letters (uppercase or lowercase).

words will have a length in the range [10, 1000].

Every words[i] will consist of lowercase letters, and have length in range [1, 15].

思路: 很暴力,找到一个word符合licensePlate的定义,并不断更新即可。只要licensePlate中出现的字母,word中必须出现,且出现的次数至少相等。采用map计数。

Java版本:

    public String shortestCompletingWord(String licensePlate, String[] words) {
        String ans = "";
        int minLen = 1111;

        int[] map = new int[32];
        for (char c : licensePlate.toLowerCase().toCharArray())
            if (c >= 'a' && c <= 'z') map[c - 'a'] ++;


        for (String word : words) {
            int len = word.length();
            if (contains(map, word)) {
                if (len < minLen) {
                    ans = word;
                    minLen = len;
                }
            }
        }
        return ans;
    }

    boolean contains(int[] map, String word) {
        int[] map2 = new int[32];
        for (char c : word.toLowerCase().toCharArray()) {
            map2[c - 'a'] ++;
        }

        for (int i = 0; i < 32; ++i) {
            if (map[i] != 0) {
                if (map2[i] < map[i]) return false;
            }
        }
        return true;
    }

Python版本:

    def shortestCompletingWord(self, licensePlate, words):
        """
        :type licensePlate: str
        :type words: List[str]
        :rtype: str
        """
        licence_plate = licensePlate.lower()
        d = dict()
        for c in licence_plate:
            if c.isalpha():
                if c not in d:
                    d[c] = 0
                d[c] += 1

        res = None
        length = 1111
        for word in words:
            freq = d.copy()
            for c in word:
                if c in freq:
                    freq[c] -= 1
                    if freq[c] == 0:
                        freq.pop(c)
                if not freq:
                    if len(word) < length:
                        res = word
                        length = len(word)

        return res 

本文参与腾讯云自媒体分享计划,欢迎正在阅读的你也加入,一起分享。

发表于

我来说两句

0 条评论
登录 后参与评论

相关文章

扫码关注云+社区