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by Sumit Chourasia | Dec 28, 2020 | Category :coding | Tags : algorithm data-structure easy leetcode string

Count the Number of Consistent Strings - String - Easy - LeetCode

Count the Number of Consistent Strings - String - Easy - LeetCode

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.

 

public class Solution {
    public int CountConsistentStrings(string allowed, string[] words) {
        var set = new HashSet<char>();
        for(int i=0;i<allowed.Length;i++){
            set.Add(allowed[i]);
        }
        
        int count = 0;
        for(int i=0;i<words.Length;i++){
            bool found = true;
            for(int j=0;j<words[i].Length;j++){
                if(!set.Contains(words[i][j])){
                    found = false;
                    break;
                }
            }
            if(found){
                count++;
            }
        }
        
        return count;
    }
}

 

Time Complexity: O(n*m)

Space Complexity: O(l)

Where n is the number of words, and m is the length of each word, while l is the length of the allowed word.

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