1160.拼写单词.py

#
# @lc app=leetcode.cn id=1160 lang=python3
#
# [1160] 拼写单词
#
# https://leetcode-cn.com/problems/find-words-that-can-be-formed-by-characters/description/
#
# algorithms
# Easy (63.37%)
# Likes:    40
# Dislikes: 0
# Total Accepted:    18.7K
# Total Submissions: 27.1K
# Testcase Example:  '["cat","bt","hat","tree"]\n"atach"'
#
# 给你一份『词汇表』（字符串数组） words 和一张『字母表』（字符串） chars。
#
# 假如你可以用 chars 中的『字母』（字符）拼写出 words 中的某个『单词』（字符串），那么我们就认为你掌握了这个单词。
#
# 注意：每次拼写时，chars 中的每个字母都只能用一次。
#
# 返回词汇表 words 中你掌握的所有单词的 长度之和。
#
#
#
# 示例 1：
#
# 输入：words = ["cat","bt","hat","tree"], chars = "atach"
# 输出：6
# 解释：
# 可以形成字符串 "cat" 和 "hat"，所以答案是 3 + 3 = 6。
#
#
# 示例 2：
#
# 输入：words = ["hello","world","leetcode"], chars = "welldonehoneyr"
# 输出：10
# 解释：
# 可以形成字符串 "hello" 和 "world"，所以答案是 5 + 5 = 10。
#
#
#
#
# 提示：
#
#
# 1 <= words.length <= 1000
# 1 <= words[i].length, chars.length <= 100
# 所有字符串中都仅包含小写英文字母
#
#
#

from typing import List


# @lc code=start
class Solution:

    def countCharacters_1(self, words: List[str], chars: str) -> int:
        from collections import Counter
        charCounts = Counter(chars)

        def fromChars(word, charCounts):
            chars = Counter(word)
            for char in chars:
                if char not in charCounts or charCounts[char] < chars[char]:
                    return False
            return True

        res = 0
        for word in words:
            if fromChars(word, charCounts):
                res += len(word)
        return res

    def countCharacters(self, words: List[str], chars: str) -> int:
        from collections import Counter
        charsCount = Counter(chars)

        res = 0
        for word in words:
            wordsCount = Counter(word)
            if all([charsCount[c] >= wordsCount[c] for c in wordsCount]):
                res += len(word)
        return res


# @lc code=end