# 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。

提示:

  • 1 <= words.length <= $10^4$
  • 1 <= allowed.length <= 26
  • 1 <= words[i].length <= 10
  • allowed 中的字符 互不相同 。
  • words [i] 和 allowed 只包含小写英文字母。

# 题解

class Solution {
public:
int countConsistentStrings(string allowed, vector<string>& words) {
int ans = 0;
int a = 0;
for (auto s : allowed) {
a |= 1 << (s - 'a');
}
for (int i = 0; i < words.size(); i++) {
int b = 0;
for (auto str : words[i]) b |= 1 << (str - 'a');
if ((a | b) == a) ans++;
}
return ans;
}
};

复杂度分析

  • 时间复杂度:$O (n+\sum m_i)$
  • 空间复杂度:$O (1)$