1684. 统计一致字符串的数目
给你一个由不同字符组成的字符串
allowed
和一个字符串数组
words
。如果一个字符串的每一个字符都在
allowed
中,就称这个字符串是 一致字符串 。
请你返回
words
数组中 一致字符串 的数目。
示例 1:
输入:allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
输出:2
解释:字符串 "aaab" 和 "baa" 都是一致字符串,因为它们只包含字符 'a' 和 'b' 。
示例 2:
输入:allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
输出:7
解释:所有字符串都是一致的。
示例 3:
输入:allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
输出:4
解释:字符串 "cc","acd","ac" 和 "d" 是一致字符串。
-
1 <= words.length <= 104
-
1 <= allowed.length <= 26
-
1 <= words[i].length <= 10
-
中的字符 互不相同 。allowed
-
和 words[i]
只包含小写英文字母。allowed
Solution
class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
cnt = len(words)
for word in words:
for letter in word:
if letter not in allowed:
cnt -= 1
break
return cnt
class Solution:
def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
mask = 0
for c in allowed:
mask |= 1 << (ord(c) - ord('a'))
res = 0
for word in words:
mask1 = 0
for c in word:
mask1 |= 1 << (ord(c) - ord('a'))
res += (mask1 | mask) == mask
return res
作者:力扣官方题解
链接:https://leetcode.cn/problems/count-the-number-of-consistent-strings/solutions/1953831/tong-ji-yi-zhi-zi-fu-chuan-de-shu-mu-by-38356/
来源:力扣(LeetCode)
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