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 <= 104
• 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。

方法一：遍历

由题意可知，字符串 allowed 和字符串数组 words 中的所有字符串都只包含小写字母，因此我们可以用一个 32 位整数来表示一个字符串出现的字母集合。如果一个字母出现了，那么将对应整数的位设为 1。使用 mask 表示字符串 allowed 出现的字母集合。依次遍历字符串数组 words，假设第 i 个字符串 words}[i] 对应出现的字母集合为 mask}_1，那么 words}[i] 是一致字符串等价于 mask}_1 是 mask 的子集，即 mask}_1 \cup \textit{mask} = \textit{mask。统计一致字符串的数目并返回结果。

[sol1-Python3]

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

[sol1-C++]

class Solution {
public:
    int countConsistentStrings(string allowed, vector<string>& words) {
        int mask = 0;
        for (auto c : allowed) {
            mask |= 1 << (c - 'a');
        }
        int res = 0;
        for (auto &&word : words) {
            int mask1 = 0;
            for (auto c : word) {
                mask1 |= 1 << (c - 'a');
            }
            if ((mask1 | mask) == mask) {
                res++;
            }
        }
        return res;
    }
};

[sol1-Java]

class Solution {
    public int countConsistentStrings(String allowed, String[] words) {
        int mask = 0;
        for (int i = 0; i < allowed.length(); i++) {
            char c = allowed.charAt(i);
            mask |= 1 << (c - 'a');
        }
        int res = 0;
        for (String word : words) {
            int mask1 = 0;
            for (int i = 0; i < word.length(); i++) {
                char c = word.charAt(i);
                mask1 |= 1 << (c - 'a');
            }
            if ((mask1 | mask) == mask) {
                res++;
            }
        }
        return res;
    }
}

[sol1-C#]

public class Solution {
    public int CountConsistentStrings(string allowed, string[] words) {
        int mask = 0;
        foreach (char c in allowed) {
            mask |= 1 << (c - 'a');
        }
        int res = 0;
        foreach (string word in words) {
            int mask1 = 0;
            foreach (char c in word) {
                mask1 |= 1 << (c - 'a');
            }
            if ((mask1 | mask) == mask) {
                res++;
            }
        }
        return res;
    }
}

[sol1-C]

int countConsistentStrings(char * allowed, char ** words, int wordsSize){
    int mask = 0;
    for (int i = 0; allowed[i] != '\0'; i++) {
        mask |= 1 << (allowed[i] - 'a');
    }
    int res = 0;
    for (int i = 0; i <wordsSize; i++) {
        int mask1 = 0;
        for (int j = 0; words[i][j] != '\0'; j++) {
            mask1 |= 1 << (words[i][j] - 'a');
        }
        if ((mask1 | mask) == mask) {
            res++;
        }
    }
    return res;
}

[sol1-Golang]

func countConsistentStrings(allowed string, words []string) (res int) {
    mask := 0
    for _, c := range allowed {
        mask |= 1 << (c - 'a')
    }
    for _, word := range words {
        mask1 := 0
        for _, c := range word {
            mask1 |= 1 << (c - 'a')
        }
        if mask1|mask == mask {
            res++
        }
    }
    return
}

[sol1-JavaScript]

var countConsistentStrings = function(allowed, words) {
    let mask = 0;
    for (let i = 0; i < allowed.length; i++) {
        const c = allowed[i];
        mask |= 1 << (c.charCodeAt() - 'a'.charCodeAt());
    }
    let res = 0;
    for (const word of words) {
        let mask1 = 0;
        for (let i = 0; i < word.length; i++) {
            const c = word[i];
            mask1 |= 1 << (c.charCodeAt() - 'a'.charCodeAt());
        }
        if ((mask1 | mask) === mask) {
            res++;
        }
    }
    return res;
};

复杂度分析

• 时间复杂度：O(n + \sum m_i)，其中 n 是字符串 allowed 的长度，m_i 是字符串 words}[i] 的长度。

• 空间复杂度：O(1)。