1684. Count the Number of Consistent Strings

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.

Approach 01:

class Solution {
public:
    int countConsistentStrings(string allowed, vector<string>& words) {
        int result = 0;
        for (auto &word : words) {
            int flag = 1;
            for (auto &ch : word) {
                if (allowed.find(ch) == -1) {
                    flag = 0;
                    break;
                }
            }
            if (flag) {
                result += 1;
            }
        }
        return result;
    }
};

class Solution:
    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
        result = 0
        
        for word in words:
            flag = 1
            for ch in word:
                if ch not in allowed:
                    flag = 0
                    break

            if flag:
                result += 1

        return result

Time Complexity

• Checking Each Word:

  The algorithm iterates over each word in the words vector. Let n be the number of words and m be the maximum length of a word. For each word, it iterates over its characters to check if they exist in the allowed string. This results in a time complexity of \(O(n \cdot m)\) in the worst case.

• Finding Characters in the Allowed String:

  The find function used to check if a character is present in the allowed string has a time complexity of \(O(|allowed|)\), where |allowed| is the length of the allowed string. Since this check is performed for every character in each word, it contributes to the overall complexity of \(O(n \cdot m \cdot |allowed|)\).

• Overall Time Complexity:

  The overall time complexity is \(O(n \cdot m \cdot |allowed|)\), where n is the number of words, m is the maximum length of a word, and |allowed| is the length of the allowed string. If |allowed| is treated as a constant (e.g., 26 lowercase English letters), the complexity can be simplified to \(O(n \cdot m)\).

Space Complexity

• Space for Variables:

  The solution uses a few local variables: result to count the number of consistent strings and flag to check if all characters in a word are allowed. This requires \(O(1)\) space.

• Space for Input Storage:

  The space required for the input words and allowed strings is already given as part of the input and does not count towards the space complexity of the algorithm itself.

• Overall Space Complexity:

  The overall space complexity is \(O(1)\) because no additional data structures or dynamic memory allocations are used.