RUST GYM v0.2.4

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.

Rust Solution

struct Solution;
use std::collections::HashSet;

impl Solution {
    fn count_consistent_strings(allowed: String, words: Vec<String>) -> i32 {
        let hs: HashSet<char> = allowed.chars().collect();
        words
            .into_iter()
            .filter(|w| w.chars().all(|c| hs.contains(&c)))
            .count() as i32
    }
}

#[test]
fn test() {
    let allowed = "ab".to_string();
    let words = vec_string!["ad", "bd", "aaab", "baa", "badab"];
    let res = 2;
    assert_eq!(Solution::count_consistent_strings(allowed, words), res);
    let allowed = "abc".to_string();
    let words = vec_string!["a", "b", "c", "ab", "ac", "bc", "abc"];
    let res = 7;
    assert_eq!(Solution::count_consistent_strings(allowed, words), res);
    let allowed = "cad".to_string();
    let words = vec_string!["cc", "acd", "b", "ba", "bac", "bad", "ac", "d"];
    let res = 4;
    assert_eq!(Solution::count_consistent_strings(allowed, words), res);
}

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