Re: [閒聊] 每日leetcode

[JerryChungYC (JerryChung)]

https://leetcode.com/problems/count-the-number-of-consistent-strings

1684. Count the Number of Consistent Strings

給一個 由不同字元組成的 allowed 字串 跟一組字串數組 words
數組中的字串 如果所有字元都在 allowed 裡面的話 代表是 consistent
求數組中 consistent 的數量

Example 1:
  Input: allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
  Output: 2
  Explanation: 只有 "aaab" 跟 "baa" 是由 'a' 和 'b' 組成的

Example 2:
  Input: allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
  Output: 7
  Explanation: 全部字串都是 consistent

Example 3:
  Input: allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
  Output: 4
  Explanation: "cc", "acd", "ac" 和 "d" 是 conststent


Python Code:

class Solution:
    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
        a_set = set(allowed)
        return sum(1 for word in words if all(w in a_set for w in set(word)))

allowed 做一次 set 就好 所以多一行
只是這成績真的好爛 :(

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→ JerryChungYC: 看了一下all應該也是遇到False就直接跳出 所以就不
→ JerryChungYC: 自己寫for了
※ 編輯: JerryChungYC (60.251.52.67 臺灣), 09/12/2024 08:50:53
→ JerryChungYC: word不轉set好像更好 哀 算了
→ JerryChungYC: count好像比sum好 哀 算了
→ sustainer123: 早早早
→ JerryChungYC: 用-=比用all生成器好 哀 算了
class Solution:
    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
        allowed_set = set(allowed)
        count = 0
        for word in words:
            c = False
            for w in word:
                if w not in allowed:
                    c = True
                    break
            if c:
                continue
            count += 1
        return count
※ 編輯: JerryChungYC (60.251.52.67 臺灣), 09/12/2024 09:09:16