README_EN.md

1684. Count the Number of Consistent Strings

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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

Python3

class Solution:
    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
        res = 0
        chars = set(allowed)
        for word in words:
            find = True
            for c in word:
                if c not in chars:
                    find = False
                    break
            if find:
                res += 1
        return res

Java

class Solution {
    public int countConsistentStrings(String allowed, String[] words) {
        boolean[] chars = new boolean[26];
        for (char c : allowed.toCharArray()) {
            chars[c - 'a'] = true;
        }
        int res = 0;
        for (String word : words) {
            boolean find = true;
            for (char c : word.toCharArray()) {
                if (!chars[c - 'a']) {
                    find = false;
                    break;
                }
            }
            if (find) {
                ++res;
            }
        }
        return res;
    }
}

C++

class Solution {
public:
    int countConsistentStrings(string allowed, vector<string>& words) {
        vector<bool> chars(26, false);
        for (char c : allowed) {
            chars[c - 'a'] = true;
        }
        int res = 0;
        for (string word : words) {
            bool find = true;
            for (char c : word) {
                if (!chars[c - 'a']) {
                    find = false;
                    break;
                }
            }
            if (find) ++res;
        }
        return res;
    }
};

Go

func countConsistentStrings(allowed string, words []string) int {
	chars := [26]bool{}
	for _, c := range allowed {
		chars[c-'a'] = true
	}
	res := 0
	for _, word := range words {
		find := true
		for _, c := range word {
			if !chars[c-'a'] {
				find = false
				break
			}
		}
		if find {
			res++
		}
	}
	return res
}

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