1684. Count the Number of Consistent Strings

Description

You are given a string allowed consisting of distinct characters and an array of strings words. A string is consistent if all characters in the string appear in the string allowed.

Return the number of consistent strings in the array words.

Example 1:

Input: allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
Output: 2
Explanation: Strings "aaab" and "baa" are consistent since they only contain characters 'a' and 'b'.

Example 2:

Input: allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
Output: 7
Explanation: All strings are consistent.

Example 3:

Input: allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
Output: 4
Explanation: Strings "cc", "acd", "ac", and "d" are consistent.

Constraints:

1 <= words.length <= 10⁴
1 <= allowed.length <=26
1 <= words[i].length <= 10
The characters in allowed are distinct.
words[i] and allowed contain only lowercase English letters.

Solutions

Solution 1: Hash Table or Array

A straightforward approach is to use a hash table or array $s$ to record the characters in allowed. Then iterate over the words array, for each string $w$, determine whether it is composed of characters in allowed. If so, increment the answer.

The time complexity is $O(m)$, and the space complexity is $O(C)$. Here, $m$ is the total length of all strings, and $C$ is the size of the character set allowed. In this problem, $C \leq 26$.

Python3JavaC++GoTypeScriptRustC

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

class Solution {
    public int countConsistentStrings(String allowed, String[] words) {
        boolean[] s = new boolean[26];
        for (char c : allowed.toCharArray()) {
            s[c - 'a'] = true;
        }
        int ans = 0;
        for (String w : words) {
            if (check(w, s)) {
                ++ans;
            }
        }
        return ans;
    }

    private boolean check(String w, boolean[] s) {
        for (int i = 0; i < w.length(); ++i) {
            if (!s[w.charAt(i) - 'a']) {
                return false;
            }
        }
        return true;
    }
}

class Solution {
public:
    int countConsistentStrings(string allowed, vector<string>& words) {
        bitset<26> s;
        for (auto& c : allowed) s[c - 'a'] = 1;
        int ans = 0;
        auto check = [&](string& w) {
            for (auto& c : w)
                if (!s[c - 'a']) return false;
            return true;
        };
        for (auto& w : words) ans += check(w);
        return ans;
    }
};

func countConsistentStrings(allowed string, words []string) (ans int) {
    s := [26]bool{}
    for _, c := range allowed {
        s[c-'a'] = true
    }
    check := func(w string) bool {
        for _, c := range w {
            if !s[c-'a'] {
                return false
            }
        }
        return true
    }
    for _, w := range words {
        if check(w) {
            ans++
        }
    }
    return ans
}

function countConsistentStrings(allowed: string, words: string[]): number {
    const set = new Set([...allowed]);
    const n = words.length;
    let ans = n;
    for (const word of words) {
        for (const c of word) {
            if (!set.has(c)) {
                ans--;
                break;
            }
        }
    }
    return ans;
}